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Maths freshers come and help with this M1gla proof :eek3:

can I just ask anybody who can recall the proof of the Cauchy-schwarz theorem, saying |x*y|<= |x|*|y|, in the m1gla's class?

Reviewing your notes, I thinkg it is pretty clear that the proof mentioned on class is logically wrong....

The lecturer deduced (x1y2-x2y1)^2>=0 from the original equation waiting to be proved, and then just reversed back the process to say the original equation is proved....
The inequality (x1y2-x2y1)^2 >= 0 is true for all real numbers x1, x2, y1 and y2. The operations that you do to it can be reversed (adding/subtracting terms from both sides - look in your lecture notes), so you basically do all the operations that you did in the lecture, but in reverse to get back to |x.y| <= |x|*|y|. Also, the proof for the 3-dimensional case (on the latest problem sheet)...

Spoiler



EDIT: I'm not a fresher but I help out at the self-help sessions for the first years so I know what to expect for the first year stuff.
Learning at Imperial College London
Imperial College London
London
Reply 2
Avatar for wudanao
wudanao
OP
Thank you for your reply. But the reverse method is clearly wrong in logic. You cannot deduce a formulae from an equation waiting to be prove, and then just reverse the process and say the equation is now proved! That is nonsense.

A silly example

If I want to prove x+1>y. I cannot say from x+1>y , we can get x=y-1
then I reverse back the process, saying since x=y-1, therefore x+1>y.

of course both statement is true, but it is clearly wrong if we use this as a proof of the original equation!

That is the same thing with what we learned from lectures.

Hope I am wrong.
The thing is, the inequality (x1y2-x2y1)^2 >= 0 IS true for all real numbers x1, x2, y1 and y2 (you'll do this in M1F soon, if you haven't already done so). So you start off with that, expand out the brackets, add on the term [(x1y1)^2 + (x2y2)^2 + 2x1x2y1y2] to both sides (I think that is right) and Cauchy-Schwarz comes right out.

The idea of "going from something you want to prove to something that you know is true, and then back in reverse order" is a bit strange at first, but you'll see it a lot in other areas. The 'something you want to prove' might be an expression that you suspect to be true (eg. through looking at numerical examples), and isn't (normally) something that is pulled out of thin air.
Reply 4
Avatar for wudanao
wudanao
OP
Wow! This makes sence. Thank you so much for your reply. Guess there are alog I need to learn!
firegalley246
The thing is, the inequality (x1y2-x2y1)^2 >= 0 IS true for all real numbers x1, x2, y1 and y2 (you'll do this in M1F soon, if you haven't already done so). So you start off with that, expand out the brackets, add on the term [(x1y1)^2 + (x2y2)^2 + 2x1x2y1y2] to both sides (I think that is right) and Cauchy-Schwarz comes right out.

The idea of "going from something you want to prove to something that you know is true, and then back in reverse order" is a bit strange at first, but you'll see it a lot in other areas. The 'something you want to prove' might be an expression that you suspect to be true (eg. through looking at numerical examples), and isn't (normally) something that is pulled out of thin air.

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