Question

Prove:
For all integers n, if n^2 is odd, then n is odd.
Use a proof by contraposition, as in Lemma 1.1.
Let n be an integer. Suppose that n is even, i.e., n = blank for some integer k. Then n^2 = blank = 2( blank) is also even.

Answer 1

Form the contrapositive of the given statement. That is,
For all integers n, if n is not odd, then n2 is not odd

But, we know that an integer is not odd if, and only if, it is even [by parity property]. So, the contrapositive becomes

For all integer n, if n is even, then n^2 is even

Answer 2

The prove the contrapositive using method of direct proof:

Suppose n is an integer. We must show that n^2 is also even. By definition of even, we have

n = 2k for some integer k.

Answer 3

Then by substitution, we have

n.n = (2k) . (2k)

n.n= 4(k^2)
n.n=2(2k^2)

Because products of integers are integers and 2 and k are both integers.] Hence, we have a form:

n . n = 2 . (some integer)

and so by definition of even is n^2 is even.

Answer 4

Therefore, the given statement is true by the logical equivalence between a statement and its contrapositve.

Answer 5

I tried using 2 and the problem said that that was incorrect

Answer 6

do you have to submit the answer electronically?

Answer 7

yeah there is a box beside n= and then another after n to the second power and then another after =2( )

Answer 8

can you screen shot it?

Answer 9

Click on the attachment

Answer 10

did you try n=2k for the first bit?

Answer 11

then n=4k^2 = 2(2k^2)

Answer 12

That worked chris00 I appreciate it

Answer 13

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