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I have to show that the $\sqrt(2/7)$ is irrational. Here is my work.

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Edward Jiang
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Bayerischer
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  • This might help: http://math.stackexchange.com/questions/448172/what-rational-numbers-have-rational-square-roots – Eitan Porat Feb 06 '16 at 10:55

2 Answers2

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You haven't actually reached a contradiction unless you've asserted at the outset that $(m,n) = 1$. But once you assert that you can stop with this line: $2n^2 = 7m^2$. If $n$ is even, $m$ must be odd and the LHS is even, the RHS isn't. Contradiction. But if $n$ is odd and $m$ is even the LHS is an odd multiple of $2$ while the RHS is an even multiple of $2$, again leading to a contradiction.

Edited to patch a hole in the proof that I just noticed after notification of the upvote.

Deepak
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  • Any rational number is written in a fraction that's irreducible. So it's understood that gcd(m,n) = 1 – Saikat Feb 06 '16 at 11:23
  • @user230452 A rational number can be represented in the form $\frac ab$. That fraction may or may not be in a reduced form, and in the latter case, a and b are not coprime. But the point is that you can always reduce that fraction by cancelling out all common factors until left with 2 coprime numbers m and n. The point I was making was that if you don't start with this assertion, you will not have a valid proof by contradiction. – Deepak Feb 06 '16 at 11:52
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You're skipping some steps.

You say that $7m^2$ is even, then you assume $m$ is even. This is true. You should say why though, you explain a little more why $n$ is even.

Also, how is the end a contradiction? You found that the gcd is even, when did you ever assume that it wasn't?

  • If a number where rational gcd(m,n) equals 1. Since this is not the case, I reached a contradiction and the number is irrational. – Bayerischer Oct 13 '15 at 01:33
  • @Bayerischer That's not what's going on here. $\frac22$ is surely rational, but $\gcd(2,2)$ isn't $1$. –  Oct 13 '15 at 01:34
  • $7m^2$ can be expanded as a product of prime factors. Since, 7 cannot be reduced further. The 2 term must one from the m^2 part, which implies m is even too. The contradiction is a gcd more than 1. – Saikat Feb 06 '16 at 11:25