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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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Note by Adam Phúc Nguyễn
10 months, 4 weeks 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 \[{ (4-\frac { 9 }{ 2 } ) }^{ 2 }={ (5-\frac { 9 }{ 2 } ) }^{ 2 }\Rightarrow 4-\frac { 9 }{ 2 } =5-\frac { 9 }{ 2 } \Rightarrow 0=1\] Trung Đặng Đoàn Đức · 10 months, 4 weeks ago

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@Trung Đặng Đoàn Đức Ohhhhhh!!! Great answer! Adam Phúc Nguyễn · 10 months, 3 weeks ago

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Great ..lol !!! Rohit Udaiwal · 10 months, 4 weeks ago

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