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I understand that there are already several answers to how to prove this question:

Prove if $n^2$ is even, then $n$ is even.

Prove that if $n^2$ is even then $n$ is even

I am trying to understand why my direct proof is wrong

Given that $a^2$ is even, prove that $a$ is even.

Proof:

We have that $a^2 = 2b$ by the definition of an even number.

  1. $a^2 = 4c^2$ by the fact that an even number times any other number results in an even number.
  2. $a = 2c$ by the definition of a square root.

Therefore $a$ is even.

My question: If I had validated step (1) by saying that any number squared is an even number would that have been circular logic?

Also, I think this proof is wrong because $c$ might be an irrational number say $\sqrt{2}$.

This makes me think that step (2) was wrong.

Can you please explain why this did not work out?

Community
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k m
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  • You might find some clarity if you write out step (1) more precisely. In particular, what are $b$ and $c$? – Travis Willse Jan 12 '15 at 02:57
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    How do you get the step $(1)$? Where did $c$ come from/ You've assumed what you were trying to prove there. – Thomas Andrews Jan 12 '15 at 02:57
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    @induktio Don't edit away the numbers of the bullet items, when several people are referencing them in comments. – Thomas Andrews Jan 12 '15 at 03:19
  • @ThomasAndrews My mistake. I accidentally forgot that an indented 1. was for enumeration and an indented hyphen was for adding bullets. Thanks for fixing it. – Daniel W. Farlow Jan 12 '15 at 03:40
  • Thanks for the feedback, would this be a better explanation for (1): $a^2$ = 4 $c^2$ where $c^2$ is an integer. This is because any integer times an even number results in an even number. So 4$c^2$ represents an arbitrary even number.

    Thus a = 2c and a is even

    I can't quite see the problem here

     – k m Jan 12 '15 at 03:47
  • @km How does $4c^2$ represent an arbitrary even number? What about the even number 6 then? Could $4c^2$ represent 6 where $c^2$ is an integer? – Daniel W. Farlow Jan 12 '15 at 03:53
  • Why does the statement "any integer times an even number result in an even number" give you $4c^2$? Also, $c^2$ can be an integer without $c$ being an integer. – Eff Jan 12 '15 at 03:53
  • Thanks! I now understand what I was doing wrong. I really appreciate it. – k m Jan 12 '15 at 03:55

2 Answers2

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Maybe you could try something like this if you want to use a similar approach to yours: $a$ has an unique representation in terms of factorization into prime factors: $a = p_1^{n_1} \cdots p_k^{n_k}$. Therefore $a^2 = p_1^{2n_1} \cdots p_k^{2n_k}$. Then if $a$ is even one of the $p_i = 2$. Hope that makes sense.

Loreno Heer
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You could not validate step (1) by saying that any number squared is an even number (e.g., $3^2 = 9$, but 3 is an odd number). As for explicitly why your proof is wrong, the statement that $a^2 = 4c^2$ is a non sequitur; that is, it does not logically follow from what you start out assuming, namely that $a^2 = 2b$. If you really want a direct proof, you could do something as follows, although it does not fall out quite nicely.

Direct proof: Suppose $a^2$ is even. Then $a^2 = 2b$, where $b\in\mathbb{Z}$. Then $\frac{a^2}{2}=b\neq 0 \rightarrow \left[\frac{a^2}{2}\right] = [0]$, where $[0]$ denotes an equivalence class representative of the even integers. This is because there are two congruence classes mod 2, namely the even and odd integers (for which $[0]$ and $[1]$ are equivalence class representatives, respectively). I doubt this is the kind of answer you were looking for though.

Alternative direct proof: Suppose $a^2$ is even. Then let us represent $a^2$ as follows: $$ a^2 = a\cdot a = a_1\cdot a_2, $$ where we know $a_1 = a_2$. Well, when two integers $c$ and $d$ are multiplied together and yield an even integer, then $c$ and $d$ must both be even or one must be even and the other odd (i.e., they cannot both be odd; if you use this fact, then you may need to prove it in case you cannot take it for granted). Well, since $a_1 = a_2$ and $a_1\cdot a_2 = a^2$, we must necessarily have that $a$ is even (this due to the fact that a single number cannot be both even and odd; if $a$ is odd, then we are multiplying two odd numbers, yielding an odd number as a product--this means $a$ can only be even then). This is kind of an ugly proof, but it may be more along the lines of what you are looking for.

A proof by contraposition seems ideal here though in terms of ease and clarity.

Daniel W. Farlow
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