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How do you prove that $\sqrt{2} + \sqrt{5}$ is irrational?

I tried to prove it by contradiction and got this equation: $a^2/b^2 = \sqrt{40}$.

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    $\begingroup$ See here math.stackexchange.com/questions/496037/… $\endgroup$
    – Supriyo
    Oct 25, 2013 at 13:08
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    $\begingroup$ I have posted answer for this two times in this forum... please try checking with some key words... $\endgroup$
    – user87543
    Oct 25, 2013 at 13:09
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    $\begingroup$ Square the number: if it's rational, then also $\sqrt{10}$ is. Is it? $\endgroup$
    – egreg
    Oct 25, 2013 at 13:10
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    $\begingroup$ @PraphullaKoushik Which key words? $\sqrt{2}$ isn't exactly eacy to search for... $\endgroup$
    – user1729
    Oct 25, 2013 at 13:55

6 Answers 6

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$\sqrt 2=\dfrac 12\left(\sqrt 2+\sqrt 5-\dfrac 3{\sqrt 2+\sqrt 5}\right)\notin\mathbb Q$, hence $\sqrt 2+\sqrt 5\notin \mathbb Q$

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Use proof by contradiction. Assume that the sum is rationial, that is $$\sqrt2 +\sqrt5 = {a\over b}$$ where $a$ and $b$ are integers with $b\neq0$. Now rewrite this as $$\sqrt5={a\over b}-\sqrt2.$$ Squaring both sides of this equation we obtain $$5={a^2\over b^2}-2\sqrt2{a\over b}+2.$$ Now, carefully solve for $\sqrt2$ and obtain $$\sqrt2={-3b\over 2a}+{a\over 2b.}$$ This implies that $\sqrt2$ is a rational number which is a contradiction. Thus $$\sqrt2+\sqrt5$$ is an irrational number.

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You have $$(\sqrt{2} + \sqrt{5})^2 = 7 + 2\sqrt{10}.$$
The square of a rational number is rational. This number, $7 + 2\sqrt{10}$, is rational iff $\sqrt{10}$ is rational. The standard argument shows that $\sqrt{10}$ is not rational. So we are done here.

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  • $\begingroup$ You're almost correct. Square of a rational number is rational, but the square of an irrational number can be rational. So the "iff" is wrong in this context, but one implication, which is the one we use, is in fact true. $\endgroup$
    – Asaf Karagila
    Oct 25, 2013 at 15:08
  • $\begingroup$ I am saying $9 + 2\sqrt{10}$ is rational iff $\sqrt{10}$ is ratioinal. $\endgroup$ Oct 25, 2013 at 15:54
  • $\begingroup$ Thanks for the comment, @Thomas $\endgroup$ Oct 25, 2013 at 19:17
  • $\begingroup$ Ah, that's better. $\endgroup$
    – Asaf Karagila
    Oct 25, 2013 at 19:38
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Let $\sqrt2+\sqrt5=a$ where $a$ is rational

$\implies\sqrt2=a-\sqrt5$

Squaring we get, $$2=a^2+5-2a\sqrt5\iff\sqrt5=\frac{a^2+3}{2a}$$ which is rational unlike $\sqrt5$

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  • $\begingroup$ can you explain how you went from the squaring to this result 2=a2+5−2a5–√5 $\endgroup$
    – A P
    Mar 8, 2022 at 13:14
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    $\begingroup$ @AviPars, en.wikipedia.org/wiki/… $\endgroup$ Mar 8, 2022 at 13:33
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If your arrival at the equation $a^2/b^2=\sqrt{40}$ was correct, you’re done, because the equation $a^4=40b^4$ for integers $a$ and $b$ is a contradiction to the Fundamental Theorem of Arithmetic, which says that the expression of an integer as product of primes can be done in only one way. But your suspect fourth-degree equation has a number of $5$’s on the left that is divisible by $4$, but not so on the right.

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If $\sqrt{2}+\sqrt{5}\in\mathbb{Q}$, i.e., $ \sqrt{2}+\sqrt{5}=\frac{b}{a}$ for $a,b\in\mathbb{Z} \mbox{ and } a\neq 0$ then $ 2-5=\frac{b}{a}(\sqrt{2}-\sqrt{5})$ and $$ \sqrt{2}-\sqrt{5} =\frac{-3\cdot a}{b} $$ Here $b$ is necessarily greater than zero by cause $\sqrt{2}+\sqrt{5}>0$.Finally we get, $$ \sqrt{2}=\frac{1}{2}\left( (\sqrt{2}+\sqrt{5})+(\sqrt{2}-\sqrt{5}) \right)= \frac{1}{2}\left( \frac{b}{a}+\frac{-3\cdot a}{b} \right)\in\mathbb{Q}. $$ But this is a contradiction since $\sqrt{2}$ is not rational.

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