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# 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
2 years, 4 months ago

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Claim is not true for $$a=1, b=1$$ Staff · 2 years, 4 months ago

Is it because 0 is not rational? · 2 years, 4 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. · 2 years, 4 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 · 2 years, 4 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. · 2 years, 4 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! · 2 years, 1 month ago

In the end, we found that if $$p = a-\sqrt{b}$$ and $$p+q , pq$$ are rational, then $$q$$ must equal to $$a+\sqrt{b}$$. · 2 years, 1 month ago