There is no legitimate way to prove that \(1=2\). Do you know why? Because \(1\neq 2\).

On one point in your "proof", you divided by zero. You can't do that. Mathematics won't take responsibility for anything that happens after you divide by zero.

You could have saved yourself the trouble by writing \(1\times 0=2\times 0\) and cancel out the zeros to get \(1=2\). But that is invalid. Any proof that allows division by zero isn't a proof at all.
–
Mursalin Habib
·
4 years ago

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@Mursalin Habib
–
1 squared=1
2 squared=2+2
3squared=3+3+3
in the same way
X squared= X+X+X+X+X+X+X+X+X+.......+X X times _{1}
differentiate both sides of 1 with respect to X
2X=1+1+1+1+1+1+1+1+1+.............+1 X times
i.e 2X=X
therefore 2=1
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Brilliant Member
·
3 years, 12 months ago

Notice that your definition of \(x^2\) is incomplete because it only works for positive integer values of \(x\). Graph it and you'll get points, not a line. Your function is not continuous. You can't even differentiate it.

What does the operator \(\frac{d}{dx}\) do? It takes a function \(f(x)\) and spits out a new function that tells you what the slope of \(f(x)\) is for a particular \(x\).

If you want to find the slope of \(f(x)\) for a particular \(x\) graphically, what do you do?

You find the point \((x, f(x))\) and draw a straight line tangential to \(f(x)\) at that point. Then you calculate \(\tan \theta\) where \(\theta\) is the angle our straight line makes with the positive \(x\) axis.

Now let's get back to your problem. Do you remember that we had points scattered on the \(xy\) plane? Let's find the slope of your defined function \(x^2\) for a particular \(x\). Now how do we do that? Can you draw a line tangential to a point? Nope.

Moral: You can't differentiate discontinuous functions.
–
Mursalin Habib
·
3 years, 12 months ago

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TopNewestThere is no legitimate way to prove that \(1=2\). Do you know why? Because \(1\neq 2\).

On one point in your "proof", you divided by zero. You can't do that. Mathematics won't take responsibility for anything that happens after you divide by zero.

You could have saved yourself the trouble by writing \(1\times 0=2\times 0\) and cancel out the zeros to get \(1=2\). But that is invalid. Any proof that allows division by zero isn't a proof at all. – Mursalin Habib · 4 years ago

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_{1} differentiate both sides of 1 with respect to X 2X=1+1+1+1+1+1+1+1+1+.............+1 X times i.e 2X=X therefore 2=1 – Brilliant Member · 3 years, 12 months agoLog in to reply

this one.

Notice that this argument is similar toSo, what is wrong with your 'proof'?

Notice that your definition of \(x^2\) is incomplete because it only works for positive integer values of \(x\). Graph it and you'll get

points, not aline. Your function is not continuous. You can't even differentiate it.What does the operator \(\frac{d}{dx}\) do? It takes a function \(f(x)\) and spits out a new function that tells you what the slope of \(f(x)\) is for a particular \(x\).

If you want to find the slope of \(f(x)\) for a particular \(x\) graphically, what do you do?

You find the point \((x, f(x))\) and draw a straight line tangential to \(f(x)\) at that point. Then you calculate \(\tan \theta\) where \(\theta\) is the angle our straight line makes with the positive \(x\) axis.

Now let's get back to your problem. Do you remember that we had points scattered on the \(xy\) plane? Let's find the slope of your defined function \(x^2\) for a particular \(x\). Now how do we do that? Can you draw a line tangential to a point? Nope.

Moral: You can't differentiate discontinuous functions. – Mursalin Habib · 3 years, 12 months ago

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– Abdhi Sharan · 2 years, 4 months ago

yeah but y=x^2 is a continous problemLog in to reply