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how to prove cubic root of 25 is irrational using mathematical induction?

I've been trying to do it for hours but can't get it, help plz guys :S

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closed as off-topic by Najib Idrissi, Davide Giraudo, Claude Leibovici, Willie Wong, Yiorgos S. Smyrlis Feb 16 at 12:12

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Do you know why the square root of two is irrational? Can you modify the proof? –  Alexander Thumm Oct 12 '12 at 7:11
Why did you tag this with the Induction tag ? Do you know about rings ? –  Belgi Oct 12 '12 at 7:19
Proofs by induction are for questions like "Prove such and such for all $n$." Since you are concerned with the single number $25$, proof by induction is not appropriate here. –  Austin Mohr Oct 12 '12 at 7:21
What does plz mean? Is that a short for "Platz" in German? –  Asaf Karagila Oct 12 '12 at 7:21
@asaf But what would "Hilfe Platz Jungs: S" then mean? –  draks ... Oct 12 '12 at 7:29

1 Answer 1

up vote 2 down vote accepted

Let's suppose $r$ is a rational such that $r^3 = 25$, then there exist $a, b\in \mathbb N$ coprimes such that $$ a^3 = 25\, b^3 \,. $$ The above relation implies the following chain $$ 5 \mid a^3 \implies 5 \mid a \implies 5^3 \mid 25\, b^3 \implies 5 \mid b^3 \implies 5 \mid b $$ but that contradicts the assumption that $a$ anb $b$ are relatively prime.

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