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

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Comments

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

Calvin Lin Staff - 6 years, 9 months ago

Is it because 0 is not rational?

Steven Zheng - 6 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.

Calvin Lin Staff - 6 years, 9 months ago

@Calvin Lin – 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.

Steven Zheng - 6 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.

Steven Zheng - 6 years, 9 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!

Abhinav Raichur - 6 years, 6 months ago

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

Steven Zheng - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$