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Given: Let $n$ be an integer. Then $n^2=3a$ for some integer $a$ if and only if $n=3b$ for some integer $b$.

Proof real number $\sqrt{3}$ is an irrational number.

Here is I have so far: Assume $\sqrt{3}=a/b$ where $a,b$ are integers. Then $a^2=3(b^2)$, which means $a^2=3c$, where $c$ is some integer.

next is what I'm confused about: With the given fact I can get $a=3d$ where $d$ is integer in order to let $a^2=3c$

I'm not quite sure how to continue this.. Can someone please help?

gt6989b
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FrankTby
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  • Your a and b are just arbitrary integers whose quotient is sqrt(3), correct? You can't assume that 3 divides a. However, you can assume that a and b are such that they share no common factors, so that a^2 and b^2 share no common factors, and hence, that a^2 divides 3. Since 3 is prime, a^2 is thus equal to 1 or 3, which implies that a is an integer that divides 1 or divides 3. See what you can get with this head-start. – Sinister Cutlass Oct 28 '15 at 17:24
  • I don't know how to connect the given fact to the proof.. – FrankTby Oct 28 '15 at 17:33
  • Does the given fact mean:only when n is divided by 3, n^2 is divided by 3? – FrankTby Oct 28 '15 at 17:36

1 Answers1

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Let $3m^2=n^2$. Then the multiplicity of the factor $3$ in the prime decomposition of the LHS is odd, while it is even for the RHS, a contradiction.