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.