Please tell me whether my proof is valid.

(1) Suppose $\sqrt{F!-1}= \frac p q$ where $p, q$ are integers $>0$ with no common factors. (If there are any common factors we cancel them in the numerator and denomintor.

(2) Squaring both sides, we get $F!-1= \frac {p^2} {q^2}$.

(3) The expression $F!-1$ is whole. Because $p$ and $q$ have no common factors, the only way $\frac {p^2} {q^2}$ can be whole is if $q^2=1$.

(4) Therefore, $F!-1=p^2$.

(5) Note that $F!-1$ is a form of $3k-1$.

(6) No square can equal any $3k-1$. A number is either $3j-1$, which squared becomes $3m+1$; or $3j$, which squared becomes $3m$; or $3j+1$, which squared becomes $3m+1$. Thus, no square can equal any $3k-1$.

(7) Therefore, $F!-1$, being a form of $3k-1$, cannot equal $p^2$, which is a direct contradiction to (4).

(8) The only resolution to the contradiction is that our supposition of (1) must be false, and $\sqrt{F!-1}$ cannot equal any $\frac p q$; consequently, it is irrational.

Thank you.

  • 3
    $\begingroup$ Looks good to me. $\endgroup$ – Adam Hughes Jun 23 '15 at 19:13
  • 5
    $\begingroup$ Do we have any restriction what $F$ can be? Note that $F=2$ makes the expression $1$. For $F\ge 3$, your proof is valid. $\endgroup$ – Berci Jun 23 '15 at 19:16
  • $\begingroup$ One should say more in step $(3).$ You seem to implicitly be using the Fundamental Theorem of Arithmetic (or some equivalent), and this invocation should be made explicit in this crucial step. $\endgroup$ – Bill Dubuque Jun 23 '15 at 19:26

If we don't have any restriction to $F$ (besides -I guess- that $F$ is a natural number) then $F=0,1,2$ produce rational numbers ($0,0,1$, respectively).

Your proof is valid for $F\ge 3$ as you are assuming $3|F!$ in step (5).

  • $\begingroup$ Yes, I should have clarified that $F \ge 3$. Thanks for pointing that out. $\endgroup$ – Robert Gross Jun 23 '15 at 22:03

I think that the following may express the second half of your proof in a slightly clearer manner:

  • $F\geq3 \implies F!\equiv0\pmod3 \implies F!-1=2\pmod3$

  • $F!-1=p^2 \implies F!-1\not\equiv2\pmod3$:

    • $p\equiv0\pmod3 \implies p^2\equiv0^2\equiv0\pmod3$

    • $p\equiv1\pmod3 \implies p^2\equiv1^2\equiv1\pmod3$

    • $p\equiv2\pmod3 \implies p^2\equiv2^2\equiv1\pmod3$


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