# Proof That 2 = 4

**Calculus**Level 3

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