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>
<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>
<em> is odd</em>
Represent the value of n as:
, where ![k \ge 0](https://tex.z-dn.net/?f=k%20%5Cge%200)
Take the cube of both sides
![n^3 = (2k + 1)^3](https://tex.z-dn.net/?f=n%5E3%20%3D%20%282k%20%2B%201%29%5E3)
Expand
![n^3 = 8k^3 + 6k^2 + 6k + 1](https://tex.z-dn.net/?f=n%5E3%20%3D%208k%5E3%20%2B%206k%5E2%20%2B%206k%20%2B%201)
Group
![n^3 =[ 8k^3 + 6k^2 + 6k] + 1](https://tex.z-dn.net/?f=n%5E3%20%3D%5B%208k%5E3%20%2B%206k%5E2%20%2B%206k%5D%20%2B%201)
Factor out 2
![n^3 =2[4k^3 + 3k^2 + 3k] + 1](https://tex.z-dn.net/?f=n%5E3%20%3D2%5B4k%5E3%20%2B%203k%5E2%20%2B%203k%5D%20%2B%201)
Assume w is an integer; where:
![w =4k^3 + 3k^2 + 3k](https://tex.z-dn.net/?f=w%20%3D4k%5E3%20%2B%203k%5E2%20%2B%203k)
So, we have:
![n^3 =2w + 1](https://tex.z-dn.net/?f=n%5E3%20%3D2w%20%2B%201)
The constant term (i.e. 1) means that
is odd.
Hence, the statement has been proved by contrapositive.
<em>i.e. If n is odd, then </em>
<em> is odd</em>
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<u>2.22 </u>
<u> is even, if and only if n is even</u>
We have:
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<u />
Assume that:
for ![k \ge 0](https://tex.z-dn.net/?f=k%20%5Cge%200)
So, we have:
![3n + 4 = 3(2k + 2) + 4](https://tex.z-dn.net/?f=3n%20%2B%204%20%3D%203%282k%20%2B%202%29%20%2B%204)
Open bracket
![3n + 4 = 6k + 6 + 4](https://tex.z-dn.net/?f=3n%20%2B%204%20%3D%206k%20%2B%206%20%2B%204)
![3n + 4 = 6k + 10](https://tex.z-dn.net/?f=3n%20%2B%204%20%3D%206k%20%2B%2010)
Factorize
![3n + 4 = 2(3k + 5)](https://tex.z-dn.net/?f=3n%20%2B%204%20%3D%202%283k%20%2B%205%29)
The factor of 2 means that
is even.
<em>Hence, </em>
<em> is even, if and only if n is even </em>
<em />
<u />
<u>2.22: </u>
<u> and </u>
<u>, then </u>
<u />
To do this, we prove by contrapositive.
The contrapositive of the above statement is:
If
and
, then ![s + t + st = -1](https://tex.z-dn.net/?f=s%20%2B%20t%20%2B%20st%20%3D%20-1)
We have:
![s + t + st = -1](https://tex.z-dn.net/?f=s%20%2B%20t%20%2B%20st%20%3D%20-1)
Substitute the values of s and t in: ![s + t + st = -1](https://tex.z-dn.net/?f=s%20%2B%20t%20%2B%20st%20%3D%20-1)
![-1 -1 -1 \times -1 = -1](https://tex.z-dn.net/?f=-1%20-1%20-1%20%5Ctimes%20-1%20%3D%20-1)
![-1 -1 + 1 = -1](https://tex.z-dn.net/?f=-1%20-1%20%2B%201%20%3D%20-1)
![-1 = -1](https://tex.z-dn.net/?f=-1%20%3D%20-1)
Hence, by contrapositive:
If
and
, then ![s + t + st = -1](https://tex.z-dn.net/?f=s%20%2B%20t%20%2B%20st%20%3D%20-1)
Read more about proofs at:
brainly.com/question/19643658