Equalities

From $a \cdot b = \frac{a}{b}$ is $b=\pm 1$ . But $b=1 \Rightarrow a=a+1$ . Absurd. Now, if $b=-1 \Rightarrow -a=a-1\Rightarrow a=\frac{1}{2}$ .

$\dfrac{a}{b} = ab \implies a(b^2-1) =0 \implies b=\pm 1$ Note that $a \not =0$ since the cases $\frac{a}{b} = a+b$ and $ab = a+b$ become false in case $a=0\implies b=0$ (not possible) too .

Now $\dfrac{a}{b}-1 = a+b -1 \implies \\ \dfrac{a-b}{b}= a+b-1 \\ \begin{cases} a = a +1 \quad 0 = 1\phantom {c} \,(\text{absurd}) \quad \text{if b=1} \\a = - a +1 \quad a = \dfrac{1}{2} \quad \text{if b =-1} \end{cases}$ Therefore the required value of $a-b = \dfrac{1}{2} +1 =\boxed{ \dfrac{3}{2}}$

a and b cannot be integers

ab = a + b (given) ⇒ a = ab + 1 ⇒ b = 1 - 1/a = 1/-2a Therefore: 1 - 1/a = 1/-2a

Let’s solve for a, 1 = 1/2a ⇒ a = 1/2

Let’s solve for b, b = 1 - 1/a ⇒ b = 1 - 1/(1/2) ⇒ b = -1

Therefore: a - b = 1/2 + 1 = 3/2

1. a*b=a/b solve for b, you get plus minus square root of 1. It can't be 1, because, then a/1=a. But then a+1 does not equal a. So it must be -1. b=-1

2. You plug in -1, and solve for a a(-1)=a=+(-1) -a=a-1 -2a=-1 a=-1/2

3. substitute into original a-b (-1/2)-(-1) 3/2

Done

this is my solution, its longer then the others but maybe someone will find it easier to understand

My reasoning went as follows:

If $a:b$ is the same as $a*b$ then b must be either $+1$ or $-1$ .

When we try $a=1$ we end up with: $a:1 = a*1=a+1$ , simplified to $a=a+1$ which is a contradiction.

So let's try $a=-1$ :

$a:-1=a*-1=a-1$ , simplified to $-a=a-1$

This is much better. If we now substract a from both sides of the equation we get:

$-2a=-1$ . Divide both sides by -2 and receive $a=1/2$

So our result is $b = -1$ , $a = 1/2$ and therefore $a - b = 1/2 - (-1) = 3/2$

a/b=ab, b^2=1, b=1,-1, Now. ab=a+b, a+b-ab=0. We have two options either b=1 or b=-1. b=1 then a=0, Won't satisfy (a/b=a*b=a+b) If b=-1 then a=1/2. This satisfying the previous. So a-b =1/2-(-1), =3/2

i hope it can help :)

First from $\begin{cases} ab=\frac{a}{b} \implies a=ab^2 \\ a+b=ab \end{cases}$ then $ab^2+b=ab \implies ab+1=a$

Next from $\begin{cases} \therefore a-b=ab+1-b \\ a+b=ab \end{cases}$ then $a-b=a+b+1-b \implies a-b=a+1 \implies a-b=a-(-1) \implies b=-1$

Finally from $\begin{cases} b=-1 \\ ab=a+b \end{cases}$ then $-a=a-1 \implies a-(-a)=1 \implies a+a=1 \implies 2a=1 \implies a=\frac{1}{2}$

And $\frac{1}{2}-(-1)=\frac{1}{2}+1=\frac{1}{2}+\frac{2}{2}=\frac{1+2}{2}=\boxed{\large{\frac{3}{2}}}$