# All numbers are equal

• amits

#### amits

All numbers are equal :D

heres the latest proof and it doesn't involve 0 anywhere. no term like (a-b) is involved unlike other similar proofs

-ab = -ab
=> a^2 - a^2 - ab = b^2 - b^2 - ab
=> a^2 - a(a+b) = b^2 - b(a+b)
=> a^2 - a(a+b) + (a+b)^2/4 = b^2 - b(a+b) + (a+b)^2/4

as (x-y)^2 = x^2 - 2xy + y^2,

[a - (a+b)/2]^2 = [b - (a+b)/2]^2

taking square roots,

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

hence proved :D

now with all numbers having been proved equal without involving "0" anywhere, what's the need to study anything :D

just noticed, this is my 1st post here after 7 years You can't take square roots. x^2 = y^2 does not imply x = y
If it did, this would have been your 7th post after 1 year.

Your second line contains an error. You added a^2 - a^2 to one side of the equation, but b^2 - b^2 to the other side. The only way for this to leave the equality unchanged is for a to equal b... so it's no surprise that you find out later that a = b.

- Warren

=> a^2 - a^2 - ab = b^2 - b^2 - ab

This is the same as
a^2 + b^2 - ab = b^2 + a^2 - ab , which don't think should create any problems.

[a - (a+b)/2]^2 = [b - (a+b)/2]^2

taking square roots,

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

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

The former yields a=b. The latter yields a+b=a+b.

Your second line contains an error. You added a^2 - a^2 to one side of the equation, but b^2 - b^2 to the other side. The only way for this to leave the equality unchanged is for a to equal b... so it's no surprise that you find out later that a = b.

- Warren
That step is correct. It's just adding zero to each side.

Your second line contains an error. You added a^2 - a^2 to one side of the equation, but b^2 - b^2 to the other side. The only way for this to leave the equality unchanged is for a to equal b... so it's no surprise that you find out later that a = b.

- Warren

it isn't an error. i added a^2 + b^2 to both sides

-ab = -ab
a^2 + b^2 - ab = a^2 + b^2 - ab

a^2 - a^2 - ab = b^2 - b^2 - ab

also a^2 - a^2 = 0 & b^2 - b^2 = 0, so adding that doesn't make a difference

You can't take square roots. x^2 = y^2 does not imply x = y
If it did, this would have been your 7th post after 1 year.

yes, you are right. a number always has 2 square roots, 1 positive & 1 negative

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You got it wrong.
2^2 = (-2)^2
when you have sqr root you have to have either + or - sign. you only considered + sign. Consider - sign and you will get a+b = a+b