Science:Math Exam Resources/Courses/MATH220/December 2010/Question 02 (a)

MATH220 December 2010

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Question 02 (a)
Let $a\in \mathbb {Z}$. Prove that $\displaystyle {}a$ is even if and only if $\displaystyle {}a^{3}$ is even.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Recall that an integer n is even if and only if it can be written in the form $\displaystyle n=2k$ for some other integer k.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since this is an if and only if proof, we have two directions to prove. First we prove that $\displaystyle a$ being even implies that $\displaystyle a^{3}$ is even. Suppose a is even. By the hint, we write $\displaystyle a=2k$ for some integer k. Cubing both sides gives $\displaystyle a^{3}=8k^{3}=2(4k^{3})$ and since $\displaystyle (4k^{3})$ is an integer, we see that $\displaystyle a^{3}$ satisfies the definition of being even. Next we prove that $\displaystyle a^{3}$ being even implies that $\displaystyle a$ is even. Suppose $\displaystyle a^{3}$ is even. We proceed by contradiction and assume that a is odd. Then $\displaystyle a=2k+1$ for some integer k, Cubing both sides yields $\displaystyle a^{3}=(2k+1)^{3}=8k^{3}+12k^{2}+6k+1=2(4k^{3}+6k^{2}+3k)+1$ and so once again by definition, we see that $\displaystyle a^{3}$ is actually an odd number. This is a contradiction. Hence $\displaystyle a$ must have been even.

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