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Rudik [331]
3 years ago
6

Problems 2.21, 2.22, 2.23

Mathematics
1 answer:
RoseWind [281]3 years ago
7 0

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

  • <em>2.21 and 2.23 are proved by contrapositive</em>
  • <em>2.22 is proved by induction</em>

<u />

<u />

<u>2.21: If </u>n^3<u> is even, then n is even (By contrapositive)</u>

The contrapositive of the above statement is that:

<em>If n is odd, then  </em>n^3<em> is odd</em>

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.

<em>i.e. If n is odd, then  </em>n^3<em> is odd</em>

<u />

<u>2.22  </u>3n + 4<u> is even, if and only if n is even</u>

We have: 3n + 4<u />

<u />

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.

<em>Hence, </em>3n + 4<em> is even, if and only if n is even </em>

<em />

<u />

<u>2.22: </u>s \ne -1<u> and </u>t \ne -1<u>, then </u>s + t + st \ne -1<u />

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

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