Is my proof of $\sqrt{2} + \sqrt{3} + \sqrt{5}$ is an irrational number valid? [duplicate]

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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.

marked as duplicate by Xander Henderson, max_zorn, user21820, Lord Shark the Unknown, Jyrki LahtonenJul 27 '18 at 5:49

• This is not clear what does it mean to say that an integer "holds a value"? – lulu Jul 26 '18 at 21:46
• 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
• 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

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.

• 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
• 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

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.

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.

• 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

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).

• 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. – gimusi Jul 26 '18 at 22:15

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.