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

Error to avoid that leads to:

Solution

We know that $2+2=4$

$$\begin{array}{rl}\Rightarrow 2+2& =4-\frac{9}{2}+\frac{9}{2}\\ \\ & =\sqrt{(4-\frac{9}{2}{)}^2}+\frac{9}{2}\\ \\ & =\sqrt{16-2\cdot 4\cdot \frac{9}{2}+(\frac{9}{2}{)}^2}+\frac{9}{2}\\ \\ & =\sqrt{16-36+(\frac{9}{2}{)}^2}+\frac{9}{2}\\ \\ & =\sqrt{-20+(\frac{9}{2}{)}^2}+\frac{9}{2}\\ \\ & =\sqrt{25-45+(\frac{9}{2}{)}^2}+\frac{9}{2}\\ \\ & =\sqrt{5^2-2\cdot 5\cdot \frac{9}{2}+(\frac{9}{2}{)}^2}+\frac{9}{2}\\ \\ & =\sqrt{(5-\frac{9}{2}{)}^2}+\frac{9}{2}=5-\frac{9}{2}+\frac{9}{2}\\ \\ & =5\end{array}$$

Many people make this mistake: $\sqrt{x^2}=x$, we should know that

$$\sqrt{x^2}=|x|=\{\begin{array}{ll}-x& \text{ if }x<0\\ x& \text{ if }x⩾0\end{array}$$

The error made in this example is from the beginning, let’s see

$$\begin{array}{rl}2+2& =4-\frac{9}{2}+\frac{9}{2}\\ \\ & =\sqrt{{(4-\frac{9}{2})}^2}+\frac{9}{2}\end{array}$$

There is no equivalence between the first and the 2nd line because

$$\sqrt{{(4-\frac{9}{2})}^2}+\frac{9}{2}=|4-\frac{9}{2}|+\frac{9}{2}$$ $$|4-\frac{9}{2}|=-(4-\frac{9}{2})=\frac{9}{2}-4$$ $$\Rightarrow \sqrt{{(4-\frac{9}{2})}^2}+\frac{9}{2}=\frac{9}{2}-4+\frac{9}{2}\ne 4$$

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

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