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# That's... impossible!

Prove $$0=1$$ $-20=-20$ $16-36=25-45$ $4^2-4\times 9=5^2-5\times9$ $4^2-4\times 9+\frac{81}{4}=5^2-5\times9+\frac{81}{4}$ $4^2-2\times4\times\frac{9}{2}+\left(\frac{9}{2}\right)^2=5^2-2\times 5\times \frac{9}{2}+\left(\frac{9}{2}\right)^2$ $\left(4-\frac{9}{2}\right)^2=\left(5-\frac{9}{2}\right)^2$ $\Rightarrow 4-\frac{9}{2}=5-\frac{9}{2}$ $\Rightarrow 0=1$

WHAT? Mistakes?

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1 year, 4 months ago

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${ (4-\frac { 9 }{ 2 } ) }^{ 2 }={ (5-\frac { 9 }{ 2 } ) }^{ 2 }\Rightarrow |4-\frac { 9 }{ 2 } |=|5-\frac { 9 }{ 2 } |\Rightarrow \frac { 9 }{ 2 } -4=5-\frac { 9 }{ 2 }$, not ${ (4-\frac { 9 }{ 2 } ) }^{ 2 }={ (5-\frac { 9 }{ 2 } ) }^{ 2 }\Rightarrow 4-\frac { 9 }{ 2 } =5-\frac { 9 }{ 2 } \Rightarrow 0=1$ · 1 year, 4 months ago

Ohhhhhh!!! Great answer! · 1 year, 4 months ago