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

Solution 1:

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.

Solution 2:

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.

Solution 3:

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 $.

Solution 4:

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.