How can 1+1=3 be possible?

Question

I've been looking at some proofs but none make any sense to me, can someone prove this is true?

One of them... Suppose $a=b$

$$ \Rightarrow a=b $$

Multiply each side by $b$

$$ \Rightarrow ab=b^2 $$

Subtract a^2 from each side

$$ \Rightarrow ab-a^2=b^2-a^2 $$

Simplify $$ \Rightarrow a=a+b $$ $$ \Rightarrow a+1=a+b+1 $$

Because a=b $$ \Rightarrow a+1=2a+1 $$

Suppose a=1 $$ \Rightarrow 1+1=2+1 $$

NOTE: I'm a 10th grader

Answer 1

OK. Let's walk through this, somewhat more nicely formatted:

Suppose $a=b$. Then \begin{align} a&=b \tag{1} \end{align} Multiply each side by $b^2$ \begin{align} ab&=b^2 \tag{2} \end{align}

Subtract $a^2$ from each side \begin{align} ab-a^2& =b^2-a^2 \tag{3} \end{align}

Simplify \begin{align} a&=a+b \tag{4}\\ a+1&=a+b+1 \tag{5} \end{align} Because $a=b$ \begin{align} a+1=2a+1 \tag{6} \end{align}

Suppose $a=1$ \begin{align} 1+1&=2+1 \tag{7} \end{align}

Where to begin? Well, right at the start, I think you're saying "Let $a$ be any real number, and let $b = a$." So let me just run with that, writing notes to the right explaining each step:

\begin{align} a & = b & \text{by assumption} \tag{1} \\ ab & = b^2 & \text{multiply both sides by $b$} \tag{2} \\ ab-a^2 & = b^2 -a ^2 & \text{subtract $a^2$ from both sides} \tag{3} \\ a(b-a) & = (b+a)(b-a)& \text{factoring, difference of squares} \tag{4} \\ a & = (b+a)& \text{Divide through by $b-a$} \tag{5} \\ a+1 & = b+a + 1& \text{Add 1 to each side} \tag{6} \\ a+1 & = a+a + 1& \text{Replace $b$ by $a$ using Eq. 1} \tag{7} \\ 1+1 & = 1+1 + 1& \text{Because $a$ was arbitrary,}\\ & & \text{we can apply equation 7 to $a = 1$} \tag{8} \\ 2 & = 3& \text{Simplification} \tag{8} \\ \end{align}

So that's the "proof". The failure is at step 5, where you say that $pq = pr$ implies that $q = r$; that's true only if you know $p$ is nonzero. For instance $0 \cdot 5 = 0 \cdot 7$, but that doesn't let you conclude that $5 = 7$, because $0$ has no multiplicative inverse. In your case, the number $b-a$ is playing the role of $p$, and because $b$ and $a$ are the same, you're really "dividing through by $0$", and that's the error.