I stumbled upon this awesome proof just now. Can you figure out what the fallacy is? (Brilliant users are brilliant, so I expect this to get solved quickly Lol)

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TopNewestYou really don't have to go through all the lines to know what went wrong. A hint for those who are struggling: if \(x^2=y^2\), does it necessarily mean that \(x=y\)? – Mursalin Habib · 2 years, 11 months ago

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– Mardokay Mosazghi · 2 years, 11 months ago

I agree, \(x^{2}=y^{2}\) doesn't mean these two variables are equalLog in to reply

– Finn Hulse · 2 years, 11 months ago

Exactly. When you use even roots, you have to put \(\pm\) in front of the expression. :DLog in to reply

– Joshua Ong · 2 years, 11 months ago

Much smart :P:):P:PLog in to reply

You cannot take positive square root on both the sides. Because if x^2 = y^2, it is not necessary that x=y only, it means x may be -y also. – Kushagra Sahni · 2 years, 11 months ago

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While taking square root there are two solutions a=b,a=-b for a^2=b^2 a=b isn't always true. in the case a+b=0 it is not necessary that a=b – Milun Moghe · 2 years, 11 months ago

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There seems to be a typo in the 4th line though. It does not affect the final result if changed accordingly. – Joshua Ong · 2 years, 11 months ago

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a clear blunder starts from the 4rth step... and agree that two variables are actually equal , thus they variables ,i.e not constant. – Mohammad Farhan · 2 years, 11 months ago

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Actually in second last step when you proved n+1 = n and then cut off 'n' from both sides and reported 1=0 is actually wrong. Since in the former equation 'n' is the variable and if the highest power of the variable is eliminated from LHS and RHS like the one above then one root of the variable (i.e. 'n' here) tends to infinity... Hence if 'n' tends to infinity

n+1 can very well be estimated to n..

This is what I think Nevertheless let's wait for others to reply... To see if I'm right or not – Vishal Sharma · 2 years, 11 months ago

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– Alessandro Passantino · 2 years, 11 months ago

When he takes the square root he simply puts the radicand even though it's negative, as n-(2n+1)/2=n-n-1/2<0; it'easy to check that using the modulo of the radicand the equation would be correct.Log in to reply

But what I said isn't wrong actually Since if you see the second last equation He wrote n+1=n and in next step concluded that 1=0. Whereas in equation like that 'n' clearly tends to infinity.. – Vishal Sharma · 2 years, 11 months ago

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– Joshua Ong · 2 years, 11 months ago

-n from both sidesLog in to reply