-1

Trying to prove or disprove this statement:

For all real numbers A, B, if $AB$ is rational and $A+B$ is rational, then $A^2$+$B^2$ is rational

Ive hit a wall as Im assuming A and B must either be rational or irrational in order for A+B to be rational (ex. A = $\sqrt2$ and B = $-\sqrt2$, or just A = 2 and B = 3), but that's not really getting me any further into proving or disproving the statement

Everstudent
  • 415
Barth
  • 15
  • 6

2 Answers2

2

Hint: $(A+B)^2=A^2+2AB+B^2,$ known as binomial theorem.

Phicar
  • 14,722
  • Im not sure I understand how that theorem helps. I see the $A^2+B^2$ when it's expanded, but doesn't the $+2AB$ make it so I can't really use it in the proof? – Barth Oct 05 '20 at 23:11
  • 2
    since $ab$ and $a+b$ are rational => $(a+b)^{2}-2ab$ is rational and $(a+b)^{2}-2ab=a^{2}+b^{2}$, hence $a^{2}+b^{2}$ is rational – Everstudent Oct 05 '20 at 23:25
0

If $A+B = r\in \mathbb Q$ and $AB = q\in \mathbb Q$.

Then $A^2 + B^2 = (A+B)^2 - 2AB = r^2 -2q$.

Can we assume that if $r, q$ are rational that $r^2 -2q$ must be rational as well?

.........

If not.

Lemma 1: if $m,n\in \mathbb Q$ then $mn \in Q$.

Pf: Let $m = \frac ab; n =\frac cd, a,b,c,d \in \mathbb Z; b\ne 0, d\ne 0$. Then $mn = \frac {ac}{bd}$ and $ac,bd \in \mathbb Z$ and $bd \ne 0$. So $mn \in \mathbb Q$.

Lemma 2: if $m,n \in \mathbb Q$ then $m + n \in \mathbb Q$.

Pf: Let $m=\frac ab,n =\frac cd, a,b,c,d\in \mathbb Z;$ and $b,d\ne 0$. Then $m +n = \frac {ad + bc}{bd}$ and $ad+bc,bd\in \mathbb Z$ and $bd \ne 0$.

So $r^2 =r\times r \in \mathbb Q$ and $(-2)q \in \mathbb Q$ by Lemma 1. And $r^2 +(-2q)=r^2-2q\in \mathbb Q$ by Lemma 2.

fleablood
  • 124,253