# Proving $1+\sqrt2+\sqrt3$ is irrational [duplicate]

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.

## marked as duplicate by Lord Shark the Unknown, Dbchatto67, Peter Foreman, Dietrich Burde, Bill Dubuque elementary-number-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 21 at 18:47

• Well, as a first step, it's sufficient to show $\sqrt{2}+\sqrt{3}$ is irrational! – Cheerful Parsnip Apr 21 at 17:45
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.
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.