I also came across some old favorites, "proofs" that clearly aren't true, but seem like they work. For a fun puzzle, can you find the "mistake" in each one that makes the fake-proof work?
a = b
a^2 = ab
a^2-b^2 = ab-b^2
(a+b)(a-b) = b(a-b)
a+b = b
2b = b
2 = 1
x = (Pi+3)/2
2x = Pi+3
2x(Pi-3) = (Pi+3)(Pi-3)
2Pix-6x = Pi^2-9
9-6x = Pi^2-2Pix
9-6x+x^2 = Pi^2-2Pix+x^2
(3-x)^2 = (Pi-x)^2
3-x = Pi-x
Pi = 3
-1 = -1
-1/1 = -1/1
-1/1 = 1/-1
sqrt(-1/1) = sqrt(1/-1)
i/1 = 1/i
i = 1/i
i * i = 1
-1 = 1
5 comments:
Want me to give all of them?
The first is you're dividing by zero in line four. a=b => (a-b) = 0.
Go ahead. And that's the second one if you count finding a set of values for which x^x = y^y and x =/= y. That one's of my own making, though. I found one, I can't figure out if there are any others.
Which solution to x^x = y^y did you find besides x=0, y=1 (or vice versa)? It gives me the feeling that the solution to Fermat's Last Theorem is going to come into play for solving that.
For the second proof, you ignored the fact that a square root has both a positive and a negative root. Looking at the LHS, you chose the positive root of (3-x)^2. If you had chosen the negative root, the equality would be intact.
For the third proof, you violated the convention that the square root operation can be distributed across a quotient if and only if both the numerator and denominator are real and positive.
I haven't done this stuff in a long time!
-A
I'm not counting 0 and 1 as a pair for which x^x = y^y, because 0^0 really isn't 1. Sometimes we use it as such because it fits a pattern or is convenient, but it's really undefined. In fact, a more simplistic proof of 0=1 is:
n^0 = 0
0^n = 1
n = 0
0^0 = 0^0
0 = 1
Hey, A, who are you? Say hi sometime (this blog has a private e-mail address you can use).
So what was your solution for x and y?
Sent you an email to say hi and give you a better idea of who I am.
-A
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