Question

Solve for `x in (0,oo)` where `[[x]+[1/x]]=1` ?

Answer 1

`x in (1/2, 1) uu (1, 2)`

Explanation:

Given:

`[[x]+[1/x]] = 1`

I will assume that you are using square brackets to denote the floor function, alternatively written as:

`floor(floor(x)+floor(1/x)) = 1`

Expanding the outer floor function, this equation can also be expressed as the inequality:

`1 <= floor(x)+floor(1/x) < 2`

Since we want to solve this for `x in (0, oo)`, both `floor(x) >= 0` and `floor(1/x) >= 0` and hence we can quickly deduce that:

`1/2 < x < 2`

Note further that:

• `floor(color(blue)(1))+floor(1/(color(blue)(1))) = 1+1 = 2`

• If `x in (1/2, 1)` then `floor(x)+floor(1/x) = 0+1 = 1`

• If `x in (1, 2)` then `floor(x)+floor(1/x) = 1+0 = 1`

So we see that the solution is:

`x in (1/2, 1) uu (1, 2)`