theoretical proof that 1 + 1 = 1 using algebra (taken from mathforum) Let a = 1 and b = 1. Therefore a = b, by substitution. If two numbers are equal, then their squares are equal, too: a^2 = b^2. Now subtract b^2 from both sides (if an equation is true, then if you subtract the same thing from both sides, the result is also a true equation) so a^2 - b^2 = 0. Now the lefthand side of the equation is a form known as the difference of two squares and can be factored into (a-b)*(a+b). If you dont believe me, then try multiplying it out carefully, and you will see that its correct. So: (a-b)*(a+b) = 0. Now if you have an equation, you can divide both sides by the same thing, right? Lets divide by (a-b), so we get: (a-b)*(a+b) / (a-b) = 0/(a-b). On the lefthand side, the (a-b)/(a-b) simplifies to 1, right? and the righthand side simplifies to 0, right? So we get: 1*(a+b) = 0, and since 1* anything = that same anything, then we have: (a+b) = 0. But a = 1 and b = 1, so: 1 + 1 = 0, or 2 = 0. Now lets divide both sides by 2, and we get: 1 = 0. Then we add 1 to both sides, and we get what your programming teacher said, namely: 1 + 1 = 1.
Posted on: Fri, 31 Oct 2014 23:28:20 +0000
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