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Show that every fraction whose square root is rational takes the form $kp^2/kq^2$.

What i've done:

We're given that $$\sqrt{\frac{a}{b}}=\frac{p}{q}$$ Take $p/q$ as a fraction in lowest terms,then by squaring we get (assuming that $a,b$ $>0$): $$\frac{a}{b}=\frac{p^2}{q^2}$$ Since $p^2/q^2$ is also a fraction in lowest terms( since if we had $p^2/ q^2 =xs/xt$ we would have a contradiction about $p/q$ being in lowest terms),we have that every other fraction,say $c/d$, which is equal to $\sqrt{a/b}$ must be of the form $kp^2/kq^2$ .

Question: Is this valid as a proof,if not where does it fail ?If you can provide also other proofs about the proof in question i'd appreciate.

P.S: i apologize if my latex syntax isn't really appropriate.

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  • $\begingroup$ The proof is somewhat valid. You have to motivate why $p^2/q^2$ is in the lowest terms too (which shouldn't be too hard given unique prime factorization). BTW using (la)tex is appropriate (at least if it works). $\endgroup$ – skyking Sep 23 '15 at 14:24
  • $\begingroup$ You should use different letters for different variables. Otherwise it's fine. $\endgroup$ – tomglabst Sep 23 '15 at 14:29
  • $\begingroup$ Is it fine now ? $\endgroup$ – Nameless Sep 23 '15 at 14:34
  • $\begingroup$ If for a number theory course, it would likely be assessed as incomplete. Note that it includes a proof that $\sqrt{2}$ is irrational. In some contexts the argument should be viewed as being adequate. $\endgroup$ – André Nicolas Sep 23 '15 at 14:43
  • $\begingroup$ the proof is only for myself.Could you tell me why it would be incomplete for a number theory course ? $\endgroup$ – Nameless Sep 23 '15 at 14:49

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