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Originally posted by smw6869If there is a general formula involved in the proof, make sure division by zero isn't happening because that isn't allowed. So for example the classic case of the fallacy of 1=2 (which is very similar to yours)
Anyone know how to prove this to be true(even though it's not). A math teacher proved it to my math class. Everything made sense and no one could find the falacy.
step 1
Let a = b
step 2 Multiply both sides by a:
aa = ab
step 3 which is the same as:
a^2 = ab ( "a squared equals a times b" )
step 4 Add the quantity ( a^2 - 2ab) to both sides:
a^2 + (a^2 - 2ab) = ab + (a^2 - 2ab)
step 5 simplifying both sides we get:
(a^2 + a^2) - 2ab = a^2 + (ab - 2ab)
2(a^2) - 2ab = a^2 - ab
2 (a^2 - ab) = a^2 - ab
step 6 divide both sides by (a^2 - ab):
2(a^2 - ab) / (a^2 - ab) = (a^2 - ab) / (a^2 -ab)
step 7 cancel out like terms in num.&denom:
2 = 1 !!!
Two equals one? Impossible! What's the catch? The catch is, our very first assumption is Let a= b, and if that's true then the quantity (a^2 - ab) = (a^2 - a^2) = 0, and we are not allowed to divide by zero which is what we do in our "proof" in step 6. The proof is not allowed. No division by zero.
I bet this is the trick that is happening in your false proof of 2+2=5.
(sorry for all the edits, but I had to work out some formatting issues)
Originally posted by wolfgang59no its not, you're wrong...
In binary there is no '4' or '256'
10+10=100 is quite correct (ie 2+2=4 in decimal)
10x10=100 is true in any base.
i was under the impression is was binary to decimal.. so you're right there, there is no 4 or 256 in binary... however you're still wrong...
in binary, it would be 10+10=10... you are wrong to say 100, its not, 1 + 1 = 1 or 0 + 1 = 1, 0 + 0 = 0.... there is no additional digit.... mr dumb!