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All numbers are equal !
Theorem: All numbers are equal.

Proof: Choose arbitrary a and b, and let t = a + b. Then

a + b = t

(a + b)(a - b) = t(a - b)

a^2 - b^2 = ta - tb

a^2 - ta = b^2 - tb

a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4

(a - t/2)^2 = (b - t/2)^2

a - t/2 = b - t/2

a = b

So all numbers are the same, and math is pointless.

Proof: Choose arbitrary a and b, and let t = a + b. Then

a + b = t

(a + b)(a - b) = t(a - b)

a^2 - b^2 = ta - tb

a^2 - ta = b^2 - tb

a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4

(a - t/2)^2 = (b - t/2)^2

a - t/2 = b - t/2

a = b

So all numbers are the same, and math is pointless.

Comments · 8

(a - t/2)^2 = (b - t/2)^2 <=> a = b OR a-t/2 = t/2-b. These are the decisions of the system of equations

a + b = t (1)

(a-b)^2=t(a+b) (2)

But it's not a proof that a=b allways, it's only one of the decisitions of this system, that all, nothing else.

a + b = t (1)

(a-b)^2=t(a+b) (2)

But it's not a proof that a=b allways, it's only one of the decisitions of this system, that all, nothing else.

Is it true that 1 = -1?

Yes, because 1^2 = 1 = (-1)^2.

That's the essence of the argument.

t=a+b. So a - t/2 = t/2 - b. And he "showed" by the above argument that a - t/2 = b - t/2, hence a = b.

Is this too serious? haha

Yes, because 1^2 = 1 = (-1)^2.

That's the essence of the argument.

t=a+b. So a - t/2 = t/2 - b. And he "showed" by the above argument that a - t/2 = b - t/2, hence a = b.

Is this too serious? haha

** laugh

;)

ohhh.. hahaha.... im trying to deal with them !

If it is... x+y = (3x + 5).(y/3) ..., how many times can a person laughs in a day?

=))))))

Yep, but you have to pay first !

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