# Proposed proof for:$b! \equiv 0 \pmod a$ $\Rightarrow$ $a \le b$

Let a,b $$\in$$ $$\mathbb{Z}$$. Obviously, a counterexample can disprove this statement.
I tried this approach to seek another method (possibly proof by contradiction):

$$a|b!$$
$$a|[b(b-1)!]$$

If $$gcd(a,(b-1)!)=1$$, then $$a|b$$ and hence: $$a \le b$$.
But I think claiming that $$a|b$$ is too restrictive because $$a$$ can be less than $$b$$ without having $$a|b$$ necessarily.

I tried to study two cases to reach a contradiction that $$gcd(a,(b-1)!)\not=1$$, yet I couldn't figure out how to continue so this might actually help out :
1) $$a=b$$
2) $$a \lt b$$

Is there any mistake that I have done? Is this proof inconclusive or are there any improvements to be made so that it is a valid proof?
EDIT: Counterexamples most certainly get the job done here. Though, I am seeking another method to disprove the statement.
EDIT 2: The original proposition is: $$a \le b \Rightarrow b! \equiv 0 \pmod a$$
I am trying to justify whether the converse of this statement is right or wrong, which is obviously wrong using counterexamples. The purpose is to find another method to disprove the converse.

• Without further restrictions this is false big time, for exmple: $\;-7=0\pmod 7\;$ , yet $\;-7<7\;$ ...This is specially important as we usually work within $\;\Bbb Z\;$ when doing this kind of problems... Jun 8, 2020 at 10:58
• Please refrain from making minor edits. Also, please take a look at the description of the tag proof-writing before adding it.
– Ѕааԁ
Jun 13, 2020 at 14:37
• Thank you for the note. The tag wiki states that:" I have written the following proof, could I somehow improve it, does it have good flow/can I improve readability? ", and i think my question was kind of similar Jun 13, 2020 at 14:53

I think there's no point to try and prove a false statement. Because $$15|5!$$ for example.

What is a true version of this statement is $$a|b! \implies \text{prime factors of }a \le b$$ You can try and prove this!

Update: I understood you seek to prove it without using counterexamples, so let's try:

Let $$p$$ and $$q$$ be prime numbers where $$p > q$$, we have $$pq|p!$$ while $$pq >p$$

Another update: in your solution of "disproving the statement" you suggested that $$\gcd(a,(b-1)!)=1$$ while $$a \le b$$. Well if $$a we have $$a|(b-1)!$$ so this directly means $$a=b$$ and this is something you wouldn't want as a step in disproving the statement. So the thing you'd first try is finding $$a>b$$ and $$a|b!$$, and I've given that in my first update.

• You are right that it is pointless, but I am trying to disprove it without using counterexamples. Appreciate your recommendation though! Jun 8, 2020 at 11:32
• @OmarS Oh, I didn't notice. I will try and edit and answer for that Jun 8, 2020 at 11:36
• @OmarS Is that ok? I think disproving a theorem doesn't require formality and cases and stuff like that. It just requires a case at which the statement doesn't work. It does work for some integers but not for all, I suggest you look closely and explain clearly what exactly you need from such a problem/proof Jun 8, 2020 at 11:39
• Perhaps you are right, i should have included what i needed exactly. I will edit the post and thank you for the head's up! Jun 8, 2020 at 12:55
• The second update is exactly what i needed. $a > b$ to be exact. I will work on it and post an updated answer. Thank you for your efforts Jun 8, 2020 at 15:55

$$b = 5$$ and $$a = 6$$ is a counterexample.

The statement is true, however, if $$a$$ is further assumed to be prime.

Note that $$b!$$ has as factors, all natural numbers $$\le b$$. If $$a$$ is prime and $$a > b$$, then $$a$$ cannot divide $$b!$$ since $$a$$ is a prime factor of $$a$$ that won't divide $$b!$$.

If $$a$$ is not prime, then it is possible for all of its prime factors (with appropriate multiplicities) to be factors of $$b!$$ even if $$b < a$$, as shown above.