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If $N$ divides $a$ and $N$ divides $b$ then $N$ divides $a \cdot b$?

Is the statement true? I mean.

$$a \equiv 0 \pmod{N}$$

$$b \equiv 0 \pmod{N}$$

$$\implies ab \equiv 0 \pmod{N}$$

But I am not sure how rigorous this is?

Bill Dubuque
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Lebes
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    It's true, and it's very easy to prove if you use the definition of divisibility. Did you try it? PS: If you want to see it with congruences, this comes from a more general property of congruences that state the following: If $a\equiv c \pmod{N}$ and $b\equiv d \pmod{N}$, then $ab\equiv cd\pmod{N}$, you can try to prove this too! – Daniel Mar 28 '15 at 07:46
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    But either $N\vert a$ or $N\vert b$ implies $N\vert ab$. – Yai0Phah Mar 28 '15 at 07:50
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    Good point @FrankScience! So you can even change the "and" in your claim by an "or" @Lebes – Daniel Mar 28 '15 at 07:51
  • "$N$ divides $a$" means that there exists $k$ such that $kN=a$. Use this and what you want follows at once. – Andrea Mori Feb 09 '23 at 18:04

1 Answers1

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If $N \mid a$ and $N \mid b$, then $N \mid ab$.

Suppose $N \mid a$ and $N \mid b$, then by the definition of divisibility there exist integers $x$ and $y$ such that $Nx = a$ and $Ny = b$. So we have that $$ab = NxNy = N^2xy$$ which is clearly divisibly by $N$.

Clyde Kertzer
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