Proof for 1=2 (Identify the error in this proof)
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Now 1 = 1 LHS = RHS 2+2 = 2^2 = 4 3+3+3 = 3^2 = 9 4+4+4+4 = 4^2 = 16 Thus by mathematical induction we can say, x+x+x+x......(x times) = x^2 Differentiating LHS and RHS with respect to x d/dx (x) =1 and d/dx (x^2) = 2x So we get, 1+1+1+1......(x times) = 2x 1 added 'x' times equals to x. ( 1 added 3 times equals 3, and so on) => x = 2x Thus, 1 = 2 So, what's wrong in this??? :-D
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Now 1 = 1 LHS = RHS 2+2 = 2^2 = 4 3+3+3 = 3^2 = 9 4+4+4+4 = 4^2 = 16 Thus by mathematical induction we can say, x+x+x+x......(x times) = x^2 Differentiating LHS and RHS with respect to x d/dx (x) =1 and d/dx (x^2) = 2x So we get, 1+1+1+1......(x times) = 2x 1 added 'x' times equals to x. ( 1 added 3 times equals 3, and so on) => x = 2x Thus, 1 = 2 So, what's wrong in this??? :-D
d/dx (x + ... + x) != 1 + ... + 1 (x times), since the number of x's depends on x itself.
d/dx (x + ... + x) = d/dx (x (1 + ... + 1))
= d/dx ( x.x )
= xcheers, Chris Maunder
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Now 1 = 1 LHS = RHS 2+2 = 2^2 = 4 3+3+3 = 3^2 = 9 4+4+4+4 = 4^2 = 16 Thus by mathematical induction we can say, x+x+x+x......(x times) = x^2 Differentiating LHS and RHS with respect to x d/dx (x) =1 and d/dx (x^2) = 2x So we get, 1+1+1+1......(x times) = 2x 1 added 'x' times equals to x. ( 1 added 3 times equals 3, and so on) => x = 2x Thus, 1 = 2 So, what's wrong in this??? :-D
Hello! We can say that '1 = 1', '2+2 = 4', '3+3+3 = 3^2', etc. is true. But it is clear that 'x+x+x+x......(x times) = x^2' is simply not true for 'x' values where x < 0, so 'x+x+x+x......(x times) = x^2' is not true in general. SlavoF "I hear and I forget. I see and I remember. I do and I understand." --Confucius
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Now 1 = 1 LHS = RHS 2+2 = 2^2 = 4 3+3+3 = 3^2 = 9 4+4+4+4 = 4^2 = 16 Thus by mathematical induction we can say, x+x+x+x......(x times) = x^2 Differentiating LHS and RHS with respect to x d/dx (x) =1 and d/dx (x^2) = 2x So we get, 1+1+1+1......(x times) = 2x 1 added 'x' times equals to x. ( 1 added 3 times equals 3, and so on) => x = 2x Thus, 1 = 2 So, what's wrong in this??? :-D
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Now 1 = 1 LHS = RHS 2+2 = 2^2 = 4 3+3+3 = 3^2 = 9 4+4+4+4 = 4^2 = 16 Thus by mathematical induction we can say, x+x+x+x......(x times) = x^2 Differentiating LHS and RHS with respect to x d/dx (x) =1 and d/dx (x^2) = 2x So we get, 1+1+1+1......(x times) = 2x 1 added 'x' times equals to x. ( 1 added 3 times equals 3, and so on) => x = 2x Thus, 1 = 2 So, what's wrong in this??? :-D
First all you're really doing is showing x * x = x ^ 2. x + x + ... { x times } = x ^ 2 => x * ( 1 + 1 + ...) = x ^ 2 {with 1 added x times } this is, x * ( x ) = x ^ 2 so d/dx of this is 1x + x1 = 2x => 2x = 2x, or x = x { or 2 = 2, if you want, as long as x<>0 } { d/dx (uv) = du/dx * v + u * dv/dx }. It seems that your error was in assuming d/dx ( x + x + ...) { x times } = x :-D
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Now 1 = 1 LHS = RHS 2+2 = 2^2 = 4 3+3+3 = 3^2 = 9 4+4+4+4 = 4^2 = 16 Thus by mathematical induction we can say, x+x+x+x......(x times) = x^2 Differentiating LHS and RHS with respect to x d/dx (x) =1 and d/dx (x^2) = 2x So we get, 1+1+1+1......(x times) = 2x 1 added 'x' times equals to x. ( 1 added 3 times equals 3, and so on) => x = 2x Thus, 1 = 2 So, what's wrong in this??? :-D
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Now 1 = 1 LHS = RHS 2+2 = 2^2 = 4 3+3+3 = 3^2 = 9 4+4+4+4 = 4^2 = 16 Thus by mathematical induction we can say, x+x+x+x......(x times) = x^2 Differentiating LHS and RHS with respect to x d/dx (x) =1 and d/dx (x^2) = 2x So we get, 1+1+1+1......(x times) = 2x 1 added 'x' times equals to x. ( 1 added 3 times equals 3, and so on) => x = 2x Thus, 1 = 2 So, what's wrong in this??? :-D
> d/dx (x) =1 and d/dx (x^2) = 2x What is this crap? D / DX * (X) = 1 // true D / DX * (X^2) != 2x // you dolt... D / DX * (X^2) = D / DX * X * X = 1 / X * X * X = (X * X) / X ... ergo equals X not 2x > 1 added 'x' times equals to x. ( 1 added 3 times equals 3, and so on) > => x = 2x 1 added X times = X * 1... 1 * X = X x = x 1 = 1 > Thus, 1 = 2 Thus 1 != 2 , 1 = 1 ===== Peace in the middle east. No><
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Now 1 = 1 LHS = RHS 2+2 = 2^2 = 4 3+3+3 = 3^2 = 9 4+4+4+4 = 4^2 = 16 Thus by mathematical induction we can say, x+x+x+x......(x times) = x^2 Differentiating LHS and RHS with respect to x d/dx (x) =1 and d/dx (x^2) = 2x So we get, 1+1+1+1......(x times) = 2x 1 added 'x' times equals to x. ( 1 added 3 times equals 3, and so on) => x = 2x Thus, 1 = 2 So, what's wrong in this??? :-D
The most basic mistake is differentiating while this is forbidden/useless here. Your induction proof is for natural numbers only, while differentialing is done over the real or complex numbers. The entire theory of differentition is built upon limit theorems which exist over the real numbers, but not over the natural numbers. Next mistake (ignoring the first) is treating x once as a constant and once as a variable. P.S. The most interesting false proofs of this kind come from integration theory, there mistakes are much harder to find. Felix.
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x+x+x+x+ ... = nx ; not x+x+x+x+ ... = x^2; d/dx 1+1+1+1+ ... = n Which is ok. I'am not sure, but you are probably mixing x's in an incompatible manner. Somehow. Anyway, I am not a math expert.
The way you've shown it (nx) is correct, except that you're forgetting that n = x in each case. I've seen this proven one other way (although I forget the exact steps), and it involved division by 0. The next proof was to show that the Earth is flat... Why do we waste our time with these things? :)
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The way you've shown it (nx) is correct, except that you're forgetting that n = x in each case. I've seen this proven one other way (although I forget the exact steps), and it involved division by 0. The next proof was to show that the Earth is flat... Why do we waste our time with these things? :)
Well, that's exactly the problem. It's true that 'x=n' in each case. But, strictly speaking can we write that down and perform derivation based on that assumption? Those variables are essentialy different. I am not sure, but it looks strange to me. Given the number of people involved in this 'math-like games', it appears that MFC programmers are quite idle falks, despite 50+ working hours/week.
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Hello! We can say that '1 = 1', '2+2 = 4', '3+3+3 = 3^2', etc. is true. But it is clear that 'x+x+x+x......(x times) = x^2' is simply not true for 'x' values where x < 0, so 'x+x+x+x......(x times) = x^2' is not true in general. SlavoF "I hear and I forget. I see and I remember. I do and I understand." --Confucius
Slavo you were very close.... The function f(x)= x+x+x+x.....(X times) = x^2 is true only for x>0 and absolute values of x. It fails between value x=1 and x=2. Thus, since this function f(x) is not continuous, it is not even differentiable, even for a limiting case. ;) Debanjan