False Proof of $7=8$

Question

I have found something really weird which compelled me to put up this post. I wish I could resolve this problem but was unsuccessful.

Consider a positive integer say 7 , now is it possible to prove that it is equal to 8 and 8 is equal to 9 and so on? Quite puzzled! Have a look at this picture.

\begin{align*} 7 &= 7-\frac{15}2+\frac{15}2 = \sqrt{\left(7-\frac{15}2\right)^2}+\frac{15}2 \\ &= \sqrt{7^2-2\times7\times\frac{15}2+\left(\frac{15}2\right)^2}+\frac{15}2 \\ &= \sqrt{49-105+\left(\frac{15}2\right)^2}+\frac{15}2 \\ &= \sqrt{-56+\left(\frac{15}2\right)^2}+\frac{15}2 \\ &= \sqrt{64-120+\left(\frac{15}2\right)^2}+\frac{15}2 \\ &= \sqrt{8^2-2\times8\times\frac{15}2+\left(\frac{15}2\right)^2}+\frac{15}2 \\ &= \sqrt{\left(8-\frac{15}2\right)^2}+\frac{15}2 \\ &= \left(8-\frac{15}2\right)+\frac{15}2 = 8 \\ \end{align*}

If this is correct, we would be able to show that all positive integers are equal.

I am sure that it is not correct but I would appreciate if someone can point out what the mistake is? Or is there something to be pondered about?

Answer 1

There is a mistake in the second equality. Note that $7-\frac{15}2=-\frac12$ is negative, so

$$7-\frac{15}{2} + \frac{15}{2}= {\bf\color{red}{-}}\sqrt{\Big(7-\frac{15}2\Big)^2} + \frac{15}2$$

Answer 2

The problem is that $7-\frac{15}{2}$ is negative.

Hence, $-0.5=7-\frac{15}{2}\neq \sqrt{(7-\frac{15}{2})^2}=|7-\frac{15}{2}|=0.5$