# How do you find two positive real numbers that differ by 1 and have a product of 1?

Mar 1, 2016

#### Explanation:

Assume $x$ and $y$ to be those positive real numbers such that $x < y$.

Condition 1 : They differ by $1$. That means $y - x = 1$ ..... (1)
Condition 2: Their product is $1$. That means $x . y = 1$ ..... (2)

Use 1 ($y = 1 + x$) to eliminate $y$ in 2
x.(1+x)=1; \qquad => x^2+x-1=0

$x = \setminus \frac{- 1 \setminus \pm \setminus \sqrt{{\left(1\right)}^{2} - 4 \left(1\right) \left(- 1\right)}}{2}$
x_1= \frac{\sqrt{5}-1}{2}; x_2=-\frac{\sqrt{5}+1}{2}
x_2 = -\frac{\sqrt{5}+1}{2}; \qquad y_2=1/x_2=-\frac{2}{\sqrt{5}+1} satisfies both conditions 1 and 2 and so is the solution.