Creative Operation

Logic Level 2

2 2 2 = ( 3 2 2 2 1 ) 2 \large 2\text{ } 2\text{ } 2=\left(3^{2^{2^2}}-1\right)^2

Is it possible to make the equation above true by inserting the appropriate operations? Any operations and functions can be used.

Yes No Not enough information

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2 solutions

Kay Xspre
Apr 2, 2016

This is another obvious cheat. d ( 2 × 2 × 2 ) d x = d ( 3 2 2 2 1 ) 2 d x \large \frac{d(2\text{×}2\text{×}2)}{dx}=\frac{d(3^{2^{2^2}}-1)^{2}}{dx}

A non-cheating is 2 × 2 × 2 = ( 3 2 2 2 1 ) 2 \large 2×2×2 =\sqrt{(3^{\sqrt{2^{\sqrt{2^2}}}}-1)^{2}}

David Molano
Apr 2, 2016

A binary operation on N \mathbb{N} is any function f : N × N N f:\mathbb{N}\times \mathbb{N}\to \mathbb{N} . So we can define a function f f such that f ( 2 , 2 ) = 3 f(2,2)=3 , f ( 3 , 2 ) = ( 3 2 2 2 1 ) 2 f(3,2)=\left(3^{2^{2^2}}-1\right)^2 and defined in any way we want for any other pair (For example f ( x , y ) = 0 f(x,y)=0 if x 2 , 3 x\neq 2,3 and y 2 y\neq 2 ), and note it like f ( x , y ) = x y f(x,y)=x\otimes y .

Then: ( 2 2 ) 2 = 3 2 = ( 3 2 2 2 1 ) 2 (2\otimes 2)\otimes 2 =3\otimes 2=\left(3^{2^{2^2}}-1\right)^2 And since any operations and functions can be used, \otimes can. And there are infinitely many other ways to do it.

So, if you cannot think of an operation that works... create one yourself.

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