In the course of proving that [itex] \sqrt{3} [/itex] is irrational, I had another question pop up. To prove that [itex] \sqrt{3} [/itex] is irrational, I first assumed 2 things: [itex] \sqrt{3}[/itex] is rational, and the rational form of [itex] \sqrt{3} [/itex] is in it's lowest form. I then broke the proof up into cases and showed that none of these cases could occur.(adsbygoogle = window.adsbygoogle || []).push({});

My question boils down to: did I actually show that [itex] \sqrt{3} [/itex] is irrational?

From a purely logical standpoint, let's say that the 2 assumptions I made were named A and B. I successfully showed that A [itex] \wedge [/itex] B is false. However, this doesn't mean that BOTH A and B are false. More specifically, A could be true and B could be false, and I would still arrive at A [itex] \wedge [/itex] B being false.

On the other hand, the second assumption that was made (the rational form of [itex] \sqrt{3} [/itex] is in it's lowest form) shouldn't (doesn't?) change the problem.

Could someone give me solace and explain this little technicality I have? Thank you very much!

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# Falsity of assumptions question.

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