Could you use the triangle inequality

1. Apr 13, 2009

Orphen89

...to prove Pythagoras' theorem? Or is the triangle inequality considered a corollary of Pythagoras theorem (so it can't be used to prove it)? I have an assingment question that asks you to prove Pythagoras' theorem, so I just wanted to know if the inequality can be used in this case.

2. Apr 13, 2009

tiny-tim

Hi Orphen89!

I don't see how you're going to use an inequality to prove an equality …

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

3. Apr 13, 2009

Orphen89

I actually thought that the triangle inequality could prove pythagoras' theorem on the condition that the 2*$$\left\left\| x\|$$ $$\left\left\|y\|$$ (from the proof of the triangle inequality) was equal to zero, because then you will be left with $$\left \left\|x+y\|$$^2 = $$\left\left\| x\|$$^2 + $$\left\left\|y\|$$^2, like in the case of two perpendicular vectors (there is no longer a $$\leq$$ in the inequality). It seems to work, but if it's mathematically incorrect to 'combine' the two theorems together then please tell me (I'm not brilliant at maths )

4. Apr 13, 2009

Dick

Yes, you can prove it that way in an inner product space. tiny-tim was pointing out that the triangle inequality is true in more general metric spaces where the Pythagorean theorem doesn't hold. Because they don't have an inner product. So it's not a consequence of the triangle inequality. You have to add other stuff to get Pythagoras.

5. Apr 14, 2009

Orphen89

So is it okay to use the Triangle Inequality when vectors are involved? I've checked up how it works in metric spaces, but I just want to make sure that I'm doing the right thing, because I'm really confused right now

6. Apr 14, 2009

Dick

I thought you were talking about proving Pythagoras by showing (x+y).(x+y)=x.x+y.y if x.y=0. I'm not sure what you are asking now. A vector space with the usual inner product IS a metric space.

7. Apr 14, 2009

tiny-tim

To prove Pythagoras?

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

8. Apr 14, 2009

Orphen89

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

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?

9. Apr 14, 2009

Dick

No, you CAN'T prove Pythagoras using the TI. Your proof is fine, but it's not directly related to the TI. TI and Pythagoras are two different things.

10. Apr 14, 2009

Orphen89

Ah ok then, I guess I'll just have to prove pythagoras theorem some other way. Thanks for all the help Dick and tiny-tim, I probably would have made a huge mistake if it weren't for you two >_>

11. Apr 14, 2009

Dick

Ok, but isn't (x+y).(x+y)=x.x+y.y if x.y=0 a proof of Pythagoras? I keep trying to tell you that it's a fine proof. It's just not the same as the TI.

12. Apr 14, 2009

Orphen89

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