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How can I prove that $1+\sqrt2+\sqrt3$ is an irrational number, without proving first $\sqrt2$ and $\sqrt3$ are irrational numbers?

Please give some hints or suggestion to proceed with this proof. Thanks in advance.

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marked as duplicate by Lord Shark the Unknown, Dbchatto67, Peter Foreman, Dietrich Burde, Bill Dubuque elementary-number-theory Apr 21 at 18:47

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  • $\begingroup$ Well, as a first step, it's sufficient to show $\sqrt{2}+\sqrt{3}$ is irrational! $\endgroup$ – Cheerful Parsnip Apr 21 at 17:45
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Another way, is to prove first that $ \sqrt{6} $ is irrational. Then you can proceed by contradiction. Assume that $ \ 1 + \sqrt{2} + \sqrt{3} $ is rational, then the square of this number is rational, and it has the value: $ 4+2(1+ \sqrt{2} + \sqrt{3}) + 2\sqrt{6} $. But the first 2 terms are rational numbers by assumption. Hence $ \sqrt{6} $ is rational.

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Hint #1: $1+\sqrt2+\sqrt3$ is irrational if and only if $\sqrt2+\sqrt3$ is irrational.

Hint #2: If $\sqrt2+\sqrt3$ was rational, then its square would be rational too.

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