Alannah Short

2022-06-21

Prove $\sqrt{2}+\sqrt{3}+\sqrt{5}$ is an irrational number.

svirajueh

Beginner2022-06-22Added 29 answers

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 $+\sqrt{5}$.

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.

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 $+\sqrt{5}$.

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.

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