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Calculating prime numbers

The question is in the title. Is there a number that is divisible only by numbers greater than its square root? If not, why? I need this because it can speed up a calculation algorithm significantly if the answer is no.

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marked as duplicate by t.b., ShreevatsaR, Zev Chonoles, Fabian, Jonas Meyer Jun 20 '11 at 20:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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No. See here. If $n$ has a divisor $d \geq \sqrt{n}$ then $1/d \leq 1/\sqrt{n}$ and thus $n/d \leq n/\sqrt{n} = \sqrt{n}$. –  t.b. Jun 19 '11 at 18:40
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If this is closed as a duplicate of the other question, its title should probably be changed — "Calculating prime numbers" is not a very illuminating title or a good search target. –  ShreevatsaR Jun 19 '11 at 18:45
    
Thank you. I think that the reason why this might have come up before is that it can significantly reduce calculation time in any prime-finding algorithm. –  TimeCoder Jun 19 '11 at 18:47
    
Note:This fact increased my program's speed exponentially –  TimeCoder Jun 19 '11 at 18:57
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Exponentially? What kind of program did you have before? –  Fabian Jun 19 '11 at 19:21
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3 Answers

up vote 7 down vote accepted

If $a$ is not a prime, $b | a$ and $\sqrt a < b < a$, then $\frac a b | a$ and $1 < \frac a b < \sqrt a$. So the answer is obviously no.

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No. Suppose that $n$ is positive. If $a > \sqrt n$ and $b > \sqrt n$ then $a b > (\sqrt n)^{2} = n$. Thus no positive number $n$ can be the product of two numbers, each of which is greater than the square root of $n$.

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No. If $n=ab$ and $a>\sqrt{n}$, or in other words $a>\sqrt{ab}$, then by squaring both sides we have $a^2>ab$. Multiplying both sides by $\frac{b}{a}$, we have $ab>b^2$. Taking the square root of both sides, we have $\sqrt{ab}>b$, i.e. $\sqrt{n}>b$. Thus, every composite number is divisible by numbers less than its square root. In fact, the divisors of $n$ greater than $\sqrt{n}$ and the divisors of $n$ less than $\sqrt{n}$ are in bijection, by sending $d\mapsto \frac{n}{d}$.

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I.e. the integer points of the hyperbola $\:xy = n\:$ are preserved by the reflection $\:(x,y)\mapsto (y,x)\:.$ –  Bill Dubuque Jun 19 '11 at 22:53
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