06 Sep '07 02:01

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.

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06 Sep '07 03:306 edits

If 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)*Originally posted by smw6869***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.

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12 Sep '07 18:38

no its not, you're wrong...*Originally posted by wolfgang59***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!