Class 12 NCERT Solutions
Chapter 1: Relations and Functions
Master the mapping of elements, the classification of equivalence relations, and the logic of bijective functions with our step-by-step logic.
Exercise 1.1
1. Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set \(A=\{1,2,3,\ldots,13,14\}\) defined as \(R=\{(x,y):3x-y=0\}\)
(ii) Relation R in the set N of natural numbers defined as \(R=\{(x,y):y=x+5 \text{ and } x<4\}\)
(iii) Relation R in the set \(A=\{1,2,3,4,5,6\}\) as \(R=\{(x,y):y \text{ is divisible by } x\}\)
(iv) Relation R in the set Z of all integers defined as \(R=\{(x,y):x-y \text{ is an integer}\}\)
(v) Relation R in the set A of human beings in a town given by:
(a) \(R=\{(x,y):x \text{ and } y \text{ work at the same place}\}\)
(b) \(R=\{(x,y):x \text{ and } y \text{ live in the same locality}\}\)
(c) \(R=\{(x,y):x \text{ is exactly 7 cm taller than } y\}\)
(d) \(R=\{(x,y):x \text{ is wife of } y\}\)
(e) \(R=\{(x,y):x \text{ is father of } y\}\)
(i) Relation R in the set \(A=\{1,2,3,\ldots,13,14\}\) defined as \(R=\{(x,y):3x-y=0\}\)
(ii) Relation R in the set N of natural numbers defined as \(R=\{(x,y):y=x+5 \text{ and } x<4\}\)
(iii) Relation R in the set \(A=\{1,2,3,4,5,6\}\) as \(R=\{(x,y):y \text{ is divisible by } x\}\)
(iv) Relation R in the set Z of all integers defined as \(R=\{(x,y):x-y \text{ is an integer}\}\)
(v) Relation R in the set A of human beings in a town given by:
(a) \(R=\{(x,y):x \text{ and } y \text{ work at the same place}\}\)
(b) \(R=\{(x,y):x \text{ and } y \text{ live in the same locality}\}\)
(c) \(R=\{(x,y):x \text{ is exactly 7 cm taller than } y\}\)
(d) \(R=\{(x,y):x \text{ is wife of } y\}\)
(e) \(R=\{(x,y):x \text{ is father of } y\}\)
Solution:
(i) Given: Set \(A=\{1,2,3,\ldots,13,14\}\) and \(R=\{(x,y):3x-y=0\}\), i.e., \(y=3x\).
Reflexive? Substitute \(y=x\) into the condition: \(x=3x\). Dividing by \(x\neq0\) gives \(1=3\), which is impossible. So \((x,x)\notin R\). Hence R is not reflexive.
Symmetric? If \((x,y)\in R\), then \(y=3x\). Testing \((y,x)\): we’d need \(x=3y=3(3x)=9x\), which forces \(x=0\). As a counterexample, \((2,6)\in R\) (since \(6=3\times2\)) but \((6,2)\notin R\) (since \(2\neq3\times6=18\)). Hence R is not symmetric.
Transitive? Suppose \((x,y)\in R\) and \((y,z)\in R\), so \(y=3x\) and \(z=3y\). Eliminating \(y\): \(z=3(3x)=9x\). For transitivity we’d need \(z=3x\), but \(9x\neq3x\) in general. So \((x,z)\notin R\). Hence R is not transitive.
(ii) Write R in roster form: \(R=\{(1,6),(2,7),(3,8)\}\).
Reflexive? \((1,1),(2,2),(3,3)\notin R\), so R is not reflexive.
Symmetric? \((1,6)\in R\) but \((6,1)\notin R\), so R is not symmetric.
Transitive? There are no pairs \((x,y)\) and \((y,z)\) both in \(R\), so the transitivity condition is vacuously satisfied. R is transitive.
Remark: When set A is small and finite, always convert R to roster form first.
(iii) Roster form: \(R=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(5,5),(6,6)\}\).
Reflexive? All pairs \((1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\in R\), so R is reflexive.
Symmetric? \((2,4)\in R\) but \((4,2)\notin R\) since 2 is not divisible by 4. Hence R is not symmetric.
Transitive? Let \((x,y)\in R\) and \((y,z)\in R\), so \(y=mx\) and \(z=ny\) for natural numbers \(m,n\). Substituting: \(z=n(mx)=(nm)x\), so \(z\) is divisible by \(x\), giving \((x,z)\in R\). Hence R is transitive.
(iv) \(R=\{(x,y):x-y\text{ is an integer}\}\).
Reflexive? \(x-x=0\) is always an integer, so \((x,x)\in R\) for all \(x\in\mathbf{Z}\). R is reflexive.
Symmetric? If \((x-y)\) is an integer, then \(-(y-x)=x-y\) means \((y-x)\) is also an integer. So \((y,x)\in R\). R is symmetric.
Transitive? If \((x-y)\) and \((y-z)\) are both integers, adding them: \((x-z)=(x-y)+(y-z)\) is an integer. So \((x,z)\in R\). R is transitive.
(v)(a) “\(x\) and \(y\) work at the same place.”
Reflexive: A person \(x\) works at the same place as themselves — trivially true. ✓
Symmetric: If \(x\) and \(y\) work at the same place, then clearly \(y\) and \(x\) do too. ✓
Transitive: If \(x,y\) share a workplace and \(y,z\) share a workplace, then \(x,z\) also share that same workplace. ✓
R is an equivalence relation.
(v)(b) Replace “work” with “live” and “place” with “locality” — the same logic applies. R is an equivalence relation.
(v)(c) “\(x\) is exactly 7 cm taller than \(y\).”
Reflexive? No person can be 7 cm taller than themselves. R is not reflexive.
Symmetric? If \(x\) is 7 cm taller than \(y\), then \(y\) is 7 cm shorter than \(x\), not taller. R is not symmetric.
Transitive? If \(x\) is 7 cm taller than \(y\) and \(y\) is 7 cm taller than \(z\), then \(x\) is 14 cm (not 7 cm) taller than \(z\). R is not transitive.
(v)(d) “\(x\) is wife of \(y\).”
Reflexive? No one can be their own wife. R is not reflexive.
Symmetric? If \(x\) is wife of \(y\), then \(y\) is the husband (not wife) of \(x\). So \((y,x)\notin R\). R is not symmetric.
Transitive? If \((x,y)\in R\), then \(y\) is a man (husband); no man can be someone’s wife, so \((y,z)\) can never belong to R. The condition is vacuously satisfied. R is transitive.
(v)(e) “\(x\) is father of \(y\).”
Reflexive? No one is their own father. R is not reflexive.
Symmetric? If \(x\) is father of \(y\), then \(y\) is son/daughter (not father) of \(x\). R is not symmetric.
Transitive? If \(x\) is father of \(y\) and \(y\) is father of \(z\), then \(x\) is the grandfather (not father) of \(z\). So \((x,z)\notin R\). R is not transitive.
(i) Given: Set \(A=\{1,2,3,\ldots,13,14\}\) and \(R=\{(x,y):3x-y=0\}\), i.e., \(y=3x\).
Reflexive? Substitute \(y=x\) into the condition: \(x=3x\). Dividing by \(x\neq0\) gives \(1=3\), which is impossible. So \((x,x)\notin R\). Hence R is not reflexive.
Symmetric? If \((x,y)\in R\), then \(y=3x\). Testing \((y,x)\): we’d need \(x=3y=3(3x)=9x\), which forces \(x=0\). As a counterexample, \((2,6)\in R\) (since \(6=3\times2\)) but \((6,2)\notin R\) (since \(2\neq3\times6=18\)). Hence R is not symmetric.
Transitive? Suppose \((x,y)\in R\) and \((y,z)\in R\), so \(y=3x\) and \(z=3y\). Eliminating \(y\): \(z=3(3x)=9x\). For transitivity we’d need \(z=3x\), but \(9x\neq3x\) in general. So \((x,z)\notin R\). Hence R is not transitive.
(ii) Write R in roster form: \(R=\{(1,6),(2,7),(3,8)\}\).
Reflexive? \((1,1),(2,2),(3,3)\notin R\), so R is not reflexive.
Symmetric? \((1,6)\in R\) but \((6,1)\notin R\), so R is not symmetric.
Transitive? There are no pairs \((x,y)\) and \((y,z)\) both in \(R\), so the transitivity condition is vacuously satisfied. R is transitive.
Remark: When set A is small and finite, always convert R to roster form first.
(iii) Roster form: \(R=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(5,5),(6,6)\}\).
Reflexive? All pairs \((1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\in R\), so R is reflexive.
Symmetric? \((2,4)\in R\) but \((4,2)\notin R\) since 2 is not divisible by 4. Hence R is not symmetric.
Transitive? Let \((x,y)\in R\) and \((y,z)\in R\), so \(y=mx\) and \(z=ny\) for natural numbers \(m,n\). Substituting: \(z=n(mx)=(nm)x\), so \(z\) is divisible by \(x\), giving \((x,z)\in R\). Hence R is transitive.
(iv) \(R=\{(x,y):x-y\text{ is an integer}\}\).
Reflexive? \(x-x=0\) is always an integer, so \((x,x)\in R\) for all \(x\in\mathbf{Z}\). R is reflexive.
Symmetric? If \((x-y)\) is an integer, then \(-(y-x)=x-y\) means \((y-x)\) is also an integer. So \((y,x)\in R\). R is symmetric.
Transitive? If \((x-y)\) and \((y-z)\) are both integers, adding them: \((x-z)=(x-y)+(y-z)\) is an integer. So \((x,z)\in R\). R is transitive.
(v)(a) “\(x\) and \(y\) work at the same place.”
Reflexive: A person \(x\) works at the same place as themselves — trivially true. ✓
Symmetric: If \(x\) and \(y\) work at the same place, then clearly \(y\) and \(x\) do too. ✓
Transitive: If \(x,y\) share a workplace and \(y,z\) share a workplace, then \(x,z\) also share that same workplace. ✓
R is an equivalence relation.
(v)(b) Replace “work” with “live” and “place” with “locality” — the same logic applies. R is an equivalence relation.
(v)(c) “\(x\) is exactly 7 cm taller than \(y\).”
Reflexive? No person can be 7 cm taller than themselves. R is not reflexive.
Symmetric? If \(x\) is 7 cm taller than \(y\), then \(y\) is 7 cm shorter than \(x\), not taller. R is not symmetric.
Transitive? If \(x\) is 7 cm taller than \(y\) and \(y\) is 7 cm taller than \(z\), then \(x\) is 14 cm (not 7 cm) taller than \(z\). R is not transitive.
(v)(d) “\(x\) is wife of \(y\).”
Reflexive? No one can be their own wife. R is not reflexive.
Symmetric? If \(x\) is wife of \(y\), then \(y\) is the husband (not wife) of \(x\). So \((y,x)\notin R\). R is not symmetric.
Transitive? If \((x,y)\in R\), then \(y\) is a man (husband); no man can be someone’s wife, so \((y,z)\) can never belong to R. The condition is vacuously satisfied. R is transitive.
(v)(e) “\(x\) is father of \(y\).”
Reflexive? No one is their own father. R is not reflexive.
Symmetric? If \(x\) is father of \(y\), then \(y\) is son/daughter (not father) of \(x\). R is not symmetric.
Transitive? If \(x\) is father of \(y\) and \(y\) is father of \(z\), then \(x\) is the grandfather (not father) of \(z\). So \((x,z)\notin R\). R is not transitive.
2. Show that the relation R in the set R of real numbers, defined as \(R=\{(a,b):a\leq b^2\}\) is neither reflexive nor symmetric nor transitive.
Solution:
Pick concrete counterexamples for each property.
Not Reflexive: Take \(a=\frac{1}{2}\). Then \(a\leq a^2\) requires \(\frac{1}{2}\leq\frac{1}{4}\), which is false. So \((a,a)\notin R\) for all \(a\).
Not Symmetric: Take \(a=1,b=2\). Now \(1\leq 2^2=4\) is true, so \((1,2)\in R\). But \(2\leq 1^2=1\) is false, so \((2,1)\notin R\).
Not Transitive: Take \(a=10,b=4,c=2\). Then \(10\leq 4^2=16\) ✓ and \(4\leq 2^2=4\) ✓ (since \(4=4\) satisfies \(\leq\)). But \(10\leq 2^2=4\) is false, so \((10,2)\notin R\).
Therefore R is neither reflexive nor symmetric nor transitive.
Pick concrete counterexamples for each property.
Not Reflexive: Take \(a=\frac{1}{2}\). Then \(a\leq a^2\) requires \(\frac{1}{2}\leq\frac{1}{4}\), which is false. So \((a,a)\notin R\) for all \(a\).
Not Symmetric: Take \(a=1,b=2\). Now \(1\leq 2^2=4\) is true, so \((1,2)\in R\). But \(2\leq 1^2=1\) is false, so \((2,1)\notin R\).
Not Transitive: Take \(a=10,b=4,c=2\). Then \(10\leq 4^2=16\) ✓ and \(4\leq 2^2=4\) ✓ (since \(4=4\) satisfies \(\leq\)). But \(10\leq 2^2=4\) is false, so \((10,2)\notin R\).
Therefore R is neither reflexive nor symmetric nor transitive.
3. Check whether the relation R defined in the set \(\{1,2,3,4,5,6\}\) as \(R=\{(a,b):b=a+1\}\) is reflexive, symmetric or transitive.
Solution:
Convert to roster form first. Substituting \(a=1,2,3,4,5,6\):
\(R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}\) (note \((6,7)\) is excluded since \(7\notin A\)).
Not Reflexive: \((1,1)\notin R\) since \(1\neq 1+1=2\). More generally, \((a,a)\notin R\) for any \(a\).
Not Symmetric: \((1,2)\in R\) but \((2,1)\notin R\), since \(1\neq 2+1=3\).
Not Transitive: \((1,2)\in R\) and \((2,3)\in R\), but \((1,3)\notin R\), since \(3\neq 1+1=2\).
Convert to roster form first. Substituting \(a=1,2,3,4,5,6\):
\(R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\}\) (note \((6,7)\) is excluded since \(7\notin A\)).
Not Reflexive: \((1,1)\notin R\) since \(1\neq 1+1=2\). More generally, \((a,a)\notin R\) for any \(a\).
Not Symmetric: \((1,2)\in R\) but \((2,1)\notin R\), since \(1\neq 2+1=3\).
Not Transitive: \((1,2)\in R\) and \((2,3)\in R\), but \((1,3)\notin R\), since \(3\neq 1+1=2\).
4. Show that the relation R in R defined as \(R=\{(a,b):a\leq b\}\) is reflexive and transitive but not symmetric.
Solution:
Reflexive: Putting \(b=a\), we get \(a\leq a\), which is always true (since \(a=a\) satisfies the \(\leq\) condition). So R is reflexive.
Not Symmetric: Take \((1,2)\in R\) since \(1\leq2\). But \((2,1)\notin R\) since \(2>1\). So R is not symmetric.
Transitive: Let \((a,b)\in R\) and \((b,c)\in R\), so \(a\leq b\) and \(b\leq c\). By transitivity of \(\leq\) on reals, \(a\leq c\), so \((a,c)\in R\). R is transitive.
Reflexive: Putting \(b=a\), we get \(a\leq a\), which is always true (since \(a=a\) satisfies the \(\leq\) condition). So R is reflexive.
Not Symmetric: Take \((1,2)\in R\) since \(1\leq2\). But \((2,1)\notin R\) since \(2>1\). So R is not symmetric.
Transitive: Let \((a,b)\in R\) and \((b,c)\in R\), so \(a\leq b\) and \(b\leq c\). By transitivity of \(\leq\) on reals, \(a\leq c\), so \((a,c)\in R\). R is transitive.
5. Check whether the relation R in R defined by \(R=\{(a,b):a\leq b^3\}\) is reflexive, symmetric or transitive.
Solution:
Not Reflexive: For \(a=\frac{1}{2}\), the condition \(a\leq a^3\) becomes \(\frac{1}{2}\leq\frac{1}{8}\), which is false.
Not Symmetric: Take \(a=1,b=2\). Then \(1\leq 2^3=8\) ✓, so \((1,2)\in R\). But \(2\leq 1^3=1\) is false, so \((2,1)\notin R\).
Not Transitive: Take \(a=10,b=4,c=2\). Then \(10\leq 4^3=64\) ✓ and \(4\leq 2^3=8\) ✓. But \(10\leq 2^3=8\) is false, so \((10,2)\notin R\).
Therefore R is neither reflexive nor symmetric nor transitive.
Not Reflexive: For \(a=\frac{1}{2}\), the condition \(a\leq a^3\) becomes \(\frac{1}{2}\leq\frac{1}{8}\), which is false.
Not Symmetric: Take \(a=1,b=2\). Then \(1\leq 2^3=8\) ✓, so \((1,2)\in R\). But \(2\leq 1^3=1\) is false, so \((2,1)\notin R\).
Not Transitive: Take \(a=10,b=4,c=2\). Then \(10\leq 4^3=64\) ✓ and \(4\leq 2^3=8\) ✓. But \(10\leq 2^3=8\) is false, so \((10,2)\notin R\).
Therefore R is neither reflexive nor symmetric nor transitive.
6. Show that the relation R in the set \(\{1,2,3\}\) given by \(R=\{(1,2),(2,1)\}\) is symmetric but neither reflexive nor transitive.
Solution:
Symmetric: The only pairs in R are \((1,2)\) and \((2,1)\), and each is the reverse of the other. So \((x,y)\in R\Rightarrow(y,x)\in R\). R is symmetric.
Not Reflexive: \(1\in A\) but \((1,1)\notin R\); similarly \((2,2)\notin R\) and \((3,3)\notin R\). R is not reflexive.
Not Transitive: \((1,2)\in R\) and \((2,1)\in R\), but \((1,1)\notin R\). R is not transitive.
Symmetric: The only pairs in R are \((1,2)\) and \((2,1)\), and each is the reverse of the other. So \((x,y)\in R\Rightarrow(y,x)\in R\). R is symmetric.
Not Reflexive: \(1\in A\) but \((1,1)\notin R\); similarly \((2,2)\notin R\) and \((3,3)\notin R\). R is not reflexive.
Not Transitive: \((1,2)\in R\) and \((2,1)\in R\), but \((1,1)\notin R\). R is not transitive.
7. Show that the relation R in the set A of all books in a library, given by \(R=\{(x,y):x \text{ and } y \text{ have the same number of pages}\}\), is an equivalence relation.
Solution:
Reflexive: Any book \(x\) has the same number of pages as itself, so \((x,x)\in R\) for all \(x\in A\). ✓
Symmetric: If books \(x\) and \(y\) have the same page count, then books \(y\) and \(x\) also have the same page count. So \((x,y)\in R\Rightarrow(y,x)\in R\). ✓
Transitive: If \(x\) and \(y\) have the same page count, and \(y\) and \(z\) also have the same page count, then \(x\) and \(z\) share the same page count. So \((x,z)\in R\). ✓
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
Reflexive: Any book \(x\) has the same number of pages as itself, so \((x,x)\in R\) for all \(x\in A\). ✓
Symmetric: If books \(x\) and \(y\) have the same page count, then books \(y\) and \(x\) also have the same page count. So \((x,y)\in R\Rightarrow(y,x)\in R\). ✓
Transitive: If \(x\) and \(y\) have the same page count, and \(y\) and \(z\) also have the same page count, then \(x\) and \(z\) share the same page count. So \((x,z)\in R\). ✓
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
8. Show that the relation R in \(A=\{1,2,3,4,5\}\) given by \(R=\{(a,b):|a-b| \text{ is even}\}\) is an equivalence relation. Show that all elements of \(\{1,3,5\}\) are related to each other and all elements of \(\{2,4\}\) are related to each other, but no element of \(\{1,3,5\}\) is related to any element of \(\{2,4\}\).
Solution:
Reflexive: \(|a-a|=0\) is even for all \(a\in A\), so \((a,a)\in R\). ✓
Symmetric: If \(|a-b|\) is even, then \(|b-a|=|-(a-b)|=|a-b|\) is also even. So \((a,b)\in R\Rightarrow(b,a)\in R\). ✓
Transitive: If \(|a-b|\) and \(|b-c|\) are both even, then \((a-b)\) and \((b-c)\) are both even. Adding: \(a-c=(a-b)+(b-c)\) is even (sum of two evens is even), so \(|a-c|\) is even and \((a,c)\in R\). ✓
Hence R is an equivalence relation.
Elements of \(\{1,3,5\}\) are all odd, and the difference of any two odd numbers is even, so they are all related to each other. Similarly, elements of \(\{2,4\}\) are both even, and differences of even numbers are even. However, the difference of an odd and an even number is always odd, so no element of \(\{1,3,5\}\) is related to any element of \(\{2,4\}\).
Reflexive: \(|a-a|=0\) is even for all \(a\in A\), so \((a,a)\in R\). ✓
Symmetric: If \(|a-b|\) is even, then \(|b-a|=|-(a-b)|=|a-b|\) is also even. So \((a,b)\in R\Rightarrow(b,a)\in R\). ✓
Transitive: If \(|a-b|\) and \(|b-c|\) are both even, then \((a-b)\) and \((b-c)\) are both even. Adding: \(a-c=(a-b)+(b-c)\) is even (sum of two evens is even), so \(|a-c|\) is even and \((a,c)\in R\). ✓
Hence R is an equivalence relation.
Elements of \(\{1,3,5\}\) are all odd, and the difference of any two odd numbers is even, so they are all related to each other. Similarly, elements of \(\{2,4\}\) are both even, and differences of even numbers are even. However, the difference of an odd and an even number is always odd, so no element of \(\{1,3,5\}\) is related to any element of \(\{2,4\}\).
9. Show that each of the relations R in the set \(A=\{x\in\mathbf{Z}:0\leq x\leq12\}\) given by
(i) \(R=\{(a,b):|a-b| \text{ is a multiple of 4}\}\)
(ii) \(R=\{(a,b):a=b\}\)
is an equivalence relation. Find the set of all elements related to 1 in each case.
(i) \(R=\{(a,b):|a-b| \text{ is a multiple of 4}\}\)
(ii) \(R=\{(a,b):a=b\}\)
is an equivalence relation. Find the set of all elements related to 1 in each case.
Solution:
(i)
Reflexive: \(|a-a|=0\) is a multiple of 4, so \((a,a)\in R\). ✓
Symmetric: If \(|a-b|\) is a multiple of 4, then \(|b-a|=|a-b|\) is also a multiple of 4. ✓
Transitive: Let \(|a-b|=4m\) and \(|b-c|=4n\). Then \(a-b=\pm4m\) and \(b-c=\pm4n\). Adding: \(a-c=\pm4m\pm4n=\pm4(m+n)\), so \(|a-c|=4(m+n)\) is a multiple of 4. ✓
R is an equivalence relation.
To find elements related to 1: we need \(|a-1|\) to be a multiple of 4. Testing \(a\in A\): only \(a=1\) (\(|0|=0\)), \(a=5\) (\(|4|=4\)), \(a=9\) (\(|8|=8\)) satisfy this. The required set is \(\{1,5,9\}\).
(ii)
Reflexive: \(a=a\) is always true, so \((a,a)\in R\). ✓
Symmetric: If \(a=b\), then \(b=a\). ✓
Transitive: If \(a=b\) and \(b=c\), then \(a=c\). ✓
R is an equivalence relation.
To find elements related to 1: we need \(a=1\). The required set is \(\{1\}\).
(i)
Reflexive: \(|a-a|=0\) is a multiple of 4, so \((a,a)\in R\). ✓
Symmetric: If \(|a-b|\) is a multiple of 4, then \(|b-a|=|a-b|\) is also a multiple of 4. ✓
Transitive: Let \(|a-b|=4m\) and \(|b-c|=4n\). Then \(a-b=\pm4m\) and \(b-c=\pm4n\). Adding: \(a-c=\pm4m\pm4n=\pm4(m+n)\), so \(|a-c|=4(m+n)\) is a multiple of 4. ✓
R is an equivalence relation.
To find elements related to 1: we need \(|a-1|\) to be a multiple of 4. Testing \(a\in A\): only \(a=1\) (\(|0|=0\)), \(a=5\) (\(|4|=4\)), \(a=9\) (\(|8|=8\)) satisfy this. The required set is \(\{1,5,9\}\).
(ii)
Reflexive: \(a=a\) is always true, so \((a,a)\in R\). ✓
Symmetric: If \(a=b\), then \(b=a\). ✓
Transitive: If \(a=b\) and \(b=c\), then \(a=c\). ✓
R is an equivalence relation.
To find elements related to 1: we need \(a=1\). The required set is \(\{1\}\).
10. Give an example of a relation which is:
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Solution:
Let \(A=\{1,2,3,4\}\).
(i) \(R=\{(1,2),(2,1),(2,4),(4,2)\}\) is symmetric (each pair and its reverse are both in R), but not reflexive (e.g., \((1,1)\notin R\)) and not transitive (\((1,2)\in R\) and \((2,4)\in R\) but \((1,4)\notin R\)).
(ii) \(R=\{(1,2),(2,3),(1,3)\}\) is transitive (\((1,2)\) and \((2,3)\) give \((1,3)\in R\)), but not reflexive (\((1,1)\notin R\)) and not symmetric (\((1,2)\in R\) but \((2,1)\notin R\)).
(iii) \(R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(1,3),(3,1)\}\) is reflexive (all diagonal pairs present) and symmetric, but not transitive (\((2,1)\in R\) and \((1,3)\in R\) but \((2,3)\notin R\)).
(iv) \(R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,3),(1,3)\}\) is reflexive (all diagonal pairs present) and transitive (all chained pairs included), but not symmetric (\((1,2)\in R\) but \((2,1)\notin R\)).
(v) \(R=\{(1,1),(1,2),(2,1),(2,2)\}\) is symmetric and transitive, but not reflexive (\(3\in A\) but \((3,3)\notin R\)).
Let \(A=\{1,2,3,4\}\).
(i) \(R=\{(1,2),(2,1),(2,4),(4,2)\}\) is symmetric (each pair and its reverse are both in R), but not reflexive (e.g., \((1,1)\notin R\)) and not transitive (\((1,2)\in R\) and \((2,4)\in R\) but \((1,4)\notin R\)).
(ii) \(R=\{(1,2),(2,3),(1,3)\}\) is transitive (\((1,2)\) and \((2,3)\) give \((1,3)\in R\)), but not reflexive (\((1,1)\notin R\)) and not symmetric (\((1,2)\in R\) but \((2,1)\notin R\)).
(iii) \(R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(1,3),(3,1)\}\) is reflexive (all diagonal pairs present) and symmetric, but not transitive (\((2,1)\in R\) and \((1,3)\in R\) but \((2,3)\notin R\)).
(iv) \(R=\{(1,1),(2,2),(3,3),(4,4),(1,2),(2,3),(1,3)\}\) is reflexive (all diagonal pairs present) and transitive (all chained pairs included), but not symmetric (\((1,2)\in R\) but \((2,1)\notin R\)).
(v) \(R=\{(1,1),(1,2),(2,1),(2,2)\}\) is symmetric and transitive, but not reflexive (\(3\in A\) but \((3,3)\notin R\)).
11. Show that the relation R in the set A of points in a plane, given by \(R=\{(P,Q):\text{distance of P from origin equals distance of Q from origin}\}\), is an equivalence relation. Further show that the set of all points related to a point \(P\neq(0,0)\) is the circle passing through P with origin as centre.
Solution:
Write the relation as \(R=\{(P,Q):OP=OQ\}\) where O is the origin.
Reflexive: \(OP=OP\) is always true, so \((P,P)\in R\). ✓
Symmetric: If \(OP=OQ\), then clearly \(OQ=OP\), so \((Q,P)\in R\). ✓
Transitive: If \(OP=OQ\) and \(OQ=OS\), then \(OP=OS\), so \((P,S)\in R\). ✓
Hence R is an equivalence relation.
For a fixed point \(P\neq(0,0)\), the set of all points Q related to P satisfies \(OQ=OP=k\) (constant). By definition, this is a circle of radius \(k\) centred at the origin O, passing through P.
Write the relation as \(R=\{(P,Q):OP=OQ\}\) where O is the origin.
Reflexive: \(OP=OP\) is always true, so \((P,P)\in R\). ✓
Symmetric: If \(OP=OQ\), then clearly \(OQ=OP\), so \((Q,P)\in R\). ✓
Transitive: If \(OP=OQ\) and \(OQ=OS\), then \(OP=OS\), so \((P,S)\in R\). ✓
Hence R is an equivalence relation.
For a fixed point \(P\neq(0,0)\), the set of all points Q related to P satisfies \(OQ=OP=k\) (constant). By definition, this is a circle of radius \(k\) centred at the origin O, passing through P.
12. Show that the relation R in the set A of all triangles, defined as \(R=\{(T_1,T_2):T_1 \text{ is similar to } T_2\}\), is an equivalence relation. Among \(T_1\) (sides 3,4,5), \(T_2\) (sides 5,12,13) and \(T_3\) (sides 6,8,10), which are related?
Solution:
Reflexive: Every triangle is similar to itself (ratios of corresponding sides all equal 1). So \((T_1,T_1)\in R\). ✓
Symmetric: If \(T_1\) is similar to \(T_2\), then by the symmetry of similarity, \(T_2\) is similar to \(T_1\). ✓
Transitive: If \(T_1\sim T_2\) and \(T_2\sim T_3\), then the ratios of corresponding sides are consistent and \(T_1\sim T_3\). ✓
R is an equivalence relation.
Check \(T_1\) vs \(T_3\): \(\dfrac{3}{6}=\dfrac{4}{8}=\dfrac{5}{10}=\dfrac{1}{2}\). All ratios equal, so \(T_1\) is related to \(T_3\).
Check \(T_1\) vs \(T_2\): \(\dfrac{3}{5},\dfrac{4}{12},\dfrac{5}{13}\) are not equal, so \(T_1\) is not related to \(T_2\).
Check \(T_2\) vs \(T_3\): \(\dfrac{5}{6},\dfrac{12}{8},\dfrac{13}{10}\) are not equal, so \(T_2\) is not related to \(T_3\).
Reflexive: Every triangle is similar to itself (ratios of corresponding sides all equal 1). So \((T_1,T_1)\in R\). ✓
Symmetric: If \(T_1\) is similar to \(T_2\), then by the symmetry of similarity, \(T_2\) is similar to \(T_1\). ✓
Transitive: If \(T_1\sim T_2\) and \(T_2\sim T_3\), then the ratios of corresponding sides are consistent and \(T_1\sim T_3\). ✓
R is an equivalence relation.
Check \(T_1\) vs \(T_3\): \(\dfrac{3}{6}=\dfrac{4}{8}=\dfrac{5}{10}=\dfrac{1}{2}\). All ratios equal, so \(T_1\) is related to \(T_3\).
Check \(T_1\) vs \(T_2\): \(\dfrac{3}{5},\dfrac{4}{12},\dfrac{5}{13}\) are not equal, so \(T_1\) is not related to \(T_2\).
Check \(T_2\) vs \(T_3\): \(\dfrac{5}{6},\dfrac{12}{8},\dfrac{13}{10}\) are not equal, so \(T_2\) is not related to \(T_3\).
13. Show that the relation R in the set A of all polygons, defined as \(R=\{(P_1,P_2):P_1 \text{ and } P_2 \text{ have the same number of sides}\}\), is an equivalence relation. What is the set of all elements in A related to the right-angle triangle T with sides 3, 4, 5?
Solution:
Reflexive: Any polygon \(P_1\) has the same number of sides as itself. ✓
Symmetric: If \(P_1\) and \(P_2\) have the same number of sides, then so do \(P_2\) and \(P_1\). ✓
Transitive: If \(P_1\) and \(P_2\) share a side count, and \(P_2\) and \(P_3\) share a side count, then \(P_1\) and \(P_3\) have the same count. ✓
R is an equivalence relation.
The given triangle T has 3 sides. All polygons with exactly 3 sides (i.e., all triangles — not only right-angled ones) in A are related to T. The required set is the set of all triangles in A.
Reflexive: Any polygon \(P_1\) has the same number of sides as itself. ✓
Symmetric: If \(P_1\) and \(P_2\) have the same number of sides, then so do \(P_2\) and \(P_1\). ✓
Transitive: If \(P_1\) and \(P_2\) share a side count, and \(P_2\) and \(P_3\) share a side count, then \(P_1\) and \(P_3\) have the same count. ✓
R is an equivalence relation.
The given triangle T has 3 sides. All polygons with exactly 3 sides (i.e., all triangles — not only right-angled ones) in A are related to T. The required set is the set of all triangles in A.
14. Let L be the set of all lines in the XY plane and R be the relation in L defined as \(R=\{(L_1,L_2):L_1 \text{ is parallel to } L_2\}\). Show that R is an equivalence relation. Find the set of all lines related to \(y=2x+4\).
Solution:
Reflexive: Every line is parallel to itself. So \((L_1,L_1)\in R\). ✓
Symmetric: If \(L_1\) is parallel to \(L_2\), then \(L_2\) is parallel to \(L_1\). ✓
Transitive: If \(L_1\parallel L_2\) and \(L_2\parallel L_3\), then \(L_1\parallel L_3\). ✓
R is an equivalence relation.
Parallel lines have the same slope and differ only in their constant term. So the set of all lines related (parallel) to \(y=2x+4\) is \(\{y=2x+c : c\in\mathbf{R}\}\), where \(c\) is any real constant.
Reflexive: Every line is parallel to itself. So \((L_1,L_1)\in R\). ✓
Symmetric: If \(L_1\) is parallel to \(L_2\), then \(L_2\) is parallel to \(L_1\). ✓
Transitive: If \(L_1\parallel L_2\) and \(L_2\parallel L_3\), then \(L_1\parallel L_3\). ✓
R is an equivalence relation.
Parallel lines have the same slope and differ only in their constant term. So the set of all lines related (parallel) to \(y=2x+4\) is \(\{y=2x+c : c\in\mathbf{R}\}\), where \(c\) is any real constant.
15. Let R be the relation in \(\{1,2,3,4\}\) given by \(R=\{(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)\}\). Choose the correct answer:
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.
Solution:
Reflexive: \((1,1),(2,2),(3,3),(4,4)\) all belong to R, so R is reflexive.
Not Symmetric: \((1,2)\in R\) but \((2,1)\notin R\), so R is not symmetric.
Transitive: Checking all chains — \((1,2),(2,2)\Rightarrow(1,2)\in R\) ✓; \((1,3),(3,2)\Rightarrow(1,2)\in R\) ✓. All such cases hold. R is transitive.
Correct answer: (B)
Reflexive: \((1,1),(2,2),(3,3),(4,4)\) all belong to R, so R is reflexive.
Not Symmetric: \((1,2)\in R\) but \((2,1)\notin R\), so R is not symmetric.
Transitive: Checking all chains — \((1,2),(2,2)\Rightarrow(1,2)\in R\) ✓; \((1,3),(3,2)\Rightarrow(1,2)\in R\) ✓. All such cases hold. R is transitive.
Correct answer: (B)
16. Let R be the relation in N given by \(R=\{(a,b):a=b-2,b>6\}\). Choose the correct answer:
(A) \((2,4)\in R\) (B) \((3,8)\in R\) (C) \((6,8)\in R\) (D) \((8,7)\in R\)
(A) \((2,4)\in R\) (B) \((3,8)\in R\) (C) \((6,8)\in R\) (D) \((8,7)\in R\)
Solution:
Check each option against \(a=b-2\) and \(b>6\):
(A) \(b=4\), not \(>6\). ✗
(B) \(a=3,b=8\): \(3\neq8-2=6\). ✗
(C) \(a=6,b=8>6\) and \(8-2=6=a\). ✓
(D) \(b=7>6\) but \(8\neq7-2=5\). ✗
Correct answer: (C)
Check each option against \(a=b-2\) and \(b>6\):
(A) \(b=4\), not \(>6\). ✗
(B) \(a=3,b=8\): \(3\neq8-2=6\). ✗
(C) \(a=6,b=8>6\) and \(8-2=6=a\). ✓
(D) \(b=7>6\) but \(8\neq7-2=5\). ✗
Correct answer: (C)
Exercise 1.2
1. Show that \(f:\mathbf{R}_*\rightarrow\mathbf{R}_*\) defined by \(f(x)=\dfrac{1}{x}\) is one-one and onto, where \(\mathbf{R}_*\) is the set of all non-zero real numbers. Is the result true if the domain \(\mathbf{R}_*\) is replaced by N with co-domain still \(\mathbf{R}_*\)?
Solution:
One-one: For \(x_1,x_2\in\mathbf{R}_*\) with \(f(x_1)=f(x_2)\): \(\dfrac{1}{x_1}=\dfrac{1}{x_2}\Rightarrow x_1=x_2\). So \(f\) is one-one.
Onto: For any \(y\in\mathbf{R}_*\), choose \(x=\dfrac{1}{y}\in\mathbf{R}_*\). Then \(f(x)=\dfrac{1}{1/y}=y\). Every element of the co-domain has a pre-image, so \(f\) is onto.
When domain is N: \(f:N\rightarrow\mathbf{R}_*\) with \(f(x)=\dfrac{1}{x}\) is still one-one (same argument). But it is not onto: for example, \(2\in\mathbf{R}_*\) would require \(x=\dfrac{1}{2}\), but \(\dfrac{1}{2}\notin N\). So no natural number maps to 2 under \(f\).
One-one: For \(x_1,x_2\in\mathbf{R}_*\) with \(f(x_1)=f(x_2)\): \(\dfrac{1}{x_1}=\dfrac{1}{x_2}\Rightarrow x_1=x_2\). So \(f\) is one-one.
Onto: For any \(y\in\mathbf{R}_*\), choose \(x=\dfrac{1}{y}\in\mathbf{R}_*\). Then \(f(x)=\dfrac{1}{1/y}=y\). Every element of the co-domain has a pre-image, so \(f\) is onto.
When domain is N: \(f:N\rightarrow\mathbf{R}_*\) with \(f(x)=\dfrac{1}{x}\) is still one-one (same argument). But it is not onto: for example, \(2\in\mathbf{R}_*\) would require \(x=\dfrac{1}{2}\), but \(\dfrac{1}{2}\notin N\). So no natural number maps to 2 under \(f\).
2. Check the injectivity and surjectivity of the following functions:
(i) \(f:N\rightarrow N\) given by \(f(x)=x^2\)
(ii) \(f:\mathbf{Z}\rightarrow\mathbf{Z}\) given by \(f(x)=x^2\)
(iii) \(f:\mathbf{R}\rightarrow\mathbf{R}\) given by \(f(x)=x^2\)
(iv) \(f:N\rightarrow N\) given by \(f(x)=x^3\)
(v) \(f:\mathbf{Z}\rightarrow\mathbf{Z}\) given by \(f(x)=x^3\)
(i) \(f:N\rightarrow N\) given by \(f(x)=x^2\)
(ii) \(f:\mathbf{Z}\rightarrow\mathbf{Z}\) given by \(f(x)=x^2\)
(iii) \(f:\mathbf{R}\rightarrow\mathbf{R}\) given by \(f(x)=x^2\)
(iv) \(f:N\rightarrow N\) given by \(f(x)=x^3\)
(v) \(f:\mathbf{Z}\rightarrow\mathbf{Z}\) given by \(f(x)=x^3\)
Solution:
(i) \(f:N\rightarrow N,\ f(x)=x^2\)
Injective: \(x_1^2=x_2^2\Rightarrow x_1=\pm x_2\). Since \(x_1,x_2\in N\) are positive, reject the negative sign: \(x_1=x_2\). So injective.
Surjective: Range \(=\{1,4,9,\ldots\}\neq N\) (e.g., 2,3 are in N but not in range). Not surjective.
(ii) \(f:\mathbf{Z}\rightarrow\mathbf{Z},\ f(x)=x^2\)
Injective: \(f(2)=4=f(-2)\) but \(2\neq-2\). Not injective.
Surjective: Range \(=\{0,1,4,9,\ldots\}\neq\mathbf{Z}\) (negative integers not in range). Not surjective.
(iii) \(f:\mathbf{R}\rightarrow\mathbf{R},\ f(x)=x^2\)
Same reasoning as (ii): neither injective nor surjective. (Range \(=[0,\infty)\neq\mathbf{R}\).)
(iv) \(f:N\rightarrow N,\ f(x)=x^3\)
Injective: \(x_1^3=x_2^3\Rightarrow x_1=x_2\). Injective.
Surjective: Range \(=\{1,8,27,\ldots\}\neq N\). Not surjective.
(v) \(f:\mathbf{Z}\rightarrow\mathbf{Z},\ f(x)=x^3\)
Injective: \(x_1^3=x_2^3\Rightarrow x_1=x_2\). Injective.
Surjective: Range \(=\{0,\pm1,\pm8,\ldots\}\neq\mathbf{Z}\) (e.g., 2 is not a perfect cube). Not surjective.
(i) \(f:N\rightarrow N,\ f(x)=x^2\)
Injective: \(x_1^2=x_2^2\Rightarrow x_1=\pm x_2\). Since \(x_1,x_2\in N\) are positive, reject the negative sign: \(x_1=x_2\). So injective.
Surjective: Range \(=\{1,4,9,\ldots\}\neq N\) (e.g., 2,3 are in N but not in range). Not surjective.
(ii) \(f:\mathbf{Z}\rightarrow\mathbf{Z},\ f(x)=x^2\)
Injective: \(f(2)=4=f(-2)\) but \(2\neq-2\). Not injective.
Surjective: Range \(=\{0,1,4,9,\ldots\}\neq\mathbf{Z}\) (negative integers not in range). Not surjective.
(iii) \(f:\mathbf{R}\rightarrow\mathbf{R},\ f(x)=x^2\)
Same reasoning as (ii): neither injective nor surjective. (Range \(=[0,\infty)\neq\mathbf{R}\).)
(iv) \(f:N\rightarrow N,\ f(x)=x^3\)
Injective: \(x_1^3=x_2^3\Rightarrow x_1=x_2\). Injective.
Surjective: Range \(=\{1,8,27,\ldots\}\neq N\). Not surjective.
(v) \(f:\mathbf{Z}\rightarrow\mathbf{Z},\ f(x)=x^3\)
Injective: \(x_1^3=x_2^3\Rightarrow x_1=x_2\). Injective.
Surjective: Range \(=\{0,\pm1,\pm8,\ldots\}\neq\mathbf{Z}\) (e.g., 2 is not a perfect cube). Not surjective.
3. Prove that the Greatest Integer Function \(f:\mathbf{R}\rightarrow\mathbf{R}\), given by \(f(x)=[x]\), is neither one-one nor onto.
Solution:
Not one-one: Take \(x_1=2.5\) and \(x_2=2.8\). Both \(f(2.5)=[2.5]=2\) and \(f(2.8)=[2.8]=2\), yet \(2.5\neq2.8\). So distinct inputs give the same output. \(f\) is not one-one.
Not onto: The range of \(f\) is the set of all integers \(\mathbf{Z}\) (since \([x]\) always yields an integer). But the co-domain is \(\mathbf{R}\), and \(\mathbf{Z}\neq\mathbf{R}\) (e.g., \(1.5\in\mathbf{R}\) is not in the range). \(f\) is not onto.
Not one-one: Take \(x_1=2.5\) and \(x_2=2.8\). Both \(f(2.5)=[2.5]=2\) and \(f(2.8)=[2.8]=2\), yet \(2.5\neq2.8\). So distinct inputs give the same output. \(f\) is not one-one.
Not onto: The range of \(f\) is the set of all integers \(\mathbf{Z}\) (since \([x]\) always yields an integer). But the co-domain is \(\mathbf{R}\), and \(\mathbf{Z}\neq\mathbf{R}\) (e.g., \(1.5\in\mathbf{R}\) is not in the range). \(f\) is not onto.
4. Show that the Modulus Function \(f:\mathbf{R}\rightarrow\mathbf{R}\), given by \(f(x)=|x|\), is neither one-one nor onto.
Solution:
Not one-one: \(f(-1)=|-1|=1=f(1)=|1|=1\), but \(-1\neq1\). \(f\) is not one-one.
Not onto: Since \(|x|\geq0\) for all \(x\in\mathbf{R}\), the range is \([0,\infty)\). This is not equal to the co-domain \(\mathbf{R}\) (e.g., \(-1\in\mathbf{R}\) has no pre-image). \(f\) is not onto.
Not one-one: \(f(-1)=|-1|=1=f(1)=|1|=1\), but \(-1\neq1\). \(f\) is not one-one.
Not onto: Since \(|x|\geq0\) for all \(x\in\mathbf{R}\), the range is \([0,\infty)\). This is not equal to the co-domain \(\mathbf{R}\) (e.g., \(-1\in\mathbf{R}\) has no pre-image). \(f\) is not onto.
5. Show that the Signum Function \(f:\mathbf{R}\rightarrow\mathbf{R}\) given by \(f(x)=\begin{cases}1,&x>0\\0,&x=0\\-1,&x<0\end{cases}\) is neither one-one nor onto.
Solution:
Not one-one: \(f(2)=1=f(3)\) since both \(2,3>0\), yet \(2\neq3\). \(f\) is not one-one.
Not onto: The range is \(\{-1,0,1\}\), which is not equal to the co-domain \(\mathbf{R}\) (e.g., \(2\in\mathbf{R}\) has no pre-image). \(f\) is not onto.
Not one-one: \(f(2)=1=f(3)\) since both \(2,3>0\), yet \(2\neq3\). \(f\) is not one-one.
Not onto: The range is \(\{-1,0,1\}\), which is not equal to the co-domain \(\mathbf{R}\) (e.g., \(2\in\mathbf{R}\) has no pre-image). \(f\) is not onto.
6. Let \(A=\{1,2,3\}\), \(B=\{4,5,6,7\}\) and \(f=\{(1,4),(2,5),(3,6)\}\) be a function from A to B. Show that \(f\) is one-one.
Solution:
From the given rule: \(f(1)=4,\ f(2)=5,\ f(3)=6\).
All three elements of the domain A have distinct images in B. Since no two different elements of A share the same image, \(f\) is one-one.
From the given rule: \(f(1)=4,\ f(2)=5,\ f(3)=6\).
All three elements of the domain A have distinct images in B. Since no two different elements of A share the same image, \(f\) is one-one.
7. In each case, state whether the function is one-one, onto or bijective:
(i) \(f:\mathbf{R}\rightarrow\mathbf{R}\) defined by \(f(x)=3-4x\)
(ii) \(f:\mathbf{R}\rightarrow\mathbf{R}\) defined by \(f(x)=1+x^2\)
(i) \(f:\mathbf{R}\rightarrow\mathbf{R}\) defined by \(f(x)=3-4x\)
(ii) \(f:\mathbf{R}\rightarrow\mathbf{R}\) defined by \(f(x)=1+x^2\)
Solution:
(i) \(f(x)=3-4x\)
One-one: \(f(x_1)=f(x_2)\Rightarrow3-4x_1=3-4x_2\Rightarrow x_1=x_2\). One-one.
Onto: For any \(y\in\mathbf{R}\), solve \(y=3-4x\): \(x=\dfrac{3-y}{4}\in\mathbf{R}\). Then \(f\!\left(\dfrac{3-y}{4}\right)=3-4\!\cdot\!\dfrac{3-y}{4}=y\). Every \(y\) has a pre-image. Onto.
Therefore \(f\) is bijective.
(ii) \(f(x)=1+x^2\)
One-one: \(f(1)=2=f(-1)\) but \(1\neq-1\). Not one-one.
Onto: Range \(=[1,\infty)\neq\mathbf{R}\) (e.g., \(0\in\mathbf{R}\) has no pre-image). Not onto.
\(f\) is neither one-one nor onto.
(i) \(f(x)=3-4x\)
One-one: \(f(x_1)=f(x_2)\Rightarrow3-4x_1=3-4x_2\Rightarrow x_1=x_2\). One-one.
Onto: For any \(y\in\mathbf{R}\), solve \(y=3-4x\): \(x=\dfrac{3-y}{4}\in\mathbf{R}\). Then \(f\!\left(\dfrac{3-y}{4}\right)=3-4\!\cdot\!\dfrac{3-y}{4}=y\). Every \(y\) has a pre-image. Onto.
Therefore \(f\) is bijective.
(ii) \(f(x)=1+x^2\)
One-one: \(f(1)=2=f(-1)\) but \(1\neq-1\). Not one-one.
Onto: Range \(=[1,\infty)\neq\mathbf{R}\) (e.g., \(0\in\mathbf{R}\) has no pre-image). Not onto.
\(f\) is neither one-one nor onto.
8. Let A and B be sets. Show that \(f:A\times B\rightarrow B\times A\) such that \(f(a,b)=(b,a)\) is bijective.
Solution:
One-one: Let \(f(a,b)=f(c,d)\). Then \((b,a)=(d,c)\), so equating components: \(b=d\) and \(a=c\), giving \((a,b)=(c,d)\). So \(f\) is injective.
Onto: For any \((b,a)\in B\times A\), the pair \((a,b)\in A\times B\) satisfies \(f(a,b)=(b,a)\). Every element of the co-domain has a pre-image. So \(f\) is surjective.
Therefore \(f\) is bijective.
One-one: Let \(f(a,b)=f(c,d)\). Then \((b,a)=(d,c)\), so equating components: \(b=d\) and \(a=c\), giving \((a,b)=(c,d)\). So \(f\) is injective.
Onto: For any \((b,a)\in B\times A\), the pair \((a,b)\in A\times B\) satisfies \(f(a,b)=(b,a)\). Every element of the co-domain has a pre-image. So \(f\) is surjective.
Therefore \(f\) is bijective.
9. Let \(f:N\rightarrow N\) be defined by \(f(n)=\begin{cases}\dfrac{n+1}{2},&n\text{ is odd}\\\dfrac{n}{2},&n\text{ is even}\end{cases}\). State whether \(f\) is onto, one-one or bijective.
Solution:
Not one-one: Take \(n=3\) (odd): \(f(3)=\dfrac{3+1}{2}=2\). Take \(n=4\) (even): \(f(4)=\dfrac{4}{2}=2\). So \(f(3)=f(4)\) but \(3\neq4\). \(f\) is not one-one.
Onto: For any \(k\in N\), take \(n=2k-1\) (odd): \(f(2k-1)=\dfrac{2k}{2}=k\). So every element \(k\) of N is achieved. Range \(=N=\) co-domain. \(f\) is onto.
Since \(f\) is onto but not one-one, it is not bijective.
Not one-one: Take \(n=3\) (odd): \(f(3)=\dfrac{3+1}{2}=2\). Take \(n=4\) (even): \(f(4)=\dfrac{4}{2}=2\). So \(f(3)=f(4)\) but \(3\neq4\). \(f\) is not one-one.
Onto: For any \(k\in N\), take \(n=2k-1\) (odd): \(f(2k-1)=\dfrac{2k}{2}=k\). So every element \(k\) of N is achieved. Range \(=N=\) co-domain. \(f\) is onto.
Since \(f\) is onto but not one-one, it is not bijective.
10. Let \(A=\mathbf{R}-\{3\}\) and \(B=\mathbf{R}-\{1\}\). Consider \(f:A\rightarrow B\) defined by \(f(x)=\dfrac{x-2}{x-3}\). Is \(f\) one-one and onto?
Solution:
One-one: Let \(f(x_1)=f(x_2)\): \(\dfrac{x_1-2}{x_1-3}=\dfrac{x_2-2}{x_2-3}\). Cross-multiplying:
\((x_1-2)(x_2-3)=(x_1-3)(x_2-2)\)
\(\Rightarrow x_1x_2-3x_1-2x_2+6=x_1x_2-2x_1-3x_2+6\)
\(\Rightarrow -3x_1-2x_2=-2x_1-3x_2\Rightarrow-x_1=-x_2\Rightarrow x_1=x_2\). So \(f\) is one-one.
Onto: Let \(y=\dfrac{x-2}{x-3}\Rightarrow xy-3y=x-2\Rightarrow x(y-1)=3y-2\Rightarrow x=\dfrac{3y-2}{y-1}\).
For every \(y\in B=\mathbf{R}-\{1\}\), \(x=\dfrac{3y-2}{y-1}\in A=\mathbf{R}-\{3\}\) (verify: \(x=3\) would require \(3y-2=3y-3\), i.e., \(-2=-3\), impossible). Then \(f(x)=y\). So \(f\) is onto.
One-one: Let \(f(x_1)=f(x_2)\): \(\dfrac{x_1-2}{x_1-3}=\dfrac{x_2-2}{x_2-3}\). Cross-multiplying:
\((x_1-2)(x_2-3)=(x_1-3)(x_2-2)\)
\(\Rightarrow x_1x_2-3x_1-2x_2+6=x_1x_2-2x_1-3x_2+6\)
\(\Rightarrow -3x_1-2x_2=-2x_1-3x_2\Rightarrow-x_1=-x_2\Rightarrow x_1=x_2\). So \(f\) is one-one.
Onto: Let \(y=\dfrac{x-2}{x-3}\Rightarrow xy-3y=x-2\Rightarrow x(y-1)=3y-2\Rightarrow x=\dfrac{3y-2}{y-1}\).
For every \(y\in B=\mathbf{R}-\{1\}\), \(x=\dfrac{3y-2}{y-1}\in A=\mathbf{R}-\{3\}\) (verify: \(x=3\) would require \(3y-2=3y-3\), i.e., \(-2=-3\), impossible). Then \(f(x)=y\). So \(f\) is onto.
11. Let \(f:\mathbf{R}\rightarrow\mathbf{R}\) be defined as \(f(x)=x^4\). Choose the correct answer:
(A) one-one onto (B) many-one onto (C) one-one but not onto (D) neither one-one nor onto
(A) one-one onto (B) many-one onto (C) one-one but not onto (D) neither one-one nor onto
Solution:
Not one-one: \(f(2)=16=f(-2)\) but \(2\neq-2\).
Not onto: Range \(=[0,\infty)\neq\mathbf{R}\) since \(x^4\geq0\).
Correct answer: (D)
Not one-one: \(f(2)=16=f(-2)\) but \(2\neq-2\).
Not onto: Range \(=[0,\infty)\neq\mathbf{R}\) since \(x^4\geq0\).
Correct answer: (D)
12. Let \(f:\mathbf{R}\rightarrow\mathbf{R}\) be defined as \(f(x)=3x\). Choose the correct answer:
(A) one-one onto (B) many-one onto (C) one-one but not onto (D) neither one-one nor onto
(A) one-one onto (B) many-one onto (C) one-one but not onto (D) neither one-one nor onto
Solution:
One-one: \(3x_1=3x_2\Rightarrow x_1=x_2\). ✓
Onto: For any \(y\in\mathbf{R}\), \(x=\dfrac{y}{3}\in\mathbf{R}\) gives \(f(x)=3\cdot\dfrac{y}{3}=y\). ✓
Correct answer: (A)
One-one: \(3x_1=3x_2\Rightarrow x_1=x_2\). ✓
Onto: For any \(y\in\mathbf{R}\), \(x=\dfrac{y}{3}\in\mathbf{R}\) gives \(f(x)=3\cdot\dfrac{y}{3}=y\). ✓
Correct answer: (A)
Miscellaneous Exercise (Chapter 1)
1. Show that \(f: \mathbb{R} \to (-1, 1)\) defined by \(f(x) = \dfrac{x}{1+|x|}\) is One-One and Onto.
Solution:
One-one (Case I, \(x\geq0\)): Here \(|x|=x\), so \(f(x)=\dfrac{x}{1+x}\). If \(f(x_1)=f(x_2)\): \(\dfrac{x_1}{1+x_1}=\dfrac{x_2}{1+x_2}\Rightarrow x_1(1+x_2)=x_2(1+x_1)\Rightarrow x_1=x_2\).
One-one (Case II, \(x<0\)): Here \(|x|=-x\), so \(f(x)=\dfrac{x}{1-x}\). If \(f(x_1)=f(x_2)\): \(\dfrac{x_1}{1-x_1}=\dfrac{x_2}{1-x_2}\Rightarrow x_1(1-x_2)=x_2(1-x_1)\Rightarrow x_1=x_2\).
In both cases \(f(x_1)=f(x_2)\Rightarrow x_1=x_2\). So \(f\) is one-one.
Onto: For \(x\geq0\): \(x<1+x\), so dividing \(\dfrac{x}{1+x}<1\), i.e., \(f(x)<1\). For \(x<0\): \(x>-(1-x)\), so \(\dfrac{x}{1-x}>-1\), i.e., \(f(x)>-1\). Thus range \(=(-1,1)=\) co-domain. \(f\) is onto.
One-one (Case I, \(x\geq0\)): Here \(|x|=x\), so \(f(x)=\dfrac{x}{1+x}\). If \(f(x_1)=f(x_2)\): \(\dfrac{x_1}{1+x_1}=\dfrac{x_2}{1+x_2}\Rightarrow x_1(1+x_2)=x_2(1+x_1)\Rightarrow x_1=x_2\).
One-one (Case II, \(x<0\)): Here \(|x|=-x\), so \(f(x)=\dfrac{x}{1-x}\). If \(f(x_1)=f(x_2)\): \(\dfrac{x_1}{1-x_1}=\dfrac{x_2}{1-x_2}\Rightarrow x_1(1-x_2)=x_2(1-x_1)\Rightarrow x_1=x_2\).
In both cases \(f(x_1)=f(x_2)\Rightarrow x_1=x_2\). So \(f\) is one-one.
Onto: For \(x\geq0\): \(x<1+x\), so dividing \(\dfrac{x}{1+x}<1\), i.e., \(f(x)<1\). For \(x<0\): \(x>-(1-x)\), so \(\dfrac{x}{1-x}>-1\), i.e., \(f(x)>-1\). Thus range \(=(-1,1)=\) co-domain. \(f\) is onto.
2. Show that \(f:\mathbf{R}\rightarrow\mathbf{R}\) given by \(f(x)=x^3\) is injective.
Solution:
Let \(f(x_1)=f(x_2)\). Then \(x_1^3=x_2^3\Rightarrow x_1=x_2\) (taking the real cube root, which is unique). So \(f\) is injective.
Let \(f(x_1)=f(x_2)\). Then \(x_1^3=x_2^3\Rightarrow x_1=x_2\) (taking the real cube root, which is unique). So \(f\) is injective.
3. Given a non-empty set X, consider P(X) the power set of X. Define R in P(X) as: for subsets A, B in P(X), ARB if and only if \(A\subset B\). Is R an equivalence relation on P(X)?
Solution:
Reflexive: Every set is a subset of itself, so \(A\subset A\) is always true. R is reflexive.
Not Symmetric: If \(A\subsetneq B\) (A is a proper subset of B), then \(B\not\subset A\). For example, \(\{1\}\subset\{1,2\}\) but \(\{1,2\}\not\subset\{1\}\). R is not symmetric.
Transitive: If \(A\subset B\) and \(B\subset C\), then \(A\subset C\). R is transitive.
Since R is not symmetric, R is not an equivalence relation.
Reflexive: Every set is a subset of itself, so \(A\subset A\) is always true. R is reflexive.
Not Symmetric: If \(A\subsetneq B\) (A is a proper subset of B), then \(B\not\subset A\). For example, \(\{1\}\subset\{1,2\}\) but \(\{1,2\}\not\subset\{1\}\). R is not symmetric.
Transitive: If \(A\subset B\) and \(B\subset C\), then \(A\subset C\). R is transitive.
Since R is not symmetric, R is not an equivalence relation.
4. Find the number of all onto functions from the set \(\{1,2,3,\ldots,n\}\) to itself.
Solution:
For a finite set A with \(n\) elements, every onto function \(A\rightarrow A\) is also one-one (and vice versa). Such functions are simply permutations of A.
The image of \(1\in A\) can be any of \(n\) elements; then the image of \(2\) can be any of the remaining \((n-1)\); and so on, by the multiplication principle.
Total number of onto functions \(=n(n-1)(n-2)\cdots1=n!\)
For a finite set A with \(n\) elements, every onto function \(A\rightarrow A\) is also one-one (and vice versa). Such functions are simply permutations of A.
The image of \(1\in A\) can be any of \(n\) elements; then the image of \(2\) can be any of the remaining \((n-1)\); and so on, by the multiplication principle.
Total number of onto functions \(=n(n-1)(n-2)\cdots1=n!\)
5. Let \(A=\{-1,0,1,2\}\), \(B=\{-4,-2,0,2\}\). Let \(f(x)=x^2-x\) and \(g(x)=2\left|x-\dfrac{1}{2}\right|-1\) for \(x\in A\). Are \(f\) and \(g\) equal?
Solution:
Two functions are equal if they have the same domain and agree at every input. Compute at each \(x\in A=\{-1,0,1,2\}\):
\(f(-1)=(-1)^2-(-1)=2\),\quad \(g(-1)=2\left|\!-\tfrac{3}{2}\right|-1=3-1=2\) ✓
\(f(0)=0\),\quad \(g(0)=2\cdot\tfrac{1}{2}-1=0\) ✓
\(f(1)=0\),\quad \(g(1)=2\cdot\tfrac{1}{2}-1=0\) ✓
\(f(2)=2\),\quad \(g(2)=2\cdot\tfrac{3}{2}-1=2\) ✓
Since \(f(x)=g(x)\) for all \(x\in A\), \(f\) and \(g\) are equal.
Two functions are equal if they have the same domain and agree at every input. Compute at each \(x\in A=\{-1,0,1,2\}\):
\(f(-1)=(-1)^2-(-1)=2\),\quad \(g(-1)=2\left|\!-\tfrac{3}{2}\right|-1=3-1=2\) ✓
\(f(0)=0\),\quad \(g(0)=2\cdot\tfrac{1}{2}-1=0\) ✓
\(f(1)=0\),\quad \(g(1)=2\cdot\tfrac{1}{2}-1=0\) ✓
\(f(2)=2\),\quad \(g(2)=2\cdot\tfrac{3}{2}-1=2\) ✓
Since \(f(x)=g(x)\) for all \(x\in A\), \(f\) and \(g\) are equal.
6. Let \(A=\{1,2,3\}\). Then the number of relations containing \((1,2)\) and \((1,3)\) which are reflexive and symmetric but not transitive is:
(A) 1 (B) 2 (C) 3 (D) 4
(A) 1 (B) 2 (C) 3 (D) 4
Solution:
Any reflexive relation on A must contain \((1,1),(2,2),(3,3)\). Including \((1,2)\) and \((1,3)\), and since R must be symmetric, we must also include \((2,1)\) and \((3,1)\). This gives:
\(R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}\).
If we try to add \((2,3)\), symmetry forces \((3,2)\in R\), and then with \((2,1),(1,3)\in R\), transitivity would force \((2,3)\in R\) — which makes R an equivalence relation. So only the R above avoids transitivity: \((2,1)\in R\) and \((1,3)\in R\) but \((2,3)\notin R\). This is the unique such relation.
Correct answer: (A)
Any reflexive relation on A must contain \((1,1),(2,2),(3,3)\). Including \((1,2)\) and \((1,3)\), and since R must be symmetric, we must also include \((2,1)\) and \((3,1)\). This gives:
\(R=\{(1,1),(2,2),(3,3),(1,2),(2,1),(1,3),(3,1)\}\).
If we try to add \((2,3)\), symmetry forces \((3,2)\in R\), and then with \((2,1),(1,3)\in R\), transitivity would force \((2,3)\in R\) — which makes R an equivalence relation. So only the R above avoids transitivity: \((2,1)\in R\) and \((1,3)\in R\) but \((2,3)\notin R\). This is the unique such relation.
Correct answer: (A)
7. Let \(A=\{1,2,3\}\). Then the number of equivalence relations containing \((1,2)\) is:
(A) 1 (B) 2 (C) 3 (D) 4
(A) 1 (B) 2 (C) 3 (D) 4
Solution:
Any equivalence relation must contain all diagonal pairs \((1,1),(2,2),(3,3)\). Including \((1,2)\) forces \((2,1)\) by symmetry.
Relation 1: \(R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}\) — this is a valid equivalence relation.
Relation 2: Try adding \((1,3)\). Symmetry forces \((3,1)\). Transitivity: \((3,1)\in R\) and \((1,2)\in R\Rightarrow(3,2)\in R\), and symmetry forces \((2,3)\). This grows to \(A\times A\), the largest equivalence relation on A.
No other equivalence relation containing \((1,2)\) is possible. There are exactly two.
Correct answer: (B)
Any equivalence relation must contain all diagonal pairs \((1,1),(2,2),(3,3)\). Including \((1,2)\) forces \((2,1)\) by symmetry.
Relation 1: \(R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}\) — this is a valid equivalence relation.
Relation 2: Try adding \((1,3)\). Symmetry forces \((3,1)\). Transitivity: \((3,1)\in R\) and \((1,2)\in R\Rightarrow(3,2)\in R\), and symmetry forces \((2,3)\). This grows to \(A\times A\), the largest equivalence relation on A.
No other equivalence relation containing \((1,2)\) is possible. There are exactly two.
Correct answer: (B)
