# Proof That 2 = 4

Calculus Level 3

What is wrong with this proof that $$2 = 4$$?

Step A:
Let $${ x }^{ {\displaystyle x }^{ {\displaystyle x }^{\displaystyle .^{.^.} } } }=y$$.
Substituting, we get $${ x }^{\displaystyle y }=y$$

Step B:
$$x=\sqrt [ y ]{ y }$$. Letting $$y = 4$$, then $$x = \sqrt[4]{4}$$ and hence
$${ \sqrt [ 4 ]{ 4 } }^{ {\displaystyle \sqrt [ 4 ]{ 4 } }^{ {\displaystyle \sqrt [ 4 ]{ 4 } }^{\displaystyle .^{.^.} } } }=4$$

Step C:
$$x=\sqrt [ y ]{ y }$$. Letting $$y = 2$$, then $$x = \sqrt[2]{2}$$ and hence
$${ \sqrt [ 2 ]{ 2 } }^{ {\displaystyle \sqrt [ 2 ]{ 2 } }^{ {\displaystyle \sqrt [ 2 ]{ 2 } }^{\displaystyle .^{.^.} } } }=2$$

Step D:
Observe that $$\sqrt { 2 } = 2^ \frac{1}{2} = \sqrt [ 4 ]{ 4 }$$. Hence, we get

$4 = { \sqrt [ 4 ]{ 4 } }^{ {\displaystyle \sqrt [ 4 ]{ 4 } }^{ {\displaystyle \sqrt [ 4 ]{ 4 } }^{\displaystyle .^{.^.} } } }= \sqrt [ 2 ]{ 2 } ^{ {\displaystyle \sqrt [ 2 ]{ 2 } }^{ {\displaystyle \sqrt [ 2 ]{ 2 } }^{\displaystyle .^{.^.} } } } = 2$

Which step is wrong?

Note: $${ x }^{ {\displaystyle x }^{ {\displaystyle x }^{\displaystyle .^{.^.} } } }=y$$ is an infinite tetration, or infinite tower of exponentiation

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