Prove $\sqrt{3} + \sqrt{5}$ is irrational

Question

If it is assumed that $\sqrt{3}$ is known to be irrational (not the case for $\sqrt{5}$), then prove that $\sqrt{3}+\sqrt{5}$ is irrational.

My approach:

Assume that $\sqrt{3}+\sqrt{5}$ is rational. Then there exist coprime integers $p$ and $q$ so that $\frac{p}{q}$ is rational and $\frac{p}{q}=\sqrt{3}+\sqrt{5}$. Thus $(\sqrt{3}+\sqrt{5})^2=2(4+\sqrt{15})=\frac{p^2}{q^2}$, which implies that $p^2$ is even, so $p$ is also even. Let $p:=2m$ for some integer $m$, then $p^2=4m^2$. Thus $q^2(4+\sqrt{15})=2m^2$, which implies that $q^2$ is even, and so $q$ is even. But this contradicts that $p$ and $q$ are coprime, and we arrive at a contradiction.

My way of proving this does not use the fact that $\sqrt{3}$ is irrational. Please let me know if my proof is correct, and how to use the above mentioned fact.

Answer 1

This is from here:

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

More generally, suppose $r =\sqrt{a}+\sqrt{b} $ is rational, where $a$ and $b$ are positive integers.

Then $r(\sqrt{a}-\sqrt{b}) =a-b $ so $\sqrt{a}-\sqrt{b} =\dfrac{a-b}{r} $ is also rational.

Adding and subtracting these, $\sqrt{a}$ and $\sqrt{b}$ are rational.

Therefore, if either or both of $\sqrt{a}$ and $\sqrt{b}$ are irrational, then $\sqrt{a}+\sqrt{b}$ is irrational (and similarly for $\sqrt{a}-\sqrt{b}$).

Answer 2

Hint:

If $\sqrt{15}=p/q$ is rational and $\sqrt3$ is irrational then $\sqrt5=\frac{\sqrt{15}}{\sqrt3}=\frac{p}{q\sqrt3}$ so $$5=\sqrt5^2=\frac{p^2}{3q^2}.$$ Hence $3$ appears an odd number of times in the prime factorisation of $5,$ a contradiction.

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Hope this helps.

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P.S. I don't know what I was thinking in putting the old Hint I here.

Answer 3

Let $x=\sqrt 3 + \sqrt 5$. Then:

$$(x - \sqrt 3)^2 = 5$$ $$x^2 - 2 \sqrt 3 x + 3 = 5$$ $$x^2 - 2 = 2 \sqrt 3 x$$ $$(x^2 - 2)^2 = 12 x^2$$ $$x^4 -16 x^2 + 4 = 0$$

By the rational root theorem, if the last equation had rational roots then they would be integer divisors of $4$. But none of $\pm1, \pm2,\pm4$ are roots, so $x$ must be irrational.

Answer 4

If $\sqrt3+\sqrt5=x$ is rational, then $5=(x-\sqrt3)^2=x^2-2\sqrt3x+3$, and

$$\sqrt3=\frac{x^2-2}{2x}$$

But then $\sqrt3$ is also rational. However, we know that $\sqrt3$ is irrational, so $x$ is also irrational.

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And we know $\sqrt3$ is irrational by the usual argument: is $\sqrt3=p/q$, with coprime integers $p$ and $q$, then $p^2=3q^2$. Hence $3$ divides $p$, but then $9$ divides $p^2$ and $3$ divides $q$, so $p$ and $q$ are not coprime, contradiction.

Answer 5

If $\sqrt 3 + \sqrt 5 = p/q; p,q \in \mathbb Z $ then...

$\sqrt 5 = p/q - \sqrt 3$

$5 = (p/q)^2 - 2 (p/q)\sqrt 3 +3$

$2 (p/q)\sqrt 3 = (p/q)^2 - 2$

$\sqrt 3 = (p/2q) - (q/p) $ which is rational. Which is a contradiction as you had already proven that.

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However if you hadn't...

This is not the easiest but

If $\sqrt 3 + \sqrt 5 = p/q $

$3 + 2\sqrt {15} + 5 = p^2/q^2$

$\sqrt {15} = p^2/2q^2 - 4 = a/b $

Where $a = p^2- 8q^2; b=2q^2$.

Let $a= 3^j5^k c; 3^l5^m d $ where $c,d $ don't have any factors of $3$ or $5$. ($j,k,l,m $ may be $0$)

$15 =\frac {3^{2j}5^{2k}c}{3^{2l}5^{2m}d}$

$3^{2l+1}5^{2m+1}c = 3^{2j}5^{2k} d$

But the LHS had an odd number of factors of 3 and 5, while the RHS has even (or 0) number of factors of 3 and 5 so they can't be equal.