Proof That 2 = 4

Calculus Level 3

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

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

Step B:
x=yyx=\sqrt [ y ]{ y } . Letting y=4 y = 4 , then x=44 x = \sqrt[4]{4} and hence
444444...=4{ \sqrt [ 4 ]{ 4 } }^{ {\displaystyle \sqrt [ 4 ]{ 4 } }^{ {\displaystyle \sqrt [ 4 ]{ 4 } }^{\displaystyle .^{.^.} } } }=4

Step C:
x=yyx=\sqrt [ y ]{ y } . Letting y=2 y = 2 , then x=22 x = \sqrt[2]{2} and hence
222222...=2{ \sqrt [ 2 ]{ 2 } }^{ {\displaystyle \sqrt [ 2 ]{ 2 } }^{ {\displaystyle \sqrt [ 2 ]{ 2 } }^{\displaystyle .^{.^.} } } }=2

Step D:
Observe that 2=212=44\sqrt { 2 } = 2^ \frac{1}{2} = \sqrt [ 4 ]{ 4 } . Hence, we get

4=444444...=222222...=2 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: xxx...=y{ x }^{ {\displaystyle x }^{ {\displaystyle x }^{\displaystyle .^{.^.} } } }=y is an infinite tetration, or infinite tower of exponentiation

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