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The question is prove $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is an irrational number.

I started by assuming the opposite that $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is a rational number. I stated that a rational number is a number made by dividing two integers. So I set $\sqrt{2} + \sqrt{3} + \sqrt{5} = i_1/i_2$, where $i_1$ and $i_2$ are two integers. I multiplied $i_2$ onto both sides and got $i_2\sqrt{2} + i_2\sqrt{3} + i_2\sqrt{5} = i_1$. I then said that in order to turn an irrational number such as $\sqrt{2}$ into a rational number you can multiply, $\sqrt{n}\sqrt{n}=n$. Meaning $i_2$ would have to hold the value of $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{5}$ which is impossible. So it is an irrational.

I think I made a mistake somewhere but I am not sure.

rogerl
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mathguy21
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  • This is not clear what does it mean to say that an integer "holds a value"? – lulu Jul 26 '18 at 21:46
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    in order to turn an irrational number such as sqrt(2) into a rational number you can multiply, sqrt(n)sqrt(n)=n. This is where the mistake is, this really makes no sense. At the very least, it seems to ignore the fact that adding irrational numbers can lead to a rational sum. – Arnaud Mortier Jul 26 '18 at 21:46
  • The sum of irrational numbers can be rational, even an integer. So , you cannot conclude this way. You need the minimal polynomial and have to show that its degree is larger than $1$ (In this case, it is $8$). – Peter Jul 26 '18 at 21:46
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    To warm up for this problem, start with something a little simpler like $\sqrt 2 +\sqrt 3$. – lulu Jul 26 '18 at 21:47
  • Thank you for the advice. I will try again and take into consideration what you all mentioned :) – mathguy21 Jul 26 '18 at 21:49
  • @Peter Can you post that as an answer, please? – Tanner Swett Jul 26 '18 at 21:58
  • Related. – Shaun Jul 26 '18 at 23:03
  • @TannerSwett No more possible because marked as duplicate. But a hint : Start with $$x-\sqrt{2}-\sqrt{3}=\sqrt{5}$$ and square the equation. Isolate a root and square twice to get the final result. To show that irrationality it is enough to show that the polynomial has no integer root. The polynomial you should get is $$x^8 - 40x^6 + 352x^4 - 960*x^2 + 576$$ – Peter Jul 28 '18 at 11:18
  • The question isn't a duplicate, since this question is asking if a given proof is valid, and the linked question is asking something different. I cast a vote to reopen. – Tanner Swett Jul 28 '18 at 14:44
  • @mathguy21 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Aug 10 '18 at 23:37

5 Answers5

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If $\sqrt{2}+\sqrt{3}$ is rational then so too is $\sqrt{2}-\sqrt{3}$ because $(\sqrt{2}+\sqrt{3})\cdot (\sqrt{2}-\sqrt{3}) = 2 - 3 = -1$

But adding the two terms, $(\sqrt{2}+\sqrt{3})+ (\sqrt{2}-\sqrt{3}) = 2\sqrt{2}$ which is irrational. Two rational numbers cannot sum to an irrational number so we have a contradiction. Therefore $\sqrt{2}+\sqrt{3}$ is irrational. We can say $\sqrt{2}+\sqrt{3}$ = I and come to the same result/conclusion for I$ + \sqrt{5}$.

In this case we reach the assumption that I$^2-5$ is rational.

But I$^2-5= (\sqrt{2}+\sqrt{3})^2-5 = 5+2\sqrt{6}-5 = 2\sqrt{6}$ which is irrational and another contradiction. Hence $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is irrational.

Phil H
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    How do you do the proof for $I+\sqrt{5}$ being irrational? One key property of the original proof would require$(I+\sqrt{5})(I-\sqrt{5})$ being rational, but it equals the irrational $I^2-5=2+3+2\sqrt{6}-5 = 2\sqrt{6}$. OK, good argument! – Ingix Jul 26 '18 at 22:57
  • @ Ingix We were working on it at the same time apparently, – Phil H Jul 26 '18 at 22:59
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    I'd suggest you correct your final evaluation of $I^2$. It doesn't change the outcome of the argument, but is confusing for somebody follwing the proof. – Ingix Jul 26 '18 at 23:02
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No, this solution is incorrect I'm afraid. You need $i_2\sqrt{2}+i_2\sqrt{3}+i_2\sqrt{5}=i_1$, that's true, but that doesn't mean that $i_2\sqrt{2}$ needs to be rational. Also, if it did need to be rational, that doesn't mean that $i_2$ must be $\sqrt{2}$. In can also be $3\sqrt{2}/5$, for example.

Alon Amit
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Consider $r$ rational and assume

$$\sqrt 2 + \sqrt 3 + \sqrt 5=r $$

then

$$\iff \sqrt 2 + \sqrt 3)=r - \sqrt 5$$

$$5+ 2\sqrt 6=r^2 + 5-2r \sqrt 5$$

$$2\sqrt 6 + 2r \sqrt 5=r^2$$

$$24+20r^2+8r \sqrt{30}=r^4$$

$$8r \sqrt{30}=r^4-24-20r^2+$$

$$\sqrt{30}=\frac{r^3}8-\frac 3 r-\frac52 r$$

then it suffices to show that $\sqrt {30}$ is not rational (which is true since $30$ is not a perfect square).

user
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  • In the last line what property is being shown that shows root(30) is irrational.? – mathguy21 Jul 26 '18 at 22:12
  • @mathguy21 30 is not a perfect square. – Phil H Jul 26 '18 at 22:14
  • @mathguy21 We are using contradiction and the initial assumption is true only if the last one is true, therfore we need to prove that $\sqrt {30}$ is irrational. – user Jul 26 '18 at 22:15
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Let $x=\sqrt 2+\sqrt 3+\sqrt 5$ and $y=x-\sqrt 5=\sqrt 2+\sqrt 3.$. We have $$0=(y-\sqrt 2-\sqrt 3)(y-\sqrt 2+\sqrt 3)=(y-\sqrt 2)^2-(\sqrt 3)^2=y^2-2y\sqrt 2-1.$$ So $y^2-1=2y\sqrt 2.$ Squaring this and re-grouping, we have $y^4-10y^2+1=0.$ Substituting $x-\sqrt 5$ for $y$ in this, expanding and re-grouping, we obtain $$A=B\sqrt 5 \quad \text {where}\quad A=x^4+4x^2-24\quad \text {and}\quad B=20x^3.$$

Now if $x$ is rational then $A$ and $B$ are rational with $B=20x^3\ne 0, $ implying $\sqrt 5=A/B\in \Bbb Q, $ which is false. So $x$ cannot be rational.

DanielWainfleet
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  • We could go further to obtain a polynomial $p(z)\in \Bbb Z[z]$ with deg $(p)=8$ and $p(x)=0$ and then prove that $p(z)$ is irreducible in $\Bbb Z[z]$ so, by a theorem of Gauss, $p(z)$ is irreducible in $\Bbb Q[z]$. But it suffices to stop at this intermediate stage. – DanielWainfleet Jul 30 '18 at 19:10
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The sum of two irrational numbers can be rational, so you cannot go from your hypothesis that $i_2\sqrt2 + i_2\sqrt 3+i_2\sqrt 5$ is rational to the claim that each of them is separately rational. An example would be $3-\sqrt 2$ and $\sqrt 2$ are both irrational but their sum is $3$.

For the sum of two square roots, like $\sqrt 2 + \sqrt 3$ you can use the fact that the square of a rational is rational. You can assume it is rational and square it to get $5+2\sqrt 6$. As you are down to one square root you can follow the usual proof that $\sqrt 2$ is irrational.

Xander Henderson
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Ross Millikan
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