Question

Problems 2.21, 2.22, 2.23

Answer 1

Statements can be proved by contrapositive, contradiction or by induction.

• 2.21 and 2.23 are proved by contrapositive
• 2.22 is proved by induction

2.21: If $n^3$ is even, then n is even (By contrapositive)

The contrapositive of the above statement is that:

If n is odd, then  $n^3$ is odd

Represent the value of n as:

$n = 2k + 1$, where $k \ge 0$

Take the cube of both sides

$n^3 = (2k + 1)^3$

Expand

$n^3 = 8k^3 + 6k^2 + 6k + 1$

Group

$n^3 =[ 8k^3 + 6k^2 + 6k] + 1$

Factor out 2

$n^3 =2[4k^3 + 3k^2 + 3k] + 1$

Assume w is an integer; where:

$w =4k^3 + 3k^2 + 3k$

So, we have:

$n^3 =2w + 1$

The constant term (i.e. 1) means that $n^3$ is odd.

Hence, the statement has been proved by contrapositive.

i.e. If n is odd, then  $n^3$ is odd

2.22  $3n + 4$ is even, if and only if n is even

We have: $3n + 4$

Assume that: $n = 2k + 2$ for $k \ge 0$

So, we have:

$3n + 4 = 3(2k + 2) + 4$

Open bracket

$3n + 4 = 6k + 6 + 4$

$3n + 4 = 6k + 10$

Factorize

$3n + 4 = 2(3k + 5)$

The factor of 2 means that $3n + 4$ is even.

Hence, $3n + 4$ is even, if and only if n is even

2.22: $s \ne -1$ and $t \ne -1$, then $s + t + st \ne -1$

To do this, we prove by contrapositive.

The contrapositive of the above statement is:

If $s = -1$ and $t=-1$, then $s + t + st = -1$

We have:

$s + t + st = -1$

Substitute the values of s and t in: $s + t + st = -1$

$-1 -1 -1 \times -1 = -1$

$-1 -1 + 1 = -1$

$-1 = -1$

Hence, by contrapositive:

If $s = -1$ and $t=-1$, then $s + t + st = -1$

Read more about proofs  at:

brainly.com/question/19643658