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How can I prove that if the cube of a number is a multiple of 2, then that number is a multiple of 2

Bill Dubuque
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Joao
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    What kind of number are you talking about? Integers? If the reals are included this is not true. Take as an example, $2^{1/3}$ whos cube is clearly a multiple of $2$ while the number itself is not. – Unknown Apr 28 '20 at 17:31
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    Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz Apr 28 '20 at 17:31

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Assume $x \in \mathbb{Z}$. We want to prove that if $x^3$ is divisible by $2$ then so is $x$.

By contradiction, if $x$ is not divisible by 2 we can write $x=2n+1$ for some $n\in \mathbb{Z}$ and get:

$x^3 = (2n+1)^3 = 8n^3 + 12n^2 + 6n +1$

which clearly is not divisible by 2. Which contradicts that $x^3$ is divisible by 2, and hence, the assumption that $x$ is not divisible by 2 must be false, which proves that $x$ is divisible by 2.

Unknown
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