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I was doing some exercises on proof techniques from the book "Proofs and Fundamentals: A first course in abstract mathematics" and then I saw this exercise which is as far as I know incorrect. The exercise says:

Let c be an integer. Suppose that c ≥ 2, and that c is not a prime number. Prove that there is an integer b such that b ≥ 2, that b|c and that b≤√c

If the number c is not prime then it can't be equal to 2 or am I missing something here? Because this affects also the rest of the proof the way that I have thought about it at least.

brahWasHere
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  • Yes... $c\geq 2$ and $c$ not prime does imply that $c>2$. – JMoravitz Nov 02 '20 at 18:54
  • As for a hint... $c$ not-prime and greater than $2$ implies that $c$ is composite and will imply that there is a smallest prime that divides $c$. – JMoravitz Nov 02 '20 at 18:57
  • It's a perfectly correct statement, just not as sharp as it could be. For example, if I said, let $c$ be an integer with $c> 1$ and $c > 0$, it wouldn't be incorrect -- the second assumption would just be unnecessary. – Cheyne H Nov 02 '20 at 19:01
  • For an entirely different approach... consider for contradictory purposes that all divisors of $c$ are strictly greater than $\sqrt{c}$. But then since $c>2$ and $c$ isn't prime, you have $c$ is composite and so can be written as $m\cdot n$ with neither $m$ nor $n$ equal to $1$. Is it possible for both $m$ and $n$ to be strictly greater than $\sqrt{c}$ while still having $m\cdot n=c$? – JMoravitz Nov 02 '20 at 19:03
  • Hint: the smallest non-prime factor of any non-prime number will be greater than or equal to two and less than or equal to the square root of the number it divides. – John Douma Nov 02 '20 at 19:29
  • @JMoravitz Okay so I tried to prove it based on your hint. Well it is not possible for m and n to be strictly greater than the square root of c because then their product would be greater than c, thus each of the factors has to be smaller or equal to the square root of c. I think this logic works or am I wrong? – brahWasHere Nov 02 '20 at 19:46
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    Thus each of the factors no. Thus at least one of the factors. – JMoravitz Nov 02 '20 at 19:53

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