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?

Source

#NumberTheory

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Comments

${ (4-\frac { 9 }{ 2 } ) }^{ 2 }={ (5-\frac { 9 }{ 2 } ) }^{ 2 }\Rightarrow |4-\frac { 9 }{ 2 } |=|5-\frac { 9 }{ 2 } |\Rightarrow \frac { 9 }{ 2 } -4=5-\frac { 9 }{ 2 }$ , 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 - 5 years, 7 months ago

Ohhhhhh!!! Great answer!

Adam Phúc Nguyễn - 5 years, 7 months ago

Great ..lol !!!

Rohit Udaiwal - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$