Chapter 1.4, Problem 5E

(a)

To determine

To prove: $x+y$ is even if $x\text{ and }y$ are even.

Explanation of Solution

Given information:

Let $x,y,\text{ and }z$ are integers.

Proof:

Let the even integer $x=2k$ , for some integer $k\in 𑨀$ , and let the even integer $y=2l$ for any integer $l\in 𑨀$ .

  $\begin{array}{l}x+y=2k+2l\\ =2(k+l)\\ =2a(\because k+l=a\in 𑨀)\end{array}$

Hence, $x+y$ is even.

(b)

To determine

To prove: $xy$ is even if $x$ is even.

Explanation of Solution

Proof:

Let the even integer $x=2k$ , for some integer $k\in 𑨀$ .

If $y$ is even, let $y=2l$ , where $l\in 𑨀$ .

  $\begin{array}{l}xy=2k\cdot 2l\\ =4kl\\ =2(2kl)\\ =2a(\because 2kl=a\in 𑨀)\end{array}$

If $y$ is odd, let $y=2l+1$ , where $l\in 𑨀$ .

  $\begin{array}{l}xy=2k\cdot (2l+1)\\ =2\left[k(2l+1)\right]\\ =2a(\because k(2l+1)=a\in 𑨀)\end{array}$

Hence, $xy$ is even.

(c)

To determine

To prove: $xy$ is devisable by $4$ if $x\text{ and }y$ are even.

Explanation of Solution

Proof:

Let the even integer $x=2k$ , for some $k\in 𑨀$ and let the even integer $y=2l$ , where $l\in 𑨀$ .

  $\begin{array}{l}xy=2k\cdot 2l\\ =4kl\\ =4a(\because kl=a\in 𑨀)\end{array}$

Hence, $xy$ is devisable by $4$ .

(d)

To determine

To prove: $3x-5y$ is even if $x\text{ and }y$ are even.

Explanation of Solution

Proof:

Let the even integer $x=2k$ , for some $k\in 𑨀$ and let the even integer $y=2l$ , where $l\in 𑨀$ .

  $\begin{array}{l}3x-5y=3\cdot 2k-5\cdot 2l\\ =2(3k-5l)\\ =2a(\because 3k-5l=a\in 𑨀)\end{array}$

Hence, $3x-5y$ is even.

(e)

To determine

To prove: $x+y$ is even if $x\text{ and }y$ are odd.

Explanation of Solution

Proof:

Let the oddinteger $x=2k+1$ , for some $k\in 𑨀$ and letthe odd integer $y=2l+1$ , where $l\in 𑨀$ .

  $\begin{array}{l}x+y=2k+1+2l+1\\ =2k+2l+2\\ =2(k+l+1)\\ =2a(\because k+l+1=a\in 𑨀)\end{array}$

Hence, $x+y$ is even.

(f)

To determine

To prove: $3x-5y$ is even if $x\text{ and }y$ are odd.

Explanation of Solution

Proof:

Let the oddinteger $x=2k+1$ , for some $k\in 𑨀$ and let the odd integer $y=2l+1$ , where $l\in 𑨀$ .

  $\begin{array}{l}3x-5y=3(2k+1)-5(2l+1)\\ =6k+3-10l-5\\ =6k-10l-2\\ =2(3k-5l-1)\\ =2a(\because 3k-5l-1=a\in 𑨀)\end{array}$

Hence, $3x-5y$ is even.

(g)

To determine

To prove: $xy$ is odd if $x\text{ and }y$ are odd.

Explanation of Solution

Proof:

Let the odd integer $x=2k+1$ , for some $k\in 𑨀$ and let the odd integer $y=2l+1$ , where $l\in 𑨀$ .

  $\begin{array}{l}xy=(2k+1)(2l+1)\\ =4kl+2k+2l+1\\ =2(2kl+k+l)+1\\ =2a+1(\because 2kl+k+l=a\in 𑨀)\end{array}$

Hence, $xy$ is odd.

(h)

To determine

To prove: $x+y$ is odd if $x$ is even and $y$ is odd.

Explanation of Solution

Proof:

Let the eveninteger $x=2k$ , for some $k\in 𑨀$ and let the odd integer $y=2l+1$ , where $l\in 𑨀$ .

  $\begin{array}{l}x+y=2k+(2l+1)\\ =2k+2l+1\\ =2(k+l)+1\\ =2a+1(\because k+l=a\in 𑨀)\end{array}$

Hence, $x+y$ is odd.

(i)

To determine

To prove: $x+y+z$ is even if exactly one of $x,y,z$ is even.

Explanation of Solution

Proof:

Let the even integer $x=2k$ , for some $k\in 𑨀$ and let the odd integers $y=2l+1$ , and $z=2m+1$ where $l,m\in 𑨀$ .

  $\begin{array}{l}x+y+z=2k+(2l+1)+(2m+1)\\ =2k+2l+1+2m+1\\ =2(k+l+m+1)\\ =2a(\because k+l+m+1=a\in 𑨀)\end{array}$

Hence, $x+y+z$ is even.

(j)

To determine

To prove: $xy+yz$ is even if exactly one of $x,y,z$ is odd.

Explanation of Solution

Proof:

If $z$ is odd:

Let the even integers $x=2k$ , and $y=2l$ for some $k,l\in 𑨀$ and let the odd integer $z=2m+1$ where $m\in 𑨀$ .

  $\begin{array}{l}xy+yz=2k\cdot 2l+2l(2m+1)\\ =2k+2l++4lm+2\\ =2(k+l+2lm+1)\\ =2a(\because k+l+2lm+1=a\in 𑨀)\end{array}$

If $y$ is odd:

Let the even integers $x=2k$ , and $z=2m$ for some $k,m\in 𑨀$ and let the odd integer $y=2l+1$ where $l\in 𑨀$ .

  $\begin{array}{l}xy+yz=2k\cdot (2l+1)+(2l+1)2m\\ =4kl+2k++4lm+2m\\ =2(2kl+k+2lm+m)\\ =2a(\because 2kl+k+2lm+m=a\in 𑨀)\end{array}$

Hence, $xy+yz$ is even.