3
$\begingroup$

This question already has an answer here:

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

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

$\endgroup$

marked as duplicate by MJD, Davide Giraudo, Dennis Gulko, Daniel Robert-Nicoud, Daniel Fischer Oct 25 '13 at 21:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 7
    $\begingroup$ See here math.stackexchange.com/questions/496037/… $\endgroup$ – Dutta Oct 25 '13 at 13:08
  • 1
    $\begingroup$ I have posted answer for this two times in this forum... please try checking with some key words... $\endgroup$ – user87543 Oct 25 '13 at 13:09
  • 1
    $\begingroup$ Square the number: if it's rational, then also $\sqrt{10}$ is. Is it? $\endgroup$ – egreg Oct 25 '13 at 13:10
  • 6
    $\begingroup$ @PraphullaKoushik Which key words? $\sqrt{2}$ isn't exactly eacy to search for... $\endgroup$ – user1729 Oct 25 '13 at 13:55
18
$\begingroup$

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.

$\endgroup$
24
$\begingroup$

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

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '13 at 15:08
  • $\begingroup$ I am saying $9 + 2\sqrt{10}$ is rational iff $\sqrt{10}$ is ratioinal. $\endgroup$ – ncmathsadist Oct 25 '13 at 15:54
  • $\begingroup$ Thanks for the comment, @Thomas $\endgroup$ – ncmathsadist Oct 25 '13 at 19:17
  • $\begingroup$ Ah, that's better. $\endgroup$ – Asaf Karagila Oct 25 '13 at 19:38
8
$\begingroup$

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 $\sqrt5$

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.