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I was asked to prove this problem by contradiction or contrapositive, but I'm not quite sure on how to proceed. I appreciate your help.

2.3.6

I assume that $c \ge 2$ and if I want to use the contradiction method I guess that I have to say that it is false that there is an integer $b\ge 2$ such that $b|c$ and $b \le \sqrt{c}$.

So, is it okay starting by saying: "Supose that $\forall b$, $b\geq 2$, $b \nmid c$ and $b > \sqrt{c}$."?

If that is so, then supose that $\forall b, b\geq 2$, $b \nmid c$, then $c \nmid c$. Wich is a contradiction.

Then it must be the case that $\forall b, b\geq 2$, $b > \sqrt{c}$. Which implies that $b^2>c$. Since it is true for all $b\geq 2$, take $b=2$, then $4>c$, so the only values that $c$ can take are 3 or 2, but that is a contradiction, because c is not a prime number.

Hence, what I supposed is false.

Is all that ok?

Alexander Navarro
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    What have you tried? – Robert Z Jul 24 '21 at 17:24
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    Hi Alexander, welcome to MSE! The community on this websites expects people to explain what they have tried when they present questions. This is not a homework service, so people in general expect that your question shows research effort (this is even shown when you hover over the upvote button). In this spirit I would suggest that you edit your question to improve it a bit. If you are not sure what is expected, check out the FAQ's. – Guenterino Jul 24 '21 at 17:31
  • Sure @Guenterino, sorry for that. – Alexander Navarro Jul 24 '21 at 18:02
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    @AlexanderNavarro Hey, no problem! Great that you came back and improved your question. Looks like you already got an answer that way! – Guenterino Jul 24 '21 at 18:10

1 Answers1

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While proving by contradiction, start by seeing what happens if the statement is false or the exact opposite of the statement is true. In your case, the statement is for a composite number $c\geq 2$, there exists an integer $b$ such that $b\geq 2$, $b\vert c$ and $b\leq \sqrt c$. But in your attempt, you started by assuming $\forall b, b\geq 2$, $b \nmid c$ and $b > \sqrt{c}$. But even if this contradicts, you can't say there is a $b$ such that $b\vert c$ when $b\leq \sqrt c$ rather you can just say that there exists a number $b$ such that $b>\sqrt c$ and $b\vert c$. But this doesn't prove the statement.

So, we assume the exact opposite of the statement which is: there exists a composite number (since $c$ is not prime is given) $c\geq 2$ such that there is no integer $b$ which satisfies $b\geq 2$, $b\vert c$ and $b\leq \sqrt c$.

Clearly, this is impossible because every composite number $n$ has a prime divisor which is greater than or equal to $2$ and less than or equal to $\sqrt n$. Hence, we get a contradiction which proves the desired statement.

Oshawott
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  • I'm a little confused. You are denying the hypothesis, aren't you? – Alexander Navarro Jul 24 '21 at 18:22
  • @AlexanderNavarro You can say that. I'm assuming that the opposite of the hypothesis is true which for the sake of contradiction. But this is same as denying the hypothesis. – Oshawott Jul 24 '21 at 19:06