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(√7) is (IR) irrational Number. How to prove it√7 .

Firstly I tried a/b =√7 and a^2 /b^2 =7 7b^2=a^2 And then I couldn't continue . How I can Prove it ?

David
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  • Do you know the classic proof that $\sqrt2$ is irrational? This proof is basically exactly the same (and you've started well). – Arthur Oct 15 '19 at 05:43
  • Yes , I aggre with you . But when I prove it in √2 I look at odd and even number . But √7 different . I can't look odd or even . I have to tried new way . Right? – David Oct 15 '19 at 05:49
  • "I look at odd and even number ..... I can't look odd or even " Even means divisible by two. So what? Divisible by $7$ just as legitimate a concept. $7$ divides evenly into $a^2$ so ..... – fleablood Oct 15 '19 at 06:13
  • ....But this does assume that Euclids lemma that if prime number $p$ divides $ab$ then $p$ divides $a$ or $p$ divides $b$. Don't know if you have proven that. – fleablood Oct 15 '19 at 06:14
  • I understand thanks for help – David Oct 15 '19 at 07:07

2 Answers2

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Suppose $\sqrt{7} = a/b$, where $a/b$ is in lowest terms, meaning $\gcd(a,b) = 1$. Then, as you showed, $7b^2 = a^2$. This forces $7 \mid a^2$, which, since $7$ is prime forces $7 \mid a$. But then $7^2$ divides $a^2$, so $7^2$ divides $7b^2$, forcing $7 \mid b^2$, forcing $7 \mid b$. Now we have shown $7 \mid a$ and $7 \mid b$, which contradicts our assumption that $\gcd(a,b) = 1$. By contradiction, $\sqrt{7}$ is not rational.

Eric Towers
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Again, suppose that $\sqrt 7=a/b$ in lowest terms. Then $a^2=7b^2$. Modulo $5$ the only possibilities for $a^2$ are $a^2\equiv0$, $1$ or $4\pmod5$. Likewise, $7b^2\equiv0$, $2$ or $3\pmod5$. Therefore $a^2\equiv 7b^2\equiv0\pmod 5$ which is only possible if $a$ and $b$ are both divisible by $5$. This contradicts $a/b$ being in lowest terms.

Angina Seng
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