Which makes me wonder... is there a 2 * 2 = 5 equivalent? and the answer is a resounding, yes! But first the 'proof' as written by Charles Seife.
Let a & b each be equal to 1. Since a ^ b are equal,
b^2 = ab ...(eq.1)
Since a equals itself, it is obvious that
a^2 = a^2 ...(eq.2)
Subtract equation 1 from equation 2. This yeilds
(a^2) - (b^2) = (a^2)-ab ...(eq. 3)
We can factor both sides of the equation; (a^2)-ab equals a(a-b). Likewise, (a^2)-(b^2) equals (a + b)(a - b) (Nothing fishy is going on here. Ths statement is perfectly true. Plug in numbers and see for yourself!) Substituting into the equation 3 , we get
(a+b)(a-b) = a (a-b) ...(eq.5)
So far, so good. Now divide both sides of the equation by (a-b) and we get
a + b = a ...(eq.5)
b = 0 ...(eq.6)
But we set b to 1 at the very beginning of this proof, so this means that
1 = 0 ...(eq.7)
...Anyways, getting that far gives us the jist of the proof, later in the proof, Charles Seife goes on to prove that Winston Churchill was a carrot! if you want to know how that is possible, I recommend you read the book.
From equation 7, add a number to either side and get it equal to any other number, one greater than itself.
Multiplying equation 7 after adding to it, and one can get: any number is equal to any other number.
Hence, conceptually, any number is equal to zero, and, theoretically, that includes infinity. But that's also the reason why when you divide by zero, it is 'Undefined.' Which, consequentially, is what is happening in this equation... just subsistute 1 into equation 3 and one will see that we are dividing by zero in equation 5.
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