Let \(a\) and \(b\) be rational numbers, and \(b\) square-free. Prove that if \(p=a+\sqrt{b}\), there exists only one value \(q\) such that \(p+q\) and \(pq\) are rational.

**Solution**

Let \(p+q=c\) and \(pq=d\). \[q=\frac{d}{p}\]

\[p+q=a+\sqrt{b}+\frac{d}{a+\sqrt{b}}=c\]

\[{a}^{2}+2a\sqrt{b}+b+d=ca+c\sqrt{b}\]

Matching each component yields:

\[2a\sqrt{b}=c\sqrt{b}\]

\[2a=c\]

\[{a}^{2}+b+d=2{a}^{2}\]

\[d={a}^{2}-b\]

Now we solve for \(q\):

\[q=\frac{{a}^{2}-b}{a+\sqrt{b}}\]

\[q=a-\sqrt{b}\]

Remark: This proposition introduces the conjugate of a number, which is reoccurs regularly in mathematics.

Check out my other notes at Proof, Disproof, and Derivation

## Comments

Sort by:

TopNewestClaim is not true for \(a=1, b=1\) – Calvin Lin Staff · 2 years, 9 months ago

Log in to reply

– Steven Zheng · 2 years, 9 months ago

Is it because 0 is not rational?Log in to reply

– Steven Zheng · 2 years, 9 months ago

Wait, I don't see what is wrong. Zero is rational, so if \(a = b = 1\), the \(p = 2\) and \(q = 0\). Therefore \(p+q =2\) and \(pq =0\) which are both rational numbers.Log in to reply

– Calvin Lin Staff · 2 years, 9 months ago

If \( p = 1 + \sqrt{1} = 2 \), then there are many values of \(q\) such that \(p+q\) and \(pq\) are both rational. For example, any integer will work.Log in to reply

– Steven Zheng · 2 years, 9 months ago

I see. I guess I should change it to \(a,b\) be rational numbers not equal to 0 and 1. Actually if \(b\) is a square-free rational number.Log in to reply

@Steven Zheng i get the derivation but how do you establish that it is the only unique way?? ...... Is it like

*p and q are roots of a quadratic *and there can be only two such roots ........ please help! – Abhinav Raichur · 2 years, 5 months agoLog in to reply

– Steven Zheng · 2 years, 5 months ago

In the end, we found that if \(p = a-\sqrt{b}\) and \(p+q , pq\) are rational, then \(q\) must equal to \(a+\sqrt{b}\).Log in to reply