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Hello.
I was wondering if this is a valid proof?
Prove that (a+b)² ≤ 2a² + 2b²
Suppose that the statement is not true i.e.
(a+b)² ≥ 2a² + 2b²
a² + 2ab + b² ≥ 2a² + 2b²
a² + b² ≤ 2ab
a² - 2ab + b² ≤ 0
(a-b)² ≤ 0
This is an absurdity, and therefore the original statement must be true.
Thanks.
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What if a = b = 0?
Do you know what values a and b can take?
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Nope, there is no other information.
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Then that does not prove it unless you state that a and b cannot be equal.
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The negation of
is , NOT .Be careful.
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Hmm.. how would you do it then?
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The same way you did it. What Jane is saying is that you should start off by assuming that (a+b)² > 2a² + 2b².
Then following the same steps, you end up with (a-b)²<0.
Which actually is an absurdity.
Why did the vector cross the road?
It wanted to be normal.
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Ohh right ok. I've never done a proof by contradiction before. Thanks.
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Alternatively, you can start with the inequality
(which is always true) and work your way to the given inequality, thereby avoiding proof by contradiction.
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I thought of that after I had finished - but unless you knew that if the statement wasn't true it would result in in (a-b)² ≤ 0 then it would be quite hard to think of.
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This is why most inequality proofs are presented backwards from the way they are derived. This is pretty much standard.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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