Questions and Answers

Example 1: Finding Sets from Cartesian Product

Question: The Cartesian product $A \times A$ has 9 elements among which are found $(-1, 0)$ and $(0, 1)$. Find the set A and the remaining elements of $A \times A$.

Solution: Since $n(A \times A) = 9$, we know that $n(A) \times n(A) = 9$, which means $n(A)=3$. The elements $(-1, 0)$ and $(0, 1)$ are in $A \times A$. By definition, this means the first and second elements of these pairs must be in A. Therefore, $-1, 0,$ and $1$ are all elements of A. Since $n(A)=3$, the set A must be $\{-1, 0, 1\}$. The full Cartesian product $A \times A$ is: $$A \times A = \{(-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)\}$$

Example 2: Verifying a Property of Cartesian Products

Question: Let $A=\{1,2\}$, $B=\{1,2,3,4\}$, $C=\{5,6\}$. Verify that $A \times (B \cap C) = (A \times B) \cap (A \times C)$.

Solution: LHS: $B \cap C = \phi$. Therefore, $A \times (B \cap C) = A \times \phi = \phi$. RHS: $A \times B = \{(1,1), (1,2), ..., (2,4)\}$ and $A \times C = \{(1,5), (1,6), (2,5), (2,6)\}$. The intersection of these two sets is $\phi$. Since LHS = RHS, the property is verified.

Example 3: Finding Domain and Range of a Relation

Question: Define a relation R on the set $\mathbb{N}$ of natural numbers by $R = \{(x, y) : y = x + 5, x \text{ is a natural number less than 4; } x, y \in \mathbb{N}\}$. Write down the domain and the range.

Solution: The natural numbers less than 4 are 1, 2, and 3. The corresponding y values are 6, 7, and 8. The relation is $R = \{(1, 6), (2, 7), (3, 8)\}$.

• Domain: $\{1, 2, 3\}$
• Range: $\{6, 7, 8\}$

Example 4: Writing a Relation in Roster Form

Question: Let $A=\{1,2,3,5\}$ and $B=\{4,6,9\}$. Define a relation R from A to B by $R = \{(x, y) : \text{the difference between x and y is odd; } x \in A, y \in B\}$. Write R in roster form.

Solution: We test each pair $(x, y)$ to see if $|x-y|$ is odd. For example, $|1-4|=3$ (odd), so $(1,4) \in R$. $|1-9|=8$ (even), so $(1,9) \notin R$. Continuing for all pairs, we get: $R = \{(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)\}$.

Example 5: Finding the Range of a Real Function

Question: Find the range of the function $f(x) = 2 - 3x$, where $x \in \mathbb{R}$ and $x > 0$.

Solution: Given $x > 0$. Multiplying by -3 reverses the inequality: $-3x < 0$. Adding 2 to both sides: $2 - 3x < 2$. Since $f(x) = 2 - 3x$, this means $f(x) < 2$. The range is $(-\infty, 2)$.

Example 6: Finding the Domain of a Rational Function

Question: Find the domain of the function $f(x) = \frac{x^2+2x+1}{x^2-8x+12}$.

Solution: The function is defined everywhere except where the denominator is zero. We solve $x^2-8x+12 = 0$. Factoring gives $(x-6)(x-2) = 0$, so the roots are $x=2$ and $x=6$. The domain is $\mathbb{R} - \{2, 6\}$.

Example 7: Finding Domain and Range of a Radical Function

Question: Find the domain and range of the real function $f(x) = \sqrt{x-1}$.

Solution: Domain: The expression inside the square root must be non-negative. $x-1 \ge 0 \implies x \ge 1$. Domain = $[1, \infty)$. Range: The output of the square root is always non-negative. The minimum value is 0 (at x=1). Range = $[0, \infty)$.

Example 8: Identifying a Function

Question: The relation g is defined by $g(x)=\begin{cases}x^2, & 0 \le x \le 2\\ 3x, & 2 \le x \le 10\end{cases}$. Show that g is not a function.

Solution: At the point $x=2$, the definition is ambiguous. From the first piece, $g(2) = 2^2 = 4$. From the second piece, $g(2) = 3(2) = 6$. Since the single input $x=2$ maps to two different outputs (4 and 6), the relation g is not a function.

Example 9: Algebra of Functions

Question: Let $f(x) = x+1$ and $g(x) = 2x-3$. Find $f+g$, $f-g$, $fg$, and $f/g$.

Solution:

• Sum: $(f+g)(x) = (x+1) + (2x-3) = 3x-2$.
• Difference: $(f-g)(x) = (x+1) - (2x-3) = -x+4$.
• Product: $(fg)(x) = (x+1)(2x-3) = 2x^2 - x - 3$.
• Quotient: $(\frac{f}{g})(x) = \frac{x+1}{2x-3}$, for $x \ne \frac{3}{2}$.

Example 10: Finding Range of a Rational Function

Question: Let $f = \{(x, \frac{x^2}{1+x^2}) : x \in \mathbb{R}\}$ be a function. Determine the range of f.

Solution: Let $y = \frac{x^2}{1+x^2}$. Since $x^2 \ge 0$ and $1+x^2 > 0$, we have $y \ge 0$. Also, since $x^2 < 1+x^2$, dividing by the positive term $1+x^2$ gives $\frac{x^2}{1+x^2} < 1$, so $y < 1$. Combining these, the range is $[0, 1)$.

Example 11: Determining a Linear Function

Question: Let $f = \{(1,1), (2,3), (0,-1), (-1,-3)\}$ be a function from $\mathbb{Z}$ to $\mathbb{Z}$ defined by $f(x) = ax+b$. Determine a and b.

Solution: Using the points $(0, -1)$ and $(1, 1)$. From $(0, -1)$: $f(0) = a(0) + b = -1 \implies b = -1$. From $(1, 1)$: $f(1) = a(1) + b = 1$. Substituting $b=-1$, we get $a - 1 = 1 \implies a = 2$. The function is $f(x) = 2x-1$. Thus, a=2, b=-1.

Example 12: Proving an Equivalence Relation

Question: Let R be a relation from $\mathbb{Q}$ to $\mathbb{Q}$ defined by $R = \{(a,b) : a,b \in \mathbb{Q} \text{ and } a-b \in \mathbb{Z}\}$. Show that R is an equivalence relation.

Solution:

1. Reflexive: For any $a \in \mathbb{Q}$, $a-a=0 \in \mathbb{Z}$, so $(a,a) \in R$. It is reflexive.
2. Symmetric: If $(a,b) \in R$, then $a-b$ is an integer. This means $b-a$ is also an integer. Therefore, $(b,a) \in R$. It is symmetric.
3. Transitive: If $(a,b) \in R$ and $(b,c) \in R$, then $a-b$ and $b-c$ are integers. Their sum, $(a-b)+(b-c) = a-c$, is also an integer. Therefore, $(a,c) \in R$. It is transitive. Since R is reflexive, symmetric, and transitive, it is an equivalence relation.

Example 13: Finding Range with Highest Prime Factor

Question: Let $A = \{9, 10, 11, 12, 13\}$ and let $f: A \to \mathbb{N}$ be defined by $f(n) =$ the highest prime factor of n. Find the range of f.

Solution: We evaluate f(n) for each n in A: $f(9)=3$, $f(10)=5$, $f(11)=11$, $f(12)=3$, $f(13)=13$. The set of all output values is the range. Range = $\{3, 5, 11, 13\}$.

Example 14: Evaluating a Function

Question: A function f is defined by $f(x) = 2x-5$. Find the values of $f(0)$, $f(7)$, and $f(-3)$.

Solution: $f(0) = 2(0) - 5 = -5$. $f(7) = 2(7) - 5 = 14 - 5 = 9$. $f(-3) = 2(-3) - 5 = -6 - 5 = -11$.

Example 15: Finding Domain of a Radical Rational Function

Question: Find the domain of the function $f(x) = \sqrt{\frac{x-2}{x+2}}$.

Solution: We need the expression inside the square root to be non-negative: $\frac{x-2}{x+2} \ge 0$. And the denominator cannot be zero: $x \ne -2$. Using the sign-chart method with critical points -2 and 2, the inequality holds for $x < -2$ and $x \ge 2$. The domain is $(-\infty, -2) \cup [2, \infty)$.

Example 16: Finding Range of $f(x) = x^2+2$

Question: Find the range of the function $f(x) = x^2+2$, where x is a real number.

Solution: We know that for any real number x, $x^2 \ge 0$. Adding 2 to both sides gives $x^2 + 2 \ge 2$. Since $f(x) = x^2+2$, this means $f(x) \ge 2$. The range is $[2, \infty)$.

Example 17: Finding subsets of a Cartesian Product

Question: Let $A=\{1,2\}$ and $B=\{3,4\}$. Write $A \times B$. How many subsets will $A \times B$ have?

Solution: $A \times B = \{(1,3), (1,4), (2,3), (2,4)\}$. The number of elements in $A \times B$ is 4. The number of subsets is $2^4 = 16$.

Example 18: Evaluating a Piecewise Function

Question: A function f is defined by $f(x) = \begin{cases} 1-x, & x<0 \\ 1, & x=0 \\ x+1, & x>0 \end{cases}$. Find $f(-2)$ and $f(3)$.

Solution: To find $f(-2)$, we use the first piece because $-2 < 0$. $f(-2) = 1 - (-2) = 3$. To find $f(3)$, we use the third piece because $3 > 0$. $f(3) = 3 + 1 = 4$.

Example 19: Determining Sets from a Product

Question: Let $n(A)=3$ and $n(B)=2$. If $(x,1), (y,2), (z,1)$ are in $A \times B$, find A and B, where x, y, z are distinct.

Solution: The first elements must be in A: $A$ contains x, y, and z. Since $n(A)=3$, $A = \{x, y, z\}$. The second elements must be in B: B contains 1 and 2. Since $n(B)=2$, $B = \{1, 2\}$.

Example 20: Function defined on $\mathbb{Z} \times \mathbb{Z}$

Question: Let $f = \{(ab, a+b) : a,b \in \mathbb{Z}\}$. Is f a function from $\mathbb{Z}$ to $\mathbb{Z}$?

Solution: For f to be a function, each input must map to exactly one output. Let the input be 4. We can form this input in multiple ways:

• Let $a=2, b=2$. The ordered pair is $(4, 2+2) = (4, 4)$.
• Let $a=4, b=1$. The ordered pair is $(4, 4+1) = (4, 5)$. Since the single input value 4 maps to two different output values (4 and 5), f is not a function.

Example 21: Domain of a Logarithmic Function

Question: Find the domain of the function $f(x) = \sqrt{\log_{10}(\frac{5x-x^2}{4})}$.

Solution: We need to satisfy two conditions:

1. The argument of the logarithm must be positive: $\frac{5x-x^2}{4} > 0 \implies 5x-x^2 > 0 \implies x(5-x) > 0$. This holds for $0 < x < 5$.
2. The argument of the square root must be non-negative: $\log_{10}(\frac{5x-x^2}{4}) \ge 0$. This means $\frac{5x-x^2}{4} \ge 10^0 = 1 \implies 5x-x^2 \ge 4 \implies x^2-5x+4 \le 0 \implies (x-1)(x-4) \le 0$. This holds for $1 \le x \le 4$. We must satisfy both conditions, so we take the intersection of $(0, 5)$ and $[1, 4]$. The domain is $[1, 4]$.

Example 22: Range of a Quadratic Rational Function

Question: Find the range of the function $f(x) = \frac{x^2 - x + 1}{x^2 + x + 1}$ for all real x.

Solution: Let $y = \frac{x^2 - x + 1}{x^2 + x + 1}$. Cross-multiply: $y(x^2+x+1) = x^2-x+1$. Rearrange into a quadratic in x: $(y-1)x^2 + (y+1)x + (y-1) = 0$. Since x is real, the discriminant of this quadratic must be greater than or equal to zero: $D = b^2 - 4ac \ge 0$. $(y+1)^2 - 4(y-1)(y-1) \ge 0$ $y^2+2y+1 - 4(y^2-2y+1) \ge 0$ $-3y^2 + 10y - 3 \ge 0 \implies 3y^2 - 10y + 3 \le 0$ $(3y-1)(y-3) \le 0$. The roots are $y=1/3$ and $y=3$. The inequality holds between the roots. The range is $[1/3, 3]$.

Example 23: Identifying Injective/Surjective Functions

Question: Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = 3x+5$. Determine if f is injective (one-to-one) and/or surjective (onto).

Solution: Injectivity: Assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in \mathbb{R}$. $3x_1+5 = 3x_2+5 \implies 3x_1 = 3x_2 \implies x_1 = x_2$. Since $f(x_1)=f(x_2)$ implies $x_1=x_2$, the function is injective. Surjectivity: Let y be an arbitrary element in the codomain $\mathbb{R}$. We need to see if there is an x in the domain such that $f(x)=y$. $y = 3x+5 \implies 3x = y-5 \implies x = \frac{y-5}{3}$. Since y is a real number, $x = \frac{y-5}{3}$ is also a real number. So for any y in the codomain, there exists a preimage x in the domain. The function is surjective.

Example 24: Solving with Greatest Integer Function

Question: Solve the equation $2[x] = x + \{x\}$, where $[x]$ is the greatest integer and $\{x\}$ is the fractional part.

Solution: We know the property $x = [x] + \{x\}$. Substitute this into the equation: $2[x] = ([x] + \{x\}) + \{x\}$ $2[x] = [x] + 2\{x\}$ $[x] = 2\{x\}$ By definition, $0 \le \{x\} < 1$. Therefore, $0 \le 2\{x\} < 2$. This means $0 \le [x] < 2$. Since $[x]$ must be an integer, the possible values for $[x]$ are 0 and 1. Case 1: If $[x] = 0$, then $0 = 2\{x\} \implies \{x\} = 0$. So, $x = [x] + \{x\} = 0+0 = 0$. Case 2: If $[x] = 1$, then $1 = 2\{x\} \implies \{x\} = 1/2$. So, $x = [x] + \{x\} = 1+1/2 = 3/2$. The solutions are $x=0$ and $x=3/2$.

Example 25: Checking Relation Properties

Question: Let a relation R be defined on the set of natural numbers $\mathbb{N}$ as $aRb$ if $|a-b| \le 1$. Check if R is reflexive, symmetric, and transitive.

Solution:

1. Reflexive: For any $a \in \mathbb{N}$, $|a-a| = 0$, and $0 \le 1$. So, $aRa$. The relation is reflexive.
2. Symmetric: If $aRb$, then $|a-b| \le 1$. This implies $|-(b-a)| \le 1$, which is the same as $|b-a| \le 1$. So, $bRa$. The relation is symmetric.
3. Transitive: We need to check if $aRb$ and $bRc$ implies $aRc$. Let's test with a counterexample. Let $a=1, b=2, c=3$. $aRb$: $|1-2|=1 \le 1$. So $(1,2) \in R$. $bRc$: $|2-3|=1 \le 1$. So $(2,3) \in R$. Now we check if $aRc$: $|1-3|=2$. Since $2$ is not less than or equal to 1, $(1,3) \notin R$. The relation is not transitive.

Example 26: Domain of a Combined Function

Question: Find the domain of the function $f(x) = \frac{\sqrt{x+1}}{x}$.

Solution: We have two conditions to satisfy:

1. The expression inside the square root must be non-negative: $x+1 \ge 0 \implies x \ge -1$.
2. The denominator cannot be zero: $x \ne 0$. Combining these, the domain is all numbers greater than or equal to -1, excluding 0. So, the Domain = $[-1, 0) \cup (0, \infty)$.

Example 27: Range of a Quadratic Function

Question: Find the range of the function $f(x) = x^2 - 6x + 5$ for $x \in \mathbb{R}$.

Solution: Let $y = x^2 - 6x + 5$. This is an upward-opening parabola. The minimum value occurs at the vertex. The x-coordinate of the vertex is $x = -b/(2a) = -(-6)/(2*1) = 3$. The minimum value of the function is $f(3) = 3^2 - 6(3) + 5 = 9 - 18 + 5 = -4$. The function can take any value greater than or equal to -4. Therefore, the Range = $[-4, \infty)$.

Example 28: Classifying a Linear Function

Question: Check if the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3x+5$ is injective and/or surjective.

Solution:

• Injective (One-one) Check: Let $f(x_1) = f(x_2)$. Then $3x_1+5 = 3x_2+5 \implies 3x_1 = 3x_2 \implies x_1 = x_2$. Since $f(x_1) = f(x_2)$ implies $x_1 = x_2$, the function is injective.
• Surjective (Onto) Check: We need to see if the range equals the codomain ($\mathbb{R}$). The range of any non-constant linear function is the entire set of real numbers. So, the range is $\mathbb{R}$. The function is surjective.
• Conclusion: Since the function is both injective and surjective, it is a bijection.

Example 29: Classifying a Quadratic Function

Question: Check if the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$ is injective and/or surjective.

Solution:

• Injective Check: Consider $f(2) = 2^2 = 4$ and $f(-2) = (-2)^2 = 4$. Since two different inputs (2 and -2) lead to the same output (4), the function is not injective.
• Surjective Check: The range of $f(x) = x^2$ is $[0, \infty)$, which contains only non-negative real numbers. The codomain is $\mathbb{R}$. Since the range is not equal to the codomain, the function is not surjective.
• Conclusion: The function is neither injective nor surjective.

Example 30: The Effect of Changing Domain and Codomain

Question: Show that the function $f: [0, \infty) \to [0, \infty)$ defined by $f(x) = x^2$ is a bijection.

Solution: This is the same function as the previous example, but with a restricted domain and codomain.

• Injective Check: The domain is now only non-negative real numbers. If we take $f(x_1) = f(x_2)$, then $x_1^2 = x_2^2$. Since $x_1$ and $x_2$ are both non-negative, this implies $x_1=x_2$. The function is injective on this restricted domain.
• Surjective Check: The range of $f(x)=x^2$ for $x \ge 0$ is $[0, \infty)$. The codomain is also given as $[0, \infty)$. Since the range equals the codomain, the function is surjective.
• Conclusion: On the specified domain and codomain, the function is bijective.

Example 31: Classifying a Rational Function

Question: Let $A = \mathbb{R} - \{3\}$ and $B = \mathbb{R} - \{1\}$. Is the function $f: A \to B$ given by $f(x) = \frac{x-2}{x-3}$ a bijection?

Solution:

• Injective Check: Let $f(x_1) = f(x_2)$. 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 $-3x_1 - 2x_2 = -3x_2 - 2x_1$, so $x_1=x_2$. The function is injective.
• Surjective Check: Let $y$ be an element in the codomain B. We set $y = \frac{x-2}{x-3}$ and solve for x. $y(x-3)=x-2 \implies yx-3y = x-2 \implies yx-x = 3y-2 \implies x(y-1)=3y-2 \implies x=\frac{3y-2}{y-1}$. Since $y \in B$, $y \ne 1$, so the denominator is never zero. This means for every y in the codomain B, we can find a corresponding x in the domain A. The function is surjective.
• Conclusion: Since the function is both injective and surjective, it is a bijection.

Example 32: Classifying a Trigonometric Function

Question: Is the function $f: \mathbb{R} \to [-1, 1]$ defined by $f(x) = \sin(x)$ injective and/or surjective?

Solution:

• Injective Check: We know that $\sin(0)=0$ and $\sin(\pi)=0$. Since different inputs (0 and $\pi$) give the same output (0), the function is not injective.
• Surjective Check: The range of the $\sin(x)$ function is $[-1, 1]$. The codomain is also given as $[-1, 1]$. Since the range equals the codomain, the function is surjective.
• Conclusion: The function is surjective but not injective.