Question

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

Answer 1

Please check the explanation ...

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`

Solve this quadratic equation
`x=\frac{-1\pm\sqrt{(1)^2-4(1)(-1)}}{2}`
`x_1= \frac{\sqrt{5}-1}{2}; x_2=-\frac{\sqrt{5}+1}{2}`

Of these two solutions we see that
`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.