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I have read several articles on math.stackexchange.com, and also this article: https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/irrationality_of_3.htm

I still can't quite understand why one of the numbers can't be even. Especially this part: "Since any choice of even values of a and b leads to a ratio a/b that can be reduced by canceling a common factor of 2, we must assume that a and b are odd, and that the ratio a/b is already reduced to smallest possible terms."

Isn't 2/3 or 3/2 smallest possible terms as well?

am1212
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  • have your read the proof about the sqrt of 2 ? –  Sep 29 '17 at 23:11
  • Are either of those numbers the square root of 3? – Nij Sep 29 '17 at 23:16
  • If $\sqrt{3} = a/b,$ is equal to a rational, then we can reduce that rational to lowest terms. So w.l.o.g. we may assume it is equal to a rational with $,a,b,$ having no common factor, which excludes the case $a,b$ both even. The rest of the proof is simpler this way:$,\bmod 4!:\ {\rm odd}^2\equiv 1$ so $, 1\equiv a^2\equiv 3b^2\equiv 3,,$ contradiction. – Bill Dubuque Sep 29 '17 at 23:16
  • I have.. from https://www.grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/irrationality_of_2.htm as well but still can't get that 2/3 or 3/2 out of the mind just from the description. – am1212 Sep 29 '17 at 23:17
  • Note that $(a/b)^2 = 3$ by assumption, so $a^2 = 3b^2$. If a and b have different parity, so do $a^2$ and $3b^2$. Thus, you may assume that both a and b have the same parity: Both are odd – H1ghfiv3 Sep 29 '17 at 23:18
  • @user3551523 The proofs on that web page of irrationality of $\sqrt 2$ and $\sqrt 3$ leave much to be desired. They are longer, less conceptual,and more poorly presented than well-known standard proofs that you will find in most elementary textbooks (and also here). If you wish to better understand these topics then study the standard proofs. – Bill Dubuque Sep 29 '17 at 23:43

2 Answers2

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Certainly, $\frac{2}{3}$ is in smallest possible terms despite $2$ being even and $3$ being odd.

The point in this proof however that they are trying to make is that given that $a$ and $b$ are both natural numbers if we were to assume that $3b^2=a^2$ we have that if at least one of the two are even then both must be even and that if at least one of them is odd that both must be odd. This has to do with this equation and does not directly have to do with that $\frac{a}{b}$ is in simplest terms. Notice that if exactly one of the two is even and the other is odd then we would have an even number equal to an odd number which is impossible.

We notice then that if we were to assume that $a$ and $b$ are both even then we run into the issue of $\gcd(a,b)\geq 2$ as $2$ would be a common divider, contradicting that $\frac{a}{b}$ was in simplest terms. This implies then that $a$ and $b$ both cannot be even, and coupling with the earlier observation that we cannot have exactly one of them even and the other odd, this implies that both must be odd.

The proof continues as written.

Really, this step about them both being odd is unnecessary to begin with. We could have directly appealed to the fact that since $3b^2=a^2$ that $3\mid a^2$. Since $3$ is prime this implies that $3\mid a$. Similarly we can then show that $3\mid b$, arriving at our contradiction directly.

JMoravitz
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Your citation establishes that $a$ and $b$ must either both be even, or both be odd:

  • $b$ is either even or odd (2 cases)
    • If $b$ is odd, then $b^2$ is odd. Hence $3b^2$ is odd, being the product of two odd numbers. But $3b^2 = a^2$, and so $a^2$ is odd. And this means $a$ must be odd, since the square of an even number would be even.
    • If $b$ is even, then $b^2$ is even, so $3b^2$ is even, so $a^2$ is even, so $a$ is even.
Managu
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