# $4=5$. Is this possible? [duplicate]

we can say 4 is the number you get when you add 1 to itself 4 times. Likewise, 5 is the number we get when we add 1 to itself 5 times. Now, let's just say 4 = 5. It would have to be the case that 4-5 = 0, however, that would imply 1 = 0. Unfortunately, one of the rules for a ring is that you cannot have 1 = 0, therefore 4 cannot equal 5 in any ring with unity. but I found a calculation.

Figure of 4 = 5 proved

Relevant snippet

\begin{eqnarray} 2+2&=&4+\frac{9}{2}-\frac{9}{2}\\&=&\sqrt{(4-\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-\frac{9}{2})^2}+\frac{9}{2}\\&=&5-\frac{9}{2}+\frac{9}{2}\\&=&5 \end{eqnarray}

Yes, I know that this is not possible but the way it is solved seems pretty much correct to me. That makes me curious that is it somehow possible anyway ?

## marked as duplicate by Daniel FischerNov 19 '16 at 12:42

• I cannot read what is in the picture – Jan Nov 19 '16 at 12:19
• This proof uses the idea that $\sqrt{u^2} = u$. That's true only when $u$ is positive, but in this case, it's not. – John Hughes Nov 19 '16 at 12:19
• Who says that you cannot have $1=0$ in a unital ring? That is a very unusual requirement. – Tobias Kildetoft Nov 19 '16 at 12:41

The mistake is: $$4-\frac92 \neq \sqrt{\left(4-\frac92\right)^2}.$$
In general you have $$\sqrt{x^2} = |x|$$ and hence $\sqrt{x^2} \neq x$ if $x$ is negative.
• They both equal $\frac{11}{5}$. Look at John Hughes comment....! – Jan Nov 19 '16 at 12:20