Proof that an equation is irrational

Question

I hope you're keeping safe and well. I stumbled across this problem and wondered whether you could help.

Show that $\left(a+\sqrt b\right)\left(a-\sqrt b\right)^3$ is irrational if $a$ and $b$ are NOT square numbers.

Thank you so much for your help in advance. Pac-Man

Hi, I looked at why √2 is irrational using the proof of √2 = a/b. But can't seem put that logic into solving this problem! — Pac-Man, Aug 05 '20 at 09:46
I also tried expanding the brackets. Which gave me a^4 -2a^3√b +2ab√b -b^2. But not sure where to go from here... Thanks — Pac-Man, Aug 05 '20 at 09:50
The wording of your question is unclear. But it is irrelevant whether $a$ is a square $-$ only $b$ has to be a square. — TonyK, Aug 05 '20 at 09:57
Also, what kind of numbers are $a$ and $b$? Integers, rationals, reals? — TonyK, Aug 05 '20 at 09:58
This was from an old textbook I found! It does not specify. I'm assuming they are integers. — Pac-Man, Aug 05 '20 at 10:02

score 3 · Accepted Answer · answered Aug 05 '20 at 09:56

Let's develop the expression: $\left(a+\sqrt b\right)\left(a-\sqrt b\right)^3=\left(a+\sqrt b\right)\left(a-\sqrt b\right)·\left(a-\sqrt b\right)^2=(a^2-b)(a^2-2a\sqrt b+b)$ We can see that the left term is rational if (but not only if) $a$ and $b$ are rational. On the other hand, the right term is rational if (again, not only if) "$-2a\sqrt b$" is rational. That holds if b is a square. As you are supposing that neither $a$ nor $b$ are squares, this term is irrational, what implies that all the expression is irrational. HOWEVER, that only holds if we suppose that both $a$ and $b$ are rational numbers.

Hope it was useful :)

Ah I see, separating the cubed expression out has helped me massively in understanding this problem. You've explained this really nicely for me. Thank you very much — Pac-Man, Aug 05 '20 at 10:02

Proof that an equation is irrational

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