# IV = V?

**Algebra**Level 2

I will attempt to prove that \(4=5\). Try and spot my mistake in the following steps:

A. Starting with the equality \[16-36=25-45,\] B. we can add \(\frac{81}{4}\) to both sides of this equality without changing anything: \[16-36+\left( \frac { 81 }{ 4 } \right) =25-45+\left( \frac { 81 }{ 4 } \right). \] C. Using the perfect square trinomial properties, we can rewrite this equality as follows: \[{ \left( 4-\frac { 9 }{ 2 } \right) }^{ 2 }={ \left( 5-\frac { 9 }{ 2 } \right) }^{ 2 }.\] D. Taking the square root of both sides, \[4-\frac { 9 }{ 2 } =5-\frac { 9 }{ 2 }. \] E. Adding \(\frac { 9 }{ 2 }\) to both sides gives \[4=5.\]

Which step contains the error?

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