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Prove that $\sqrt{2}+\sqrt{3}$ is irrational.

So far, I have this:

If $\sqrt{2}+\sqrt{3}$ is rational, then it can be written as $\frac{p}{q}$ with integral $p, q$ and in lowest terms.

$$\sqrt{2}+\sqrt{3}=\frac{p}{q}$$ $$2\sqrt{6}+5 =\frac{p^2}{q^2}$$ $$(2\sqrt{6}+5)q^2=p^2$$

And that's about where I get stuck. In a similar question (prove that $\sqrt{2}+\sqrt{6}$ is irrational,) I was able to show that both $p,q$ had to be even which is impossible. I obviously can't apply this trick here. Any hints?


marked as duplicate by MJD, Magdiragdag, J.R., John Habert, Andrew D. Hwang Mar 25 '14 at 21:22

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.

  • $\begingroup$ Try squaring again after moving around some terms. $\endgroup$ – user45878 Mar 25 '14 at 17:26
  • $\begingroup$ See the following post. $\endgroup$ – Mauro ALLEGRANZA Mar 25 '14 at 17:33
  • 3
    $\begingroup$ If you see on the right side of the screen, under "Related", your question appears twice ... $\endgroup$ – Mauro ALLEGRANZA Mar 25 '14 at 17:35
  • $\begingroup$ There are still twice as many who've answered than who've up-voted the question. $\endgroup$ – Michael Hardy Mar 25 '14 at 17:36
  • $\begingroup$ Good catch, @MauroALLEGRANZA . +1 $\endgroup$ – DonAntonio Mar 25 '14 at 17:52

$(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)=3-2=1$ is rational, so if the first factor were rational, so would be the second, and their difference $2\sqrt2$. But surely you know that the latter isn't rational.

  • $\begingroup$ A very clever answer $\endgroup$ – Jose Antonio Mar 25 '14 at 17:54
  • $\begingroup$ This is clever and very simple. I barely missed this idea. I was trying to do stuff with conjugates, but didn't quite connect the dots. $\endgroup$ – Josh Infiesto Mar 25 '14 at 18:01
  • $\begingroup$ @Josh See my answer here for a much more general application of this idea. I didn't mention it in my answer since I thought you were interested only in completing your approach. Genrally there are many ways to do such irrationality proofs. $\endgroup$ – Bill Dubuque Mar 25 '14 at 18:18

Hint $\ \sqrt2 +\sqrt3 = r\in\Bbb Q\,\overset{\large\rm square}\Rightarrow\ 5+2\sqrt 6 = r^2 \,\Rightarrow\ \sqrt 6 = (r^2-5)/2 \in \Bbb Q,\,$ contradiction,

since $\,\sqrt{6} = a/b\,\Rightarrow\, 6b^2 = a^2\,$ has an odd number of $2$'s on the LHS, but an even number on RHS (by uniqueness of prime factorizations).

There are also many other ways, e.g. using the Rational Root Test on its minimal polynomial, or using Bezout, etc see prior posts here.


Let $x=\sqrt2+\sqrt3$. Then this is a solution to the equation $0=x^4-10x^2+1$. By rational root theorem, this equation has no rational solution. Therefore, $x$ must be irrational.

  • 4
    $\begingroup$ Could you please share how you came up with that equation? $\endgroup$ – vonbrand Mar 25 '14 at 17:45
  • $\begingroup$ @vonbrand He brought the radical 2 to the left, then squared both sides. Then he isolated the radical 2 term and squared both sides again... $\endgroup$ – imranfat Mar 25 '14 at 18:57
  • $\begingroup$ @vonbrand There are several answer to that question here. $\endgroup$ – MJD Mar 25 '14 at 19:05


$$\sqrt2+\sqrt3=\frac pq\implies 3=\frac{p^2}{q^2}-2\frac pq\sqrt2+2\implies$$

$$\sqrt2=\left(\frac{p^2}{q^2}-1\right)\frac q{2p}\in\Bbb Q\;\ldots$$

  • $\begingroup$ Three answer so far, but I'm the only one who's up-voted the question. $\endgroup$ – Michael Hardy Mar 25 '14 at 17:33
  • $\begingroup$ That's because you are sooooo nice, @MichaelHardy...:) Here, I'm upvoting it, too. $\endgroup$ – DonAntonio Mar 25 '14 at 17:34
  • $\begingroup$ Oh: I intended this comment to be under the question itself. $\endgroup$ – Michael Hardy Mar 25 '14 at 17:36

If $2\sqrt{6} + 5 = p^2/q^2$, then $\displaystyle \sqrt{6} = \frac 1 2 \left(\frac{p^2}{q^2} - 5\right)$, so $\sqrt{6}$ is rational.

Can you take it from there?


We can solve this problem by contradiction. Suppose that it is rational, so it is a non-zero rational number such as $\frac{a}{b}$.

$(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=1\rightarrow \sqrt{3}-\sqrt{2}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{b}{a}$

So we conclude that $\sqrt{3}-\sqrt{2}$ is a rational number as well.

$\sqrt{3}-\sqrt{2}=\frac{b}{a}\rightarrow 5-2\sqrt{6}=(\frac{b}{a})^2$

$\sqrt{3}+\sqrt{2}=\frac{a}{b}\rightarrow 5+2\sqrt{6}=(\frac{a}{b})^2$

Subtract the former equation from the latter one, we conclude $4\sqrt{6}=(\frac{a}{b})^2-(\frac{b}{a})^2$

From above equation we conclude $\sqrt{6}$ is rational, but we can prove the irrationality of $\sqrt{6}$ quite simple(It has been brought in following) which shows contradiction and hence $\sqrt{2}+\sqrt{3}$ is irrational.

proof for irrationality of $\sqrt{6}$

If it is rational then it is equal to $\frac{p}{q}$ such that $p$ and $q$ are relatively prime natural numbers. then $p^2=6q^2\rightarrow q^2|p^2\rightarrow$ $p$ and $q$ are relatively prime if and only if $q=1$ which implies 6 is a square integer which is a contradiction.


From your very last step: $$\sqrt{6} = \frac{p^2 - 5q^2}{2q^2}$$ This a rational number- a contradiction, hence the answer.


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