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I have two numbers. $x, p$

This numbers, have a integer root, then: $\sqrt{x} \in \mathbb Z, \sqrt{p} \in \mathbb Z$. And also:

$x = c^2$, $p = d^2$, because it have a integer root.

So, prove that $xp = k$, where $k$ have a integer root.

My development was:

If, $xp = k$ , that is: $c^2d^2 = k$ and since $c,d$ are integer numbers, so:

$cd = \sqrt{k}$, And here I have not been able to continue, because now I need to prove that:

The product of two integers, will always be integer

¿How i can prove it?

kub0x
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ESCM
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  • You only need $\sqrt{xp}=\sqrt{x}\cdot \sqrt{p}\in \mathbb Z$ , hence $xp$ must be a perfect square. Your approach works as well. – Peter Jun 21 '18 at 11:33
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    The product of two integers is an integer by definition, so you are done. – 5xum Jun 21 '18 at 11:38

2 Answers2

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This is a really general question about multiplying numbers. If the course is not too basic, it is ok to continue by stating that it is trivial. If you want to have a proof, see http://foolproof.pbworks.com/w/page/8978143/The%20Product%20Of%20Two%20Integers%20Is%20An%20Integer

Stan Tendijck
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Let $a,b \in \mathbb{Z}$. Since the set of integers forms a semigroup under multiplication:

$a,b \in (\mathbb{Z}, *)$ then $a*b \in (\mathbb{Z},*)$ and $b*a \in (\mathbb{Z},*)$ (Satisfies closure).

Note: it is a semigroup because it doesn't satisfy inverse axiom.

kub0x
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  • This is a tautology, not an answer. The OP wants to know why the integers are closed under multiplication. S/he is not likely to know what a semigroup is. – Ethan Bolker Jun 21 '18 at 12:32
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    Take $a,b \in \mathbb{Z}$. Add $a$ $b$ times $a+a+a+a...=ab$. Since $a$ is an integer added $b$ times to itself, ab is an integer. Is it better approach? – kub0x Jun 21 '18 at 12:37