Beatty’s Theorem

Posted on January 23, 2015 by Sumant Sumant

In 1926, Sam Beatty of the University of Toronto made a remarkable discovery concerning sequences of irrational numbers. For a given irrational number X and \frac{1}{X}. The sequence n(1+X) and n(1+\frac{1}{X}) is a complementary sequence.

Proof: We see that

\frac{n}{1+X} -1 < \left [ \frac{n}{1+X} \right ] < \frac{n}{1+X}

\frac{n}{1+Y} -1 < \left [ \frac{n}{1+Y} \right ] < \frac{n}{1+Y}

\frac{n}{1+X} -1+\frac{n}{1+Y} -1 <\left [ \frac{n}{1+X} \right ]+\left [ \frac{n}{1+Y} \right ] <\frac{n}{1+X}+\frac{n}{1+Y}

\frac{n}{1+X}+\frac{nX}{1+X} -2 <\left [ \frac{n}{1+X} \right ]+\left [ \frac{n}{1+Y} \right ] <\frac{n}{1+X}+\frac{nX}{1+X}

n-2<\left [ \frac{n}{1+X} \right ]+\left [ \frac{n}{1+Y} \right ] <n

Since \left [ \frac{n}{1+X} \right ]+\left [ \frac{n}{1+Y} \right ] is Integer, we conclude that \left [ \frac{n}{1+X} \right ]+\left [ \frac{n}{1+Y} \right ] = n-1

Also the total number of terms up to n+1 is n. That is to say, if we increase the integer n by 1, another term of the sequence is admitted, implying that exactly one term lies between n and n+1.

About Sumant Sumant

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