Home -> Solved problems -> 2 + 2 = 5

## Error to avoid that leads to:

### Solution

We know that $$2+2=4$$
\begin{aligned} \Rightarrow 2+2 &=4-\frac{9}{2}+\frac{9}{2} \\\\ &=\sqrt{(4-\frac{9}{2})^{2}}+\frac{9}{2} \\\\ &=\sqrt{16-2\cdot4\cdot\frac{9}{2}+(\frac{9}{2})^{2}}+\frac{9}{2} \\\\ &=\sqrt{16-36+(\frac{9}{2})^{2}}+\frac{9}{2} \\\\ &=\sqrt{-20+(\frac{9}{2})^{2}}+\frac{9}{2} \\\\ &=\sqrt{25-45+(\frac{9}{2})^{2}}+\frac{9}{2} \\\\ &=\sqrt{5^{2}-2\cdot5 \cdot\frac{9}{2}+(\frac{9}{2})^{2}}+\frac{9}{2} \\\\ &=\sqrt{(5-\frac{9}{2})^{2}}+\frac{9}{2}=5-\frac{9}{2}+\frac{9}{2} \\\\ &=5 \end{aligned}
Many people make this mistake: $$\sqrt{x^{2}}=x$$, we should know that
$\sqrt{x^{2}}=|x|= \begin{cases}-x & \text { if } x<0 \\ x & \text { if } x \geqslant 0\end{cases}$
The error made in this example is from the beginning, let’s see
\begin{aligned} 2+2 &=4-\frac{9}{2}+\frac{9}{2} \\\\ &=\sqrt{\left(4-\frac{9}{2}\right)^{2}}+\frac{9}{2} \end{aligned}
There is no equivalence between the first and the 2nd line because
$\sqrt{\left(4-\frac{9}{2}\right)^{2}}+\frac{9}{2}=|4-\frac{9}{2}|+\frac{9}{2}$ $|4-\frac{9}{2}|=-(4-\frac{9}{2})=\frac{9}{2}-4$ $\Rightarrow\sqrt{\left(4-\frac{9}{2}\right)^{2}}+\frac{9}{2}=\frac{9}{2}-4+\frac{9}{2}\neq4$
Home -> Solved problems -> 2 + 2 = 5

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Home -> Solved problems -> 2 + 2 = 5