Uniqueness of Conjugate Numbers

Let aa and bb be rational numbers, and bb square-free. Prove that if p=a+bp=a+\sqrt{b}, there exists only one value qq such that p+qp+q and pqpq are rational.

Solution

Let p+q=cp+q=c and pq=dpq=d. q=dpq=\frac{d}{p}

p+q=a+b+da+b=cp+q=a+\sqrt{b}+\frac{d}{a+\sqrt{b}}=c

a2+2ab+b+d=ca+cb{a}^{2}+2a\sqrt{b}+b+d=ca+c\sqrt{b}

Matching each component yields:

2ab=cb2a\sqrt{b}=c\sqrt{b}

2a=c2a=c

a2+b+d=2a2{a}^{2}+b+d=2{a}^{2}

d=a2bd={a}^{2}-b

Now we solve for qq:

q=a2ba+bq=\frac{{a}^{2}-b}{a+\sqrt{b}}

q=abq=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
4 years, 12 months ago

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

Calvin Lin Staff - 4 years, 12 months ago

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Is it because 0 is not rational?

Steven Zheng - 4 years, 12 months ago

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If p=1+1=2 p = 1 + \sqrt{1} = 2 , then there are many values of qq such that p+qp+q and pqpq are both rational. For example, any integer will work.

Calvin Lin Staff - 4 years, 12 months ago

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@Calvin Lin I see. I guess I should change it to a,ba,b be rational numbers not equal to 0 and 1. Actually if bb is a square-free rational number.

Steven Zheng - 4 years, 12 months ago

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

Steven Zheng - 4 years, 12 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 - 4 years, 8 months ago

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

Steven Zheng - 4 years, 8 months ago

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