I'm new here (on Mathematics Stack Exchange). Also, I'm a 10th grade student not a math expert.

So, my question is whether,

$$\sqrt {n!} $$

comes in the set of the Natural Numbers.

There were some answers when I tried to search on Google. But, it was out of my knowledge, that is, I didn't read/learnt the methods, functions, etc. they were using. So, if possible give a detailed explanation. And pardon me if there is some good answer there and if if it's there then please share a link.

  • 6
    $\begingroup$ Usually not, but if $n=1$ then $\sqrt{1!}=1$. Another example is $n=0$ and $\sqrt{0!}=1$. $\endgroup$
    – vadim123
    Jun 4 '16 at 17:56
  • 5
    $\begingroup$ Are you asking if it always is an integer (it's not), or sometimes an integer, or infinitely often an integer? $\endgroup$
    – Clement C.
    Jun 4 '16 at 18:00
  • $\begingroup$ @vadim123 How can $$\sqrt {0!} $$ be 1? 0! means 0 (I suppose if my concepts are clear) and $$\sqrt{0} $$ is 0. Please correct me if I'm wrong. $\endgroup$
    – Mayank M.
    Jun 4 '16 at 18:11
  • 24
    $\begingroup$ No, $0!$ is equal to $1$. The reason is that you have $n!=\prod_{k=1}^n k$, i.e. you multiply all integers $1\leq k \leq n$. When $n=0$, the set of integers $\{k\mid 1\leq k\leq n\}$ is empty, so you "multiply nothing together": and an empty product is equal to $1$ (like an empty sum is equal to $0$) to be consistent with the axioms defining product (and addition). @MayankM. $\endgroup$
    – Clement C.
    Jun 4 '16 at 18:13
  • 1
    $\begingroup$ Oh I'm really sorry the above comment was a mistype, I've delected it now. I meant $$x^{0}=1$$. Is that related with it? $\endgroup$
    – Mayank M.
    Jun 5 '16 at 17:26

For any $n \gt 1$ there will be some prime in the range $(n/2,n]$ which will only occur once in the factorization of $n!$ by Bertrand's Postulate. This will ensure that $\sqrt{n!}$ is not an integer.

  • $\begingroup$ Wow. How did you think of a proof so quickly and neatly? $\endgroup$
    – Saikat
    Jun 5 '16 at 4:45
  • 14
    $\begingroup$ @user230452: it depends on what you have in your toolbox. If you know Bertrand's Postulate, it is easy. If you don't, it is hard. There are many tools not in my toolbox, but I have this one. $\endgroup$ Jun 5 '16 at 4:55
  • $\begingroup$ Coincidentally, I was recently studying Erdos' proof of the very same proof! Are you a professor? $\endgroup$
    – Saikat
    Jun 5 '16 at 5:02
  • 2
    $\begingroup$ @user230452: no, I am an interested amateur $\endgroup$ Jun 5 '16 at 5:03
  • 1
    $\begingroup$ Even if you don't immediately have a proof in mind a perfect square most have an even number power of every prime factor. thus each and every prime between 1..n most occur in factors an even number of times. That should seem extremely unlikely and should set you to think how frequently do primes occur and in particular if p is the largest prime < n and it occurs at least twice so n > 2p and from p to 2p there must be no primes. That seems unlikely because... well... now you have something to think about. $\endgroup$
    – fleablood
    Jun 5 '16 at 6:10

Look at the prime factors of $n!$. If the square root of $n!$ was an integer, then $n!$ would be the square of an integer, and in the square of an integer, all prime factors occur an even number of times.

For example, if you take $100!$, which ends with $\cdots 95\times 96\times 97\times 98\times 99\times 100$, you see the prime number $97$. That prime number only occurs once in the factorisation of $100!$. All primes from $51$ to $100$ occur only once in the factorisation of $100!$. Therefore $100!$ is not a square.

There is a theorem that there is always a prime number between $n$ and $2n$, and therefore any factorial starting with $2$! has one prime factor that only comes up once.


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