We will do a proof by contradiction:

Lets assume, that the square root of 2 is rational. Then you can write the square root of 2 as a fraction a over b. Lets assume that a and b are hole numbers and that you can not reduce a over b. Then you can do this:

It follows, that a is even.

Now we prove that b is even. For a we can write 2k.

It follows that b is even. That means, that a/2 and b/2 are hole numbers. It follows, that you can reduce a over b with 2. This is a contradiction, since we assumed that a over b can not be reduced.

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seriouscephalopodIt’s always nice to reread this good old proof but I think your presentation would be a lot more lucid if you clarified what the contradiction is and how it relates to the assumptions rather than just stated ‘This is a contradiction’ at the end of it all. It’s good to provoke the reader to think for himself but I found some of the wording used such as ‘smallest’ and ‘where that works’ to be a little vague so either that part or the final part could benefit from some clarification so that the connection becomes more obvious.

Have a nice one!

mathisawsomePost authorOk I will change that