Show that if S=a+b√2 : a,b are rational numbers and T=r+s√3 :r,s are rational numbers, then$S \cap T$ = rational

Question

Someone please correct a formatting error in the problem [still a newbie] ; "S&T" (And = upside down U)

Here's a bonus question that was on a test we received that I couldn't figure out. I'd really appreciate some help with where/how to begin to solve this. Apolo,gies for the lack of context - This was on our unit in "sets" (unions and intersections) The question was stated exactly as I put it in the Title.

Sorry for the inconvenience!

I asked this question already and it was closed due to lack of specificity so I ammended it and it wasn't opened up.. So i decided to add my specificity and open it up again I hope this isn't a problem.

Answer 1

You're basically trying to prove that if $a+b\sqrt2=c+d\sqrt3$, for rational $a,b,c,d$, then $b=d=0$.

Try placing the square roots on one side, then square both sides. Rearrange, and try to prove that $b$ or $d$ must be equal to 0.

Answer 2

We want to show that $$a + b\sqrt 2 = c + d\sqrt 3 \tag 1$$ with $a$, $b$, $c$, $d$ rational implies $b=d=0$

From (1) $$ 2 b^2 = (c + d\sqrt 3 -a)^2 = (c-a)^2 + 2 (c-a) d \sqrt 3 + 3 d^2$$

or $$ 2 (c-a) d \sqrt{3} = 2 b^2 - (c-a)^2 - 3 d^2 $$ since $\sqrt{3}$ is irrational we need both sides to be identically zero.

If $d \ne 0$ then $$ a = c, ~\text{and}~ 2b^2 = 3 d^2 $$

This says that $\sqrt{2/3}$ is rational which is not true. So $b=0$ and $d=0$.

Additional comments based on OP's question $$ 2b^2= 3 d^2 \Rightarrow \sqrt{3/2} =\frac{b}{d} \Rightarrow \sqrt{3/2} ~\text{is rational}$$

Finally, having shown that $$a + b\sqrt 2 = c + d\sqrt 3 \Rightarrow b = d 0$$ we can conclude that numbers in the intersection must be $a=c$, i.e. rational numbers. Clearly all rationals can be formed with $b=d=0$.