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I'm working on some high school geometry homework, and I'm having some trouble with a problem about proofs and counterexamples. The question posses the statement

  • $n$ is divisible by $4$ if and only if $n^2$ is even

and asks if that is a true statement (and to provide a counter example if it is not). My understanding of the statement is that "a prerequisite of divisibility by $4$ is that a number is even when squared." Since the square root of an even number is also even (even $\cdot$ even = even), and the definition of an even number is even divisibility by $2$, the statement can be reduced to "a prerequisite to divisibility by $4$ is divisibility by $2$", which is clearly true. However, I'm concerned that my understanding of the statement is fundamentally flawed. Is the statement true or false, and why?

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The square root of an even square number is even, but not for all of them... ($\sqrt(2)$) –  rschwieb Oct 4 '12 at 1:34
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You should check to be sure that the problem is not $\rm\:4\:|\:n^2\!\! \iff 2\:|\:n\ \ $ –  Bill Dubuque Oct 4 '12 at 1:39
    
Think very carefully. Can you think of a number $n$ which is NOT divisible by 4, but where $n^2$ is even? Just think through some numbers, like 1, 2, 3 and so on. EITHER you'll come to such a number, OR you'll start to gain an understanding of why such numbers mightn't exist. I don't think the problem will take you very long if you adopt this approach. –  user22805 Oct 4 '12 at 2:02
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4 Answers 4

It is false. Let $n=2$.${}{}{}{}{}{}{}{}$

The sentence has an "if and only if."

So it is claiming that (i) if $n$ is divisible by $4$, then $n^2$ is even (true) and (ii) if $n^2$ is even, then $n$ is divisible by $4$ (false, easy counterexample).

If I claim that Ottawa is the capital of Canada and elephants fly, then I am uttering a falsehood.

Remark: In general, if you are trying to prove that $A$ if and only if $B$, the first step is to separate the assertion into its two component parts: if $A$ then $B$, and if $B$ then $A$. One part may be correct and the other not, in which case the if and only if assertion is false. Or both may be true, but the proof for one may be very different from the proof for the other.

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@David did you examine a single concrete case at all? –  rschwieb Oct 4 '12 at 1:23
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@rschwieb: If David has correctly described how he understood the statement, every example that he examined would have been evidence that the statement was true. His problem, as he suspected, was in his understanding of the statement. –  Brian M. Scott Oct 4 '12 at 1:25
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You’ve correctly understood half of the statement. A statement of the form $A$ if and only if $B$ means both that $B$ is a prerequisite for $A$ and that $A$ is a prerequisite for $B$. Equivalently, it means that $A$ implies $B$ and $B$ implies $A$.

In you specific case, the statement means that if $n$ is divisible by $4$, then $n^2$ is even, which is true, and that if $n^2$ is even, then $n$ is divisible by $4$, which is not true: just take $n=2$.

However, the following statement is true, and it might be instructive to convince yourself of this:

$n^2$ is divisible by $4$ if and only if $n$ is even.

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$\rm\begin{eqnarray}{\bf Hint}\ \ prime\ \ p\:|\:nm&\iff&\rm p\:|\:n\ \ or\ \ p\:|\:m \\ \rm hence\ \ prime\ \ p\:|\: n^2 &\iff&\rm p\:|\:n\end{eqnarray} $

Equivalently, $\rm\ mod\ p\!:\ nm\equiv 0\!\iff\! n\equiv 0\ \ or\ \ m\equiv 0,\ \ thus\ \ n^2\equiv 0\!\iff\! n\equiv 0.$

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I think the appropriate question is:

Is "$n^{2}$ is divisible by 4 if and only if $n$ is even" a true statement?

and the answer is sure.

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That was already mentioned in my comment above, and also by Brian (as I just now noticed). –  Bill Dubuque Oct 4 '12 at 1:52
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