Proof That All Numbers Are Equal?
Date: 12/06/2005 at 17:00:34 From: RV Subject: 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. 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.
Date: 12/06/2005 at 20:08:07 From: Doctor Minter Subject: Re: All numbers are equal Hi RV! You pose a VERY interesting point. But before I give up my career as a mathematician and concede that all numbers are equal, let me demonstrate that there are two paths to be taken from the sixth line of your proof. On one side, you have (a - t/2)^2, and on the other, (b - t/2)^2. If you want to eliminate the exponent by taking the square root of both sides, remember that the square root of the square of a number is the "absolute value" of that number, since we cannot take the square root of a negative number without delving into the realm of complex (a.k.a. "imaginary") numbers. For example, if we know that sqrt(x^2) = 4 we can simplify this to |x| = 2 and then x can be equal to +/- 2. Either value makes the equations true. x does not have to be BOTH positive AND negative 2 simultaneously. It can be either. In regards to your example (which I may use on my friends, regardless of this reply), the aforementioned line of your proof would simplify to |a - t/2| = |b - t/2| Having absolute value signs on both sides of the equation is redundant, so we can simplify again to |a - t/2| = b - t/2 (or a - t/2 = |b - t/2|, it makes no difference) So then, +(a - t/2) = b - t/2 OR -(a - t/2) = b - t/2 The first indeed does show your result. The second, however, simply shows that a + b = t, which was the original assumption, anyway. To know if these results are correct, you simply have to plug in values for a and b. If they are equal, then you can show it with this method. If you want to show that a + b = t, you can show that also. But it's not required that a = b, because there are two "paths" that you can take after eliminating the exponents. It is not necessary that both of these "paths" SIMULTANEOUSLY lead to logical results. I hope that this has been of some help. I did thoroughly enjoy the proof that you provided, and I will likely show others for laughs. Please feel free to write again if you need further assistance, or if you have any other questions. Thanks for using Dr. Math! - Doctor Minter, The Math Forum http://mathforum.org/dr.math/
Date: 12/07/2005 at 07:26:26 From: RV Subject: Thank you (All numbers are equal) Dr. Minter - Thanks a lot for the explanation, I emailed it to my friends and they are all amazed.
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