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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
3 years, 4 months ago

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Comments

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Claim is not true for \(a=1, b=1\)

Calvin Lin Staff - 3 years, 4 months ago

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Is it because 0 is not rational?

Steven Zheng - 3 years, 4 months ago

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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 - 3 years, 4 months ago

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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 - 3 years, 4 months ago

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@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 - 3 years, 4 months ago

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@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 - 3 years, 1 month ago

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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 - 3 years, 1 month ago

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