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Statement to be proved:

Prove that if $n > 1$ is not divisible by any prime number $p$ where $p \le \sqrt{n}$ then $n$ is a prime number.

Suppose we assume that $n$ is composite. We then prove that $n$ is divisible by a prime number $\le \sqrt{n}$.

We proved that a composite number is divisible by a prime number $\le \sqrt{n}$. Can we fairly deduce from this that a prime number is not divisible by a number $\le \sqrt{n}$? I doubt this. As far as this proof is concerned, for all we know, both composite and prime numbers could be divisible by a prime number $\le \sqrt{n}$. Where is the contradiction?

ForgotALot
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sammy
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  • If $n$ is prime then $n$ is the only prime dividing $n$. And it is not $\leq \sqrt{n}$ – Asinomás Jul 23 '16 at 22:12
  • We've proven if composite => then divisible by som number < sqrt root. Therefore if NOT divisible by seme number < sqrt root => not composite => prime. – fleablood Jul 23 '16 at 23:11
  • Things are either alive or inert. Prove if it doesn't have skin it is inert. Proof by contradiction: If it is alive then it has skin. Done. Is is possible for something with skin to be intert. Sure. But if it doesn't have skin it can't be alive and it therefore inert. We don't care about things with skin. We only care that things without skin must be intert. – fleablood Jul 23 '16 at 23:16
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    We haven't prove a composite may be divisible. We've proven every composite must be divisible. So if it isn't divisible it can not be composite. And if it isn't composite it must be prime. – fleablood Jul 23 '16 at 23:56
  • See Proof by contrapositive: Relation to proof by contradiction: "Any proof by contrapositive can also be trivially formulated in terms of a Proof by contradiction". – Mauro ALLEGRANZA Jul 24 '16 at 08:44

4 Answers4

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Your question really concerns the logic of contrapositives. You are proving that a sufficient condition for $n$ to be prime is that no prime number less than or equal to $\sqrt n$ divides $n$. One way of doing that is to show that a necessary condition for $n$ to be composite (negation of $n>1$ being prime) is the negation of the condition given above.

So, the logic works as follows. You want to show $P$ implies $Q$ (so $P$ is the sufficient condition). The method is to prove the contrapositive of the implication, that $\neg Q$ implies $\neg P$ (i.e. proving $\neg P$ is a necessary condition for $\neg Q$). Given that the contrapositive has been proven, one can be certain of $P$ implies $Q$ as well. One way to see that is to compare truth tables. But, one may be convinced the following argument: $P\implies Q$ follows from $\neg Q\implies \neg P$ since $P$ and $\neg Q$ would entail $\neg P$ a contradiction.

Karl Kroningfeld
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  • Prime if condition a, composite if condition b. Conditions a and b do not exist at the same time. If condition b, then not condition a. – sammy Jul 23 '16 at 23:20
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I am not sure if I understand your question. But if you want to prove that if $n$ is composite there must be at least one prime $p\leq\sqrt{n}$ that divides it:

  1. Each composite number has at least two prime factors.
  2. Assume the smallest prime that divides $n$ is $p$, and the next smallest $q$

Now if $p \gt \sqrt{n}$, that means $q \gt \sqrt{n}$, since if it wasn't, then it would not be larger than $p$.

Two numbers multiplied together which both are larger than $\sqrt{n}$ can not be equal to $n$, because the square root function is strict increasing. Therefore the smallest prime dividing $n$ must be $\leq \sqrt{n}$.

mathreadler
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" Can we fairly deduce from this that a prime number is not divisible by a number <= sqrt(n)?"

Well, obviously[#], but that's not at issue at all. We were never asked to prove any such thing at all and we don't need to.

We were asked to prove "If $n >1$ is not divisible by any number, $k$ $1 < k \le \sqrt{n}$ then $n$ is prime".

We did a proof by contradiction "If $n$ is composite, then $k|n$ for some $k$"

THEREFORE "If $n$ is not divisible $\implies$ $n$ is not composite $\implies$ $n$ is prime."

Did we prove that $n$ prime $\implies$ $n$ is not divisible by any number less than $\sqrt{n}$? No, we did not. Why should we have? That was not what we were trying to show

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[#] Well, duh. If $n$ is prime it's not divisible by anything other than itself and 1. Duh!

fleablood
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we know any positive integer is always greater than its sq.root of itself like $25>\sqrt{25}$ then $n>\sqrt{n}$. now given $p\le \sqrt{n}$ therefore $n > \sqrt{n}\ge p$. then $n>p$ and $p>1$. therefore $p$ not equal to $n$ nor $1$.

let $p$ divides $n$. now $p$ is neither $n$ nor $1$ but divides $n$.

now if $n$ is prime then it would only be divisible by $n$ and $1$. but $p$ divides $n$ though it is neither $n$ nor $1$.

therefore $n$ is not prime , i.e $n$ will be prime if it is not divisible by $p$ such that $p\le\sqrt{n}$. (proved)

riz
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