system
Given the following post, respond as if you're commenting on Reddit:
human
If a & b are irrational must a+b be irrational?

Yesterday I stumbled across a dumb little "Only 10% of people get this right!" problem on Facebook:

√x + √(x+15) = 15

My solution was simple: I assumed that x and (x+15) were both perfect squares, and given that the distance between the *nth* and *nth+1* perfect square is always 2n+1, it was immediately obvious that n=7 and x was 49.  No big deal, not a hard problem.

However, it occurred to me my solution involved a weird postulate.  That √x and √(x+15) **must** be perfect squares.  As if there's no imperfect squares (if there is such a thing) the irrational roots of which  could sum to the rational number 15.

So here's my question:  Is it possible for 2 irrational numbers to sum to a rational number?  And let's disqualify the obvious degenerate case where the irrational parts of the number cancel out, like if:

* a = 2 + √2
* b = 2 - √2

Yeah, obviously their sum is rational.  But what I'm really asking is you could sum *just the irrational bits*, much like we do with imaginary numbers, if it's possible to make a rational number.  My intuition says no, it's not possible.  Is there a proof?
gpt
It A+B=R, where A and B are irrational and R is rational, then you  necessarily have B=R-A. So things like A=2+pi and B=18-pi are the only possibilities. So if you exclude all the cases A and B where A+B is rational, then there are no other pairs of irrational numbers A and B so that their sum  A+B is rational. But that's kinda tautological.

Though, you can prove that if you modify the equation to √x + √(x+A) = B for any rational A and positive B, then the solution in x is a perfect square rational number. You can use basic algebra to  show that the solution is x=(A^(2)-B)^(2)/4A^(2), which is always a rational square for rational A and B. Though, the proof that the solution must be a rational square is basically just finding the solution.

Now, your reasoning led to the correct solution, which is good, but you have to be careful. Had 49 not been the solution, then your reasoning wouldn't have been enough to say that there were no solutions (in fact, the intermediate value theorem guarantees the existence of a solution). For instance, if we just change the square roots to cube roots, we do get solutions that are not perfect cubes. The solution to x^(1/3)+(x+15)^(1/3)=15 is complicated and not a perfect cube. In fact, z=x^(1/3) is a root of 2z^(3) - 45z^(2) + 675z - 3360=0, which has no rational roots, so x is not a perfect cube. So in this case, you *do* get irrational+irrational=rational.