Identify the statements that are true. (a) The expression always gives odd numbers. (b) All even numbers can be expressed as . (c) Both expressions and describe all odd numbers. (d) The expression gives both even and odd numbers.

Understand the Question An even number is of the form , where is an integer. Parity properties: To evaluate whether a statement is true or false, analyze the algebraic structure of the expression for integer values of the variable. Part (a) The expression always gives odd numbers. Step 1 · Analyze For any integer , is always an even number. Subtracting (an odd number) from an even number always gives an odd number. Therefore, always gives odd numbers. Answer > (a) True Part (b) All even numbers can be expressed as . Step 1 · Analyze The expression always gives an even number for any integer , but it does not give all even numbers. For example, testing if (an even number) can be obtained: Since is not an integer, cannot be expressed in this form. Answer > (b) False Part (c) Both expressions and describe all odd numbers. Step 1 · Analyze and For any integer , is even, so generates all odd integers: Similarly, for any integer , generates all odd integers: Both expressions describe the set of all odd numbers. Answer > (c) True Part (d) The expression gives both even and odd numbers. Step 1 · Analyze For any integer , is always even and is odd. Since , the expression only produces odd numbers: It never gives an even number. Answer > (d) False Common Mistakes Generating Some vs. Generating All: An expression like produces only even numbers, but it does not produce all even numbers (multiples of plus ). Parity Arithmetic Rules: Misapplying parity rules—remember that adding an odd number to an even number always results in an odd number (e.g., is always odd).

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Question 25

Identify the statements that are true.

(a) The expression $4m - 1$ always gives odd numbers. (b) All even numbers can be expressed as $6j - 4$ . (c) Both expressions $2p + 1$ and $2q - 1$ describe all odd numbers. (d) The expression $2f + 3$ gives both even and odd numbers.

Solution

Understand the Question

An even number is of the form $2k$ , where $k$ is an integer.
Parity properties:
- $\text{Even} \pm \text{Odd} = \text{Odd}$
- $\text{Even} \pm \text{Even} = \text{Even}$
To evaluate whether a statement is true or false, analyze the algebraic structure of the expression for integer values of the variable.

(a) The expression $4m - 1$ always gives odd numbers.

Step 1 · Analyze $4m - 1$

For any integer $m$ , $4m = 2(2m)$ is always an even number.

Subtracting $1$ (an odd number) from an even number always gives an odd number.

\begin{aligned} \text{For } m = 1: &\quad 4(1) - 1 = 4 - 1 = 3 \\ \text{For } m = 2: &\quad 4(2) - 1 = 8 - 1 = 7 \\ \text{For } m = 0: &\quad 4(0) - 1 = 0 - 1 = -1 \end{aligned}

Therefore, $4m - 1$ always gives odd numbers.

Answer

(a) True

(b) All even numbers can be expressed as $6j - 4$ .

Step 1 · Analyze $6j - 4$

The expression $6j - 4 = 2(3j - 2)$ always gives an even number for any integer $j$ , but it does not give all even numbers.

For example, testing if $0$ (an even number) can be obtained:

\begin{aligned} 6j - 4 &= 0 \\[0.6em] 6j &= 4 \\[0.6em] j &= \dfrac{4}{6} \\[0.6em] j &= \dfrac{2}{3} \end{aligned}

Since $j = \dfrac{2}{3}$ is not an integer, $0$ cannot be expressed in this form.

Answer

(b) False

(c) Both expressions $2p + 1$ and $2q - 1$ describe all odd numbers.

Step 1 · Analyze $2p + 1$ and $2q - 1$

For any integer $p$ , $2p$ is even, so $2p + 1$ generates all odd integers:

\begin{aligned} p = 0 &\implies 2(0) + 1 = 1 \\ p = 1 &\implies 2(1) + 1 = 3 \\ p = -1 &\implies 2(-1) + 1 = -1 \end{aligned}

Similarly, for any integer $q$ , $2q - 1$ generates all odd integers:

\begin{aligned} q = 1 &\implies 2(1) - 1 = 1 \\ q = 2 &\implies 2(2) - 1 = 3 \\ q = 0 &\implies 2(0) - 1 = -1 \end{aligned}

Both expressions describe the set of all odd numbers.

Answer

(c) True

(d) The expression $2f + 3$ gives both even and odd numbers.

Step 1 · Analyze $2f + 3$

For any integer $f$ , $2f$ is always even and $3$ is odd.

Since $\text{Even} + \text{Odd} = \text{Odd}$ , the expression $2f + 3$ only produces odd numbers:

\begin{aligned} \text{For } f = 1: &\quad 2(1) + 3 = 2 + 3 = 5 \\ \text{For } f = 2: &\quad 2(2) + 3 = 4 + 3 = 7 \\ \text{For } f = 0: &\quad 2(0) + 3 = 0 + 3 = 3 \end{aligned}

It never gives an even number.

Answer

(d) False

Common Mistakes

Generating Some vs. Generating All: An expression like $6j - 4$ produces only even numbers, but it does not produce all even numbers (multiples of $6$ plus $2$ ).
Parity Arithmetic Rules: Misapplying parity rules—remember that adding an odd number to an even number always results in an odd number (e.g., $2f + 3$ is always odd).

Question 25

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