This interesting proof is from a theorem published by RV 12/06/2005 at 17:00:34
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. What is the error
in this proof? I think that if a = b then a - b = 0 and we cannot
multiply both sides of the equation by (a - b) because when a number
is multiplied by zero the answer is always zero.


Comments: 20
I would like to say sorry for taking a while to get to your article. I have been away from gather for a while and I am finally getting to the 3000 plus emails I have awaiting me on here to go through. so I am starting from the most recent received to the first I ever received.... So now I am finally able to read your piece. Thank you for sending me the link to this article.
Now second thing:
I am a math lover and this is very interesting to look at...
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You should get together with my kid. Math Guys Unite! (The rest of us will wait here)
When applying this in algebra, a +/- is required in font of one of the expressions. One must take note that the extraneous solution is produced. It is actually introduced by the presence of the squaring operator.
Unlike a combinative operator, the power operator is not required to yield exactly one value.
For example: (-3)^2=3^2, but -3 does not equal 3. The '^2' introduces the extraneous relationship. Raising both sides to the ½ power produces the false '-3=3', along with the correct '3=3'.
In the above bogus "proof", the a - t/2 = b - t/2 expression is the extraneous solution. The correct one is a - t/2 = -(b - t/2), which equals
a - t/2 = -b + t/2), or a+b=t.
Incidentally, if you have any more of these, please feel free to submit to
Fun Math.