Solved Examples: Relations, Functions, and Composition

Example 1: Checking Equivalence Relation (Divisibility)

Question: Show that the relation R on the set of integers $\mathbb{Z}$ given by $R = \{(a, b) : 2 \text{ divides } a-b\}$ is an equivalence relation.

Solution: To be an equivalence relation, R must be reflexive, symmetric, and transitive.

• Reflexive: We check if $aRa$ for any integer $a$. This means we check if 2 divides $a-a$. Since $a-a = 0$ and $0 = 2 \times 0$, 2 divides 0. Thus, R is reflexive.

• Symmetric: We check if $aRb$ implies $bRa$.

  Assume $aRb$, so $a-b$ is divisible by 2. This means $a-b = 2k$ for some integer $k$.

  Multiplying by -1, we get $b-a = -2k = 2(-k)$. Since $-k$ is also an integer, $b-a$ is divisible by 2.

  Thus, $bRa$ is true. R is symmetric.

• Transitive: We check if $aRb$ and $bRc$ implies $aRc$.

Assume $a-b = 2k_1$ and $b-c = 2k_2$.

Adding these equations gives $(a-b) + (b-c) = 2k_1 + 2k_2$,

which simplifies to $a-c = 2(k_1+k_2)$.

Since $k_1+k_2$ is an integer, $a-c$ is divisible by 2. Thus, $aRc$ is true. R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Example 2: Non-Equivalence Relation

Question: Let R be the relation on the set of real numbers defined as $aRb$ if $a \le b^2$. Check if R is reflexive, symmetric, or transitive.

Solution:

• Reflexive: We need to check if $a \le a^2$ for all real numbers $a$. This is not always true. Counterexample: Let $a=0.5$. Then $a^2 = 0.25$. The inequality $0.5 \le 0.25$ is false. Thus, R is not reflexive.

• Symmetric: We need to check if $aRb$ implies $bRa$. Counterexample:

  Let $a=2$ and $b=5$. $a \le b^2$ is $2 \le 25$, which is true.

  But $b \le a^2$ is $5 \le 4$, which is false. Thus, R is not symmetric.

• Transitive: We need to check if $aRb$ and $bRc$ implies $aRc$.

  Counterexample: Let $a=2, b=-2, c=0$.

  • $aRb$: $2 \le (-2)^2 = 4$. (True)

  • $bRc$: $-2 \le 0^2 = 0$. (True)

  • But is $aRc$? $2 \le 0^2=0$. (False)

  Thus, R is not transitive.

Example 3: Relation of Parallel Lines

Question: Show that the relation 'is parallel to' on the set L of all lines in a plane is an equivalence relation.

Solution: Let R be the relation 'is parallel to' ($||$).

• Reflexive: For any line $l$, $l || l$ (a line is parallel to itself). R is reflexive.

• Symmetric: If line $l_1 || l_2$, then it follows that $l_2 || l_1$. R is symmetric.

• Transitive: If line $l_1 || l_2$ and $l_2 || l_3$, then it follows that $l_1 || l_3$. R is transitive.

Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Example 4: Smallest Equivalence Relation

Question: Let $A=\{1,2,3\}$. Find the smallest equivalence relation on A containing the element (1,2).

Solution: We start with the required element and add the minimum pairs needed to satisfy all three properties.

1. Start: $R$ must contain $\{(1,2)\}$.

2. Reflexivity: To be reflexive on A={1,2,3}, R must contain $\{(1,1), (2,2), (3,3)\}$.

3. Symmetry: We have (1,2), so we must add its reverse, $\{(2,1)\}$.

4. Transitivity: Now we check for chains. We have $(1,2)$ and $(2,1)$, which requires $(1,1)$ to be in R. We have $(2,1)$ and $(1,2)$, which requires $(2,2)$ to be in R. Both (1,1) and (2,2) were already added for reflexivity.

The smallest set containing all required elements is $R = \{(1,1), (2,2), (3,3), (1,2), (2,1)\}$.

Example 5: Counting Relations

Question: Let $A=\{1,2\}$. How many reflexive relations can be defined on A?

Solution:

1. Total Pairs: The set $A \times A = \{(1,1), (1,2), (2,1), (2,2)\}$ has $n^2=4$ elements.

2. Reflexive Condition: A reflexive relation on A must contain all diagonal elements, which are (1,1) and (2,2). These are mandatory.

3. Optional Pairs: The non-diagonal elements are (1,2) and (2,1). For each of these, we have a choice: either include it in the relation or not.

4. Counting:

   • Choice for (1,2): In or Out (2 options)

   • Choice for (2,1): In or Out (2 options)

   The total number of reflexive relations is $2 \times 2 = 2^2 = 4$. In general, for a set with n elements, there are $n^2-n$ non-diagonal elements, so the number of reflexive relations is $2^{n^2-n}$.

Answer: There are 4 reflexive relations.

Example 6: Relation on Ordered Pairs

Question: Let R be a relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a,b)R(c,d)$ if $ad=bc$. Prove R is an equivalence relation.

Solution: The condition $ad=bc$ is equivalent to $\frac{a}{b}=\frac{c}{d}$, representing pairs with the same ratio.

• Reflexive: Check $(a,b)R(a,b)$. This requires $a \cdot b = b \cdot a$, which is true by the commutative property of multiplication. R is reflexive.

• Symmetric: Assume $(a,b)R(c,d)$, which means $ad=bc$. We need to check if $(c,d)R(a,b)$, which requires $cb=da$. Since $ad=bc$ is the same equation, this is true. R is symmetric.

• Transitive: Assume $(a,b)R(c,d)$ and $(c,d)R(e,f)$. This means (1) $ad=bc$ and (2) $cf=de$. We need to prove $(a,b)R(e,f)$, i.e., $af=be$.

  From (1), $d=\frac{bc}{a}$. From (2), $c=\frac{de}{f}$.

  Substitute $c$ into the first equation: $$d = b(\frac{de}{f})/a \implies d = \frac{bde}{af} \implies adf = bde$$

  Since $d \in \mathbb{N}$, $d eq 0$, so we can cancel $d$ from both sides to get $af=be$. R is transitive.

Conclusion: It is an equivalence relation.

Example 7: Checking One-one (Injective)

Question: Show that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x)=x^3$ is one-one.

Solution: To prove a function is one-one (injective), we must show that for any two elements $x_1, x_2$ in the domain, if their outputs are equal, their inputs must also be equal. That is, $$f(x_1) = f(x_2) \implies x_1=x_2$$

1. Assume $f(x_1) = f(x_2)$: Let $x_1^3 = x_2^3$.

2. Solve for $x_1$: Taking the cube root of both sides gives $x_1=x_2$.

3. Conclusion: Since the condition holds, the function is one-one.

Example 8: Checking Onto (Surjective)

Question: Show that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x)=x^3$ is onto.

Solution: To prove a function is onto (surjective), we must show that for any element $y$ in the codomain, there exists at least one element $x$ in the domain (a pre-image) such that $f(x)=y$. The range must equal the codomain.

1. Set $y=f(x)$: Let $y$ be any number in the codomain $\mathbb{R}$. We have $y = x^3$.

2. Solve for x: To find a potential pre-image, we solve for $x$: $x = \sqrt[3]{y}$.

3. Verify: For any real number $y$, its real cube root $x$ is also a unique real number. This $x$ is in the domain $\mathbb{R}$.

4. Conclusion: Since every element $y$ in the codomain has a pre-image $x=\sqrt[3]{y}$, the function is onto.

Example 9: Many-one and Into Function

Question: Let $f: \mathbb{Z} \to \mathbb{Z}$ be defined by $f(x)=x^2$. Classify the function.

Solution:

• One-one vs. Many-one: The function is many-one because we can find different inputs that produce the same output. For example, $f(-2)=4$ and $f(2)=4$.

• Onto vs. Into: The function is into because its range is not equal to its codomain. The codomain is all integers ($\mathbb{Z}$), but the range is only the set of non-negative perfect squares $\{0, 1, 4, 9, ...\}$. For example, the integer 3 is in the codomain but has no integer pre-image (since $\sqrt{3}$ is not an integer).

Example 10: Bijective Function

Question: Is the function $f: \mathbb{R}-\{3\} \to \mathbb{R}-\{1\}$ defined by $f(x) = \frac{x-2}{x-3}$ bijective?

Solution: A function is bijective if it is both one-one and onto.

• One-one: Assume $f(x_1)=f(x_2)$. Then $$\frac{x_1-2}{x_1-3}=\frac{x_2-2}{x_2-3}$$

  Cross-multiplying gives $$(x_1-2)(x_2-3)=(x_2-2)(x_1-3)$$ which simplifies to $$x_1x_2-3x_1-2x_2+6 = x_1x_2-3x_2-2x_1+6$$

  This further simplifies to $-x_1=-x_2$, or $x_1=x_2$.

  So, it is one-one.

• Onto: Let $y \in \mathbb{R}-\{1\}$ and set $y = \frac{x-2}{x-3}$.

  Solving for $x$ gives $$y(x-3)=x-2$$ $$\implies xy-3y=x-2$$ $$\implies x(y-1)=3y-2$$ $$\implies x = \frac{3y-2}{y-1}$$

  Since $y \ne 1$, a valid pre-image $x$ exists for every $y$ in the codomain. So, it is onto. Conclusion: Since it is both one-one and onto, the function is bijective.

Example 11: Modulus Function

Question: Classify the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = |x|$.

Solution:

• One-one vs. Many-one: The function is many-one. For example, $f(-1)=1$ and $f(1)=1$. Different inputs (-1 and 1) lead to the same output.

• Onto vs. Into: The function is into. The codomain is all real numbers ($\mathbb{R}$), but the range is only the set of non-negative real numbers, $[0, \infty)$. No negative number in the codomain has a pre-image.

Example 12: Signum Function

Question: Classify the function $f: \mathbb{R} \to \mathbb{Z}$ defined by $f(x) = sgn(x)$.

Solution: The Signum function is defined as

$sgn(x) = 1$ if $x>0$, $0$ if $x=0$, and $-1$ if $x<0$.

• One-one vs. Many-one: The function is many-one. For example, $f(2)=1$ and $f(3)=1$. All positive numbers map to 1.

• Onto vs. Into: The function is into. The codomain is the set of all integers ($\mathbb{Z}$), but the range is only the set $\{-1, 0, 1\}$. For example, the integer 2 has no pre-image.

Example 13: One-one but not Onto

Question: Let $f: \mathbb{N} \to \mathbb{N}$ be defined by $f(n) = n+1$. Classify the function.

Solution:

• One-one: Assume $f(n_1)=f(n_2)$.

  Then $n_1+1=n_2+1$, which implies $n_1=n_2$. The function is one-one.

• Onto: The codomain is $\mathbb{N}=\{1, 2, 3, ...\}$. The range is the set of outputs: $f(1)=2, f(2)=3, ...$. So the range is $\{2, 3, 4, ...\}$.

  Since the range does not include the number 1 (which is in the codomain), the function is not onto.

Example 14: Many-one and Onto

Question: Let $f: \mathbb{R} \to [0, \infty)$ be defined by $f(x) = x^2$. Classify the function.

Solution:

• One-one vs. Many-one: The function is many-one because $f(-2)=4$ and $f(2)=4$.

• Onto vs. Into: The codomain is specified as $[0, \infty)$. The range of $f(x)=x^2$ is also the set of all non-negative real numbers, $[0, \infty)$.

  Since the range is equal to the codomain, the function is onto.

Example 15: Proving a function is NOT one-one

Question: Show that $f(x) = \cos x$ from $\mathbb{R} \to \mathbb{R}$ is not one-one.

Solution: To show a function is not one-one, we only need to find a single counterexample of two different inputs that produce the same output.

• Let $x_1=0$ and $x_2=2\pi$. Clearly $x_1 \ne x_2$.

• Calculate the outputs: $f(x_1)=\cos(0)=1$ and $f(x_2)=\cos(2\pi)=1$.

• Since $f(x_1) = f(x_2)$ but $x_1 \ne x_2$, the function is not one-one.

Example 16: Finding Compositions

Question: Let $f(x) = \sin x$ and $g(x) = x^2$. Find $(g \circ f)(x)$ and $(f \circ g)(x)$.

Solution:

• To find $(g \circ f)(x) = g(f(x))$: We substitute the entire function $f(x)$ into the variable $x$ of the function $g(x)$. $$g(f(x)) = g(\sin x) = (\sin x)^2 = \sin^2x$$

• To find $(f \circ g)(x) = f(g(x))$: We substitute the entire function $g(x)$ into the variable $x$ of the function $f(x)$.

  $f(g(x)) = f(x^2) = \sin(x^2)$.

Note that in general, $g \circ f \neq f \circ g$.

Example 17: Finding the Inverse of a Linear Function

Question: Find the inverse of the bijective function $f(x)=2x-3$.

Solution:

1. Replace $f(x)$ with y: $y = 2x-3$.

2. Swap x and y: $x = 2y-3$. This step represents the inversion.

3. Solve for y: $x+3 = 2y \implies y = \frac{x+3}{2}$.

4. State the Inverse: $f^{-1}(x) = \frac{x+3}{2}$.

Example 18: Finding Inverse of a Quadratic Function

Question: Let $f: [0, \infty) \to [4, \infty)$ be a function defined by $f(x)=x^2+4$. Find the inverse of f.

Solution: The restricted domain makes the function bijective.

1. Replace $f(x)$ with y: $y=x^2+4$.

2. Swap x and y: $x=y^2+4$.

3. Solve for y: $y^2=x-4 \implies y=\pm\sqrt{x-4}$.

4. Choose the Correct Root: The range of the inverse function $f^{-1}$ must be the domain of the original function $f$, which is $[0, \infty)$. To ensure our output $y$ is non-negative, we must choose the positive root.

   Therefore, $f^{-1}(x) = \sqrt{x-4}$.

Example 19: Composition with Itself

Question: If $f(x) = \frac{x}{\sqrt{1+x^2}}$, find $(f \circ f)(x)$.

Solution: We need to compute $f(f(x))$.

$$(f \circ f)(x) = f(f(x)) = \frac{f(x)}{\sqrt{1+(f(x))^2}} = \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{1+\left(\frac{x}{\sqrt{1+x^2}}\right)^2}}$$

Simplify the term in the denominator's square root: $$1+\frac{x^2}{1+x^2} = \frac{1+x^2+x^2}{1+x^2} = \frac{1+2x^2}{1+x^2}$$

Substitute this back: $$(f \circ f)(x) = \frac{\frac{x}{\sqrt{1+x^2}}}{\sqrt{\frac{1+2x^2}{1+x^2}}} = \frac{\frac{x}{\sqrt{1+x^2}}}{\frac{\sqrt{1+2x^2}}{\sqrt{1+x^2}}}$$

Cancel the $\sqrt{1+x^2}$ terms: $(f \circ f)(x) = \frac{x}{\sqrt{1+2x^2}}$.

Example 20: Finding a Value in a Composition

Question: If $f(x)=x^2-1$ and $g(x)=2x+1$, find a value of x for which $(f \circ g)(x) = (g \circ f)(x)$.

Solution:

1. Calculate $(f \circ g)(x)$: $$f(g(x)) = f(2x+1) = (2x+1)^2-1 = (4x^2+4x+1)-1 = 4x^2+4x$$

2. Calculate $(g \circ f)(x)$: $$g(f(x)) = g(x^2-1) = 2(x^2-1)+1 = 2x^2-2+1 = 2x^2-1$$

3. Set them equal and solve:

   $4x^2+4x = 2x^2-1 \implies 2x^2+4x+1=0$.

   Using the quadratic formula, $$x = \frac{-4 \pm \sqrt{4^2-4(2)(1)}}{2(2)} = \frac{-4 \pm \sqrt{8}}{4} = \frac{-4 \pm 2\sqrt{2}}{4} = -1 \pm \frac{\sqrt{2}}{2}.$$

Example 21: Inverse of a Rational Function

Question: Find the inverse of $f(x) = \frac{x-1}{x-2}$ where $f: \mathbb{R}-\{2\} \to \mathbb{R}-\{1\}$.

Solution:

1. Replace f(x) with y: $$y = \frac{x-1}{x-2}$$

2. Swap x and y: $$x = \frac{y-1}{y-2}$$

3. Solve for y: $$x(y-2)=y-1$$ $$\implies xy-2x=y-1$$ $$\implies xy-y=2x-1$$ $y(x-1)=2x-1 \implies y=\frac{2x-1}{x-1}$.

4. State the inverse: $f^{-1}(x) = \frac{2x-1}{x-1}$.

Example 22: Finding Domain of a Composition

Question: Let $f(x) = \sqrt{x}$ and $g(x) = x-2$. Find the domain of $f \circ g$.

Solution: The domain of $f \circ g$ is the set of all $x$ in the domain of $g$ such that the output $g(x)$ is in the domain of $f$.

1. Find Domains of f and g: Domain of $g(x)=x-2$ is $\mathbb{R}$. Domain of $f(x)=\sqrt{x}$ is $[0, \infty)$.

2. Apply Condition: The output of $g$, which is $g(x)=x-2$, must be in the domain of $f$. So, $x-2 \ge 0$.

3. Solve: $x \ge 2$. The domain is $[2, \infty)$.

Alternatively: First find the composite function $$(f \circ g)(x) = f(g(x)) = f(x-2) = \sqrt{x-2}$$ The domain of this new function is the set of x for which the expression is defined, which is $x-2 \ge 0$, or $x \ge 2$.

Example 23: Composition of a Function and its Inverse

Question: Let $f(x)=4x+3$. Verify that $(f \circ f^{-1})(x) = x$.

Solution:

1. Find the inverse $f^{-1}(x)$: Let $y=4x+3$. Swap variables: $x=4y+3$. Solve for y: $$y=(x-3)/4$$

   So, $$f^{-1}(x) = \frac{x-3}{4}$$

2. Perform the composition: $$(f \circ f^{-1})(x) = f(f^{-1}(x))$$

   $$f(\frac{x-3}{4}) = 4(\frac{x-3}{4}) + 3 = (x-3)+3 = x$$

3. Conclusion: The composition results in the identity function, as expected.

Example 24: Verifying Invertibility

Question: Is the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x)=\cos(x)$ invertible?

Solution: A function is invertible if and only if it is bijective (both one-one and onto).

• One-one: The function is not one-one. For example, $f(0)=\cos(0)=1$ and $f(2\pi)=\cos(2\pi)=1$. Different inputs (0 and $2\pi$) give the same output.

• Onto: The function is not onto. The codomain is $\mathbb{R}$, but the range of $\cos(x)$ is the closed interval $[-1,1]$. Values in the codomain like 2 have no pre-image. Since the function is neither one-one nor onto, it is not invertible.

Example 25: Inverse of a Cubic Function

Question: Find the inverse of the bijective function $f(x) = x^3+1$.

Solution:

1. Replace f(x) with y: $y=x^3+1$.

2. Swap x and y: $x=y^3+1$.

3. Solve for y: $x-1=y^3 \implies y=\sqrt[3]{x-1}$.

4. State the inverse: $f^{-1}(x) = \sqrt[3]{x-1}$.

Example 26: Composition of Three Functions

Question: Let $f(x)=x+1, g(x)=x^2, h(x)=1/x$. Find $(h \circ g \circ f)(x)$.

Solution: We work from the inside out.

1. First, find $(g \circ f)(x)$: $$(g \circ f)(x) = g(f(x)) = g(x+1) = (x+1)^2$$

2. Then, apply h to the result: $$(h \circ g \circ f)(x) = h((g \circ f)(x)) = h((x+1)^2) = \frac{1}{(x+1)^2}$$

Answer: $(h \circ g \circ f)(x) = \frac{1}{(x+1)^2}$.

Example 27: Finding g when f and gof are known

Question: Let $f(x)=x+1$ and $(g \circ f)(x) = x^2+2x+1$. Find $g(x)$.

Solution: We are given $g(f(x)) = x^2+2x+1$.

Substitute the expression for $f(x)$: $$g(x+1) = x^2+2x+1$$

We can see that the right side is a perfect square: $$g(x+1) = (x+1)^2$$

Let $u = x+1$. The equation becomes $g(u) = u^2$.

Replacing the variable $u$ with $x$, we get the function $g(x) = x^2$.

Example 28: Finding f when g and gof are known

Question: Let $g(x)=x+1$ and $(g \circ f)(x) = 2x+3$. Find $f(x)$.

Solution: We start with the definition of the composition: $$(g \circ f)(x) = g(f(x))$$

To find $g(f(x))$, we substitute the entire function $f(x)$ into the variable $x$ of $g(x)$.

$$g(f(x)) = f(x) + 1$$

We are given that this is equal to $2x+3$.

So, $f(x)+1 = 2x+3$.

Solving for $f(x)$: $$f(x) = 2x+3-1 = 2x+2$$

Answer: $f(x)=2x+2$.

Example 29: Inverse Property $(g \circ f)^{-1}$

Question: Let $f(x)=2x$ and $g(x)=x+1$. Find $(g \circ f)^{-1}(x)$.

Solution: Method 1: Using the Inverse Property

1. Property: $(g \circ f)^{-1}(x) = (f^{-1} \circ g^{-1})(x)$.

2. Find the inverses:

   • For $f(x)=2x$, the inverse is $f^{-1}(x)=x/2$.

   • For $g(x)=x+1$, the inverse is $g^{-1}(x)=x-1$.

3. Compose the inverses: $(f^{-1} \circ g^{-1})(x) = f^{-1}(g^{-1}(x)) = f^{-1}(x-1) = \frac{x-1}{2}$.

Method 2: Find the Composition First

1. Find $(g \circ f)(x)$: $$(g \circ f)(x) = g(f(x)) = g(2x) = 2x+1$$

2. Find the inverse of the result: Let $h(x) = 2x+1$. To find $h^{-1}(x)$, let $y=2x+1$.

   Swap variables: $x=2y+1$. Solve for y: $y=(x-1)/2$.

Both methods give the same answer: $(g \circ f)^{-1}(x) = \frac{x-1}{2}$.

Example 30: Identity Function

Question: Let $f: \mathbb{R} \to \mathbb{R}$ be the identity function, $f(x)=x$. Find $(f \circ f)(x)$ and $f^{-1}(x)$.

Solution:

• Composition: $(f \circ f)(x) = f(f(x)) = f(x) = x$.

The composition of the identity function with itself is the identity function.

• Inverse: The function $f(x)=x$ is a straight line with slope 1, passing through the origin. It is both one-one and onto, so it is bijective and has an inverse.

  1. Let $y=x$.

  2. Swap variables: $x=y$.

  3. Solve for y: $y=x$.

  The function is its own inverse: $f^{-1}(x)=x$.