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Thanks in advance.

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- Thread starter Orphen89
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- #1

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Thanks in advance.

- #2

tiny-tim

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I don't see how you're going to use an

it's not precise enough, is it?

Anyway, the triangle inequality works on the surface of a hemisphere, say, where Pythagoras *doesn't*, so the first *definitely* doesn't prove the second

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Dick

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Dick

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tiny-tim

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So is it okay to use the Triangle Inequality when vectors are involved?

To prove Pythagoras?

A vector space (with a vector basis an' all)

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I thought you were talking about proving Pythagoras by showing (x+y).(x+y)=x.x+y.y if x.y=0.

Yup, that's what I was trying to prove. Sorry if I got the terminology mixed up (I'm new to all this stuff).

To prove Pythagoras?

A vector space (with a vector basis an' all)assumesPythagoras.

Does that assumption mean that you *can't* prove Pythagoras theorem using the triangle inequality since the TI is a sort of derivation of Pythagoras in a vector space?

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Dick

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Dick

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Ah ****, I think I finally understand what you mean now about how the proof is not directly related to the TI (I hadn't read your previous post properly). I know what you mean now, but I think I worded my posts poorly earlier to the point that even I didn't understand it

Thanks again for all the help

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