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equality..

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posted on 5 May 2007 09:41:02 IST

friends before i hav shown that 4 = 5......now we will see general procedure to be adopted to show that any 2 numbers are equal....

let us take 2 numbers..let them be "a" and "b"

now say,,,

- ab = - ab

(a)2 - a*(a+b) = (b)2 - b*(a+b) {take a square and b square...}

(a)2 - 2*a*(a+b/2) = (b)2 - 2*b*(a+b/2)

now add (a+b/2)2 on both sides..{say (a+b) whole divided by 2.}....

(a)2- 2*a*(a+b/2)+(a+b/2)2 =(b)2- 2*b*(a+b/2)2

+ (a+b/2)2

{ a - (a+b/2)}2 = { b - (a+b/2)}2

that implies in one case =>

a = b

this proof is developed by me...well u may get doubt in one step that is why i mentioned in 1 case....

interesting right?

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comment by iitianwannabe (posted on 5 May 2007 10:36:05 IST)

let me first give an example...
x=2
squaring
x^2=4
square root...
x=+/-2
BUT obviously -2 does NOt satisfy as x=2 only...ie this method always gives an extra root...
now u have done a similar thing..by expressing in terms of a square and then rooting...uve created an extra root...by taking '-' sign only...the equality is satisfied but not otherwise...

comment by raghuram369 (posted on 5 May 2007 10:38:23 IST)

thats y i mentioned in one case...

comment by iitianwannabe (posted on 5 May 2007 10:39:13 IST)

but the 1 case u mentioned is the extra case...the incorrect case...which is y its not right

comment by raghuram369 (posted on 5 May 2007 10:44:56 IST)

the wase where both are positive or negative isnt an incorrect one right?

comment by Prakriteesh (posted on 5 May 2007 11:23:26 IST)

How did you get the step { a - (a+b/2)}2 = { b - (a+b/2)}2 ?

comment by raghuram369 (posted on 5 May 2007 17:51:37 IST)

it is in the form a2 - 2ab - b2

comment by Prakriteesh (posted on 5 May 2007 22:37:06 IST)

iitianwannabe is correct. Raghuram, your steps upto {a-(a+b)/2}square ={b-(a+b)/2}square is of course correct, and this particular step is an identity. It can be derived in a much simpler way. Let a and b be two nonequal numbers. Then {a/2 - b/2} not= {b/2-a/2}. But {a/2 - b/2}sq = {b/2 - a/2}sq, since x sq = (-x)sq.
So, {a - (a+b)/2}sq = {b - (a+b)/2}sq.
but now you can't equate the first pair of sqroots of the L.H.S. and R.H.S. because of the given condition {a/2 - b/2} not= {b/2 - a/2}. You have to take the other combination of square roots which give:
{a - (a+b)/2} = -{b - (a+b)/2}
or, a/2 - b/2 = a/2 - b/2 which is obviously true.
What you have done unknowingly is essentially this:
(a-b)sq = (b-a)sq
sq.rooting: (this is the wrong step)
a-b = b-a
so, 2a = 2b or a=b

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