# Prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational [duplicate]

How can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?

I know that $\sqrt{3}, \sqrt{5}$ and $\sqrt{7}$ are all irrational and that $\sqrt{3}+\sqrt{5}$, $\sqrt{3}+\sqrt{7}$, $\sqrt{5}+\sqrt{7}$ are all irrational, too. But how can I prove that $\sqrt{3}+ \sqrt{5}+ \sqrt{7}$ is irrational?

## marked as duplicate by Watson, Namaste, Gibbs, Lord Shark the Unknown, Lord_FarinNov 28 '18 at 19:17

Suppose $\sqrt{3}+\sqrt{5}+\sqrt{7} = r$ for some rational $r$. Then, $$(\sqrt{3}+\sqrt{5})^2 = (r-\sqrt{7})^2 \implies 8+2\sqrt{15} = 7+r^2-2r\sqrt{7}$$ So, $$1-r^2+2\sqrt{15} =-2r\sqrt{7}$$ Let $1-r^2 = k$, which will be a rational number. So, $$(k+2\sqrt{15})^2 = k^2+ 60+4k\sqrt{15} = 28r^2$$ The LHS is irrational while the RHS is rational. Hence, we have a contradiction.

You can easily show (probably you have already done so) $\sqrt{5},\sqrt{7} \in \mathbb Q(\sqrt{5}+\sqrt{7})$.

If your sum would be rational, we would deduce $\sqrt{5},\sqrt{7} \in \mathbb Q(\sqrt{5}+\sqrt{7}) \subset \mathbb Q(\sqrt{3})$, clearly a contradiction.

Assume $$\sqrt{3}+ \sqrt{5}+ \sqrt{7}= \frac{a}{b}$$ for some integers $a,b$. Multiblying both sides by $\sqrt{3}+ \sqrt{5}- \sqrt{7}$ yields $$(\sqrt{3}+\sqrt{5})^2 - 7 = \frac{a(\sqrt{3}+ \sqrt{5}- \sqrt{7})}{b}=\frac{a\left(a/b-2\sqrt{7}\right)}{b} \\ 2\sqrt{15}+8-7=\frac{a^2/b-2a\sqrt{7}}{b} \\ 2b\sqrt{15} = \frac{a^2}{b}-2a\sqrt{7}-b \\ 60b^2=\frac{a^4}{b^2}+28a^2+b^2-\frac{4a^3\sqrt{7}}{b}+4ab\sqrt{7}-2a^2 \\ 59b^2-26a^2-\frac{a^4}{b^2}=4a\sqrt{7}\left(b-\frac{a^2}{b}\right).$$ We can divide both sides by $4a(b^2−a^2)=4a(b-a)(b+a)$ because $a$ can't be $0$ by definition, $b+a$ is a positive integer and $b−a$ can't be $0$ because that would imply $a/b=1$, which is impossible. Thus we get $$\frac{b\left(59b^4-26a^2b^2-a^4\right)}{4a(b^2-a^2)}=\sqrt{7},$$ contradiction.

• Contradiction only if the numerator and the denominator of that last fraction are not zero... – lhf Mar 23 '15 at 19:18
• If that last fraction is undefined, then the RHS is too, but it isn't; why wouldn't that be a contradiction anyway? – Vincenzo Oliva Mar 23 '15 at 19:30
• Strictly speaking, you only have ${b\left(59b^4-26a^2b^2-a^4\right)}=\sqrt{7}\cdot{4a(b^2-a^2)}$ – lhf Mar 23 '15 at 19:32
• We can divide both sides by $4a(b^2-a^2)$ because $a$ can't be $0$ by definition, $b+a$ is a positive integer and $b-a$ can't be $0$ because that would imply $a/b=1$, which is impossible. Or am I missing something? – Vincenzo Oliva Mar 23 '15 at 19:35
• That's much better. Consider adding this to your answer. – lhf Mar 23 '15 at 19:37

Let $\alpha=\sqrt{3}+ \sqrt{5}+ \sqrt{7}$.

Then $\alpha$ is a root of $x^8-60 x^6+782 x^4-3180 x^2+3481=0$. Since this is a monic polynomial with integer coefficients, the rational root theorem tells you that $\alpha$ is either irrational or an integer.

Now

$\quad 1.7 < \sqrt 3 < 1.8$

$\quad 2.2 < \sqrt 5 < 2.3$

$\quad 2.6 < \sqrt 7 < 2.7$

gives

$\quad 6.5 < \alpha < 6.8$

which proves that $\alpha$ is not an integer.

We can avoid even these fine estimates: if $\alpha$ is an integer then it must divide $3481=59^2$, but clearly $1 < \alpha < 3\sqrt 7 < 9 < 59$.

According to Galois, if $\sqrt3 +\sqrt5+\sqrt7$ is rational, then it must be invariant after changing the sign of any or all of the square roots. But the expression is clearly positive, so it cannot be equal to its negative.

• Please clarify. $\sqrt{4}$ is rational, but we generally don't say $\sqrt{4} = -\sqrt{4}$. – DanielV Mar 23 '15 at 21:08
• In this case, the Galois transformations involve changing the signs of any or all of the roots of the three primes. Total of eight transformations, including unity. Your suggestion is not included. – Bill Kleinhans Mar 24 '15 at 3:33