How does \( x^2 = 2y^2\) imply that x is an even number?
Stacey Warren - Expert brainly.com
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I think this is under the assumption that y is an integer, right?
Assuming these are integers, whether or not y^2 is even or odd, 2*y^2 will be even. An even times an odd is even.
If this is the case, then it's obvious that \(y^2\) would have to be an integer too, and then 2 times any integer is an even number (that would be your x).
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Yes, I agree if it was \(x^2\) but only the \(x\).
Oh right, didn't look carefully. If the assumption was that BOTH \(x\) and \(y\) are integers, then both \(x^2\) and \(y^2\) have to be perfect squares. Add to that \(x^2\) is even then we can easily say that it follows that \(x\) is also even because the square root of a "perfect even square" is an even number.
Ahhhhh, yes your absolutely right :) Thanks a lot. Since \(x^2 = 2k\) that also means that you have to be able to do a square root of the even number, to be able to do that, the factor has to be 4 not 2. The k just has to have a whole square root.