If $a-b = \frac{a}{b} = \frac{1}{b} - \frac{1}{a}$ , comment on $(a,b)$ .

Question

If $a$ and $b$ are two distinct and non-zero real numbers such that $$ a - b = \frac{a}{b} = \frac{1}{b} - \frac{1}{a}$$ Determine the quadrant of $\mathrm {XY}$ plane in which the point $(a,b)$ lies.

Given, $$a-b = \frac{a-b}{ab} \implies ab=1>0$$Hence, $a$ and $b$ are of same signs and ($a,b$) lies in either $1^{st}$ or $3^{rd}$ quadrant.

Also, $$a-b = \frac{a}{b} \implies1-a= b^2> 0$$

Hence, $a<1 \implies b>1$ Or $b<0$.

However, I can't proceed further. Inserting $ab$ in place of $1$ at multiple positions is also not giving any new information. Please give a hint/solution that how I can solve it further.

Answer 1

$a-b =\frac {a-b}{ab}$ implies $ab = 1$ or $b =\frac {1}{a}$`

$a-b = \frac {a}{b}$

Substituting $b = \frac {1}{a}$ from above...

$a - \frac 1a = a^2\\ a^3 - a^2+ 1 = 0$

Where is the root of this cubic?
If $a\ge 1$ then $a^3\ge a^2$ and $a^3-a^2+1>0$

If $0\le a \le 1$ then $-1 \le a^3 - a^2 \le 0$

That is $a^3 - a^2 = a^2(a-1)$ and the absolute value of each factor must be less than 1.

Since every cubic with real coefficients has a real root, it must be that $a<0$

Answer 2

Alternative approach:

Neither $~a~$ nor $~b~$ can equal $~0.~$

I agree with the OP's (i.e. original poster's) analysis that $~a~$ and $~b~$ must have the same sign, that $~a < 1,~$ and that $~ab = 1.$

Also, $~0 < a < 1 \implies 1 < b \implies (a-b) < 0.~$

This is a contradiction because

$$a - b = \frac{a}{b} > 0.$$

Therefore, you must have that $~a < 0 \implies b < 0.$