Statements can be proved by contrapositive, contradiction or by induction.
2.21: If [tex]n^3[/tex] is even, then n is even (By contrapositive)
The contrapositive of the above statement is that:
If n is odd, then [tex]n^3[/tex] is odd
Represent the value of n as:
[tex]n = 2k + 1[/tex], where [tex]k \ge 0[/tex]
Take the cube of both sides
[tex]n^3 = (2k + 1)^3[/tex]
Expand
[tex]n^3 = 8k^3 + 6k^2 + 6k + 1[/tex]
Group
[tex]n^3 =[ 8k^3 + 6k^2 + 6k] + 1[/tex]
Factor out 2
[tex]n^3 =2[4k^3 + 3k^2 + 3k] + 1[/tex]
Assume w is an integer; where:
[tex]w =4k^3 + 3k^2 + 3k[/tex]
So, we have:
[tex]n^3 =2w + 1[/tex]
The constant term (i.e. 1) means that [tex]n^3[/tex] is odd.
Hence, the statement has been proved by contrapositive.
i.e. If n is odd, then [tex]n^3[/tex] is odd
2.22 [tex]3n + 4[/tex] is even, if and only if n is even
We have: [tex]3n + 4[/tex]
Assume that: [tex]n = 2k + 2[/tex] for [tex]k \ge 0[/tex]
So, we have:
[tex]3n + 4 = 3(2k + 2) + 4[/tex]
Open bracket
[tex]3n + 4 = 6k + 6 + 4[/tex]
[tex]3n + 4 = 6k + 10[/tex]
Factorize
[tex]3n + 4 = 2(3k + 5)[/tex]
The factor of 2 means that [tex]3n + 4[/tex] is even.
Hence, [tex]3n + 4[/tex] is even, if and only if n is even
2.22: [tex]s \ne -1[/tex] and [tex]t \ne -1[/tex], then [tex]s + t + st \ne -1[/tex]
To do this, we prove by contrapositive.
The contrapositive of the above statement is:
If [tex]s = -1[/tex] and [tex]t=-1[/tex], then [tex]s + t + st = -1[/tex]
We have:
[tex]s + t + st = -1[/tex]
Substitute the values of s and t in: [tex]s + t + st = -1[/tex]
[tex]-1 -1 -1 \times -1 = -1[/tex]
[tex]-1 -1 + 1 = -1[/tex]
[tex]-1 = -1[/tex]
Hence, by contrapositive:
If [tex]s = -1[/tex] and [tex]t=-1[/tex], then [tex]s + t + st = -1[/tex]
Read more about proofs at:
https://brainly.com/question/19643658
