# Uniqueness of Conjugate Numbers

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 Note by Steven Zheng
6 years, 3 months ago

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Claim is not true for $a=1, b=1$

Staff - 6 years, 3 months ago

Is it because 0 is not rational?

- 6 years, 3 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.

Staff - 6 years, 3 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.

- 6 years, 3 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.

- 6 years, 3 months ago

@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!

- 5 years, 11 months ago

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

- 5 years, 11 months ago