π \pi = 3

Level 1

In which step is the mistake made? \text{In which step is the mistake made?}

x = 3 + π 2 x=\frac{3 + \pi}{2}

2 x = 3 + π 2x= 3 + \pi

2 x ( 3 − π ) = ( 3 + π ) ( 3 − π ) 2x(3 - \pi) = (3 + \pi)(3 - \pi)

6 x − 2 π x = ( 3 2 ) − ( π 2 ) 6x - 2\pi x = (3^{2}) - (\pi^{2})

π 2 − 2 π x = 3 2 − 6 x \pi^{2} - 2\pi x = 3^{2} - 6x

π 2 − 2 π x + x 2 = 3 2 − 6 x + x 2 \pi^{2} - 2 \pi x + x^{2} = 3^{2} - 6x + x^{2}

( π − x ) 2 = ( 3 − x ) 2 ( \pi - x)^{2} = (3 - x)^{2}

π − x = 3 − x \pi - x = 3 - x

π = 3 \pi = 3

π − x = 3 − x \pi - x = 3 - x ( π − x ) 2 = ( 3 − x ) 2 ( \pi - x)^{2} = (3 - x)^{2} 6 x − 2 π x = ( 3 2 ) − ( π 2 ) 6x - 2\pi x = (3^{2}) - (\pi^{2}) 2 x = 3 + π 2x= 3 + \pi

Shriniketan Ruppa by Shriniketan Ruppa , 15, India

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1 solution

If a 2 a^2 = b 2 b^2 it is not necessary that a=b. As an example, ( − 2 ) 2 (-2)^2 = ( 2 ) 2 (2)^2 this does not mean that -2=2. In this case ( π − x ) 2 = ( 3 − x ) 2 ( \pi - x)^{2} = (3 - x)^{2} but π − x \pi - x ≠ \neq 3 − x 3 - x . As a result we get a really strange answer. Actually π − x \pi - x = -( 3 - x )

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