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Would this be the same thing as saying "Prove that if $a|b$ and $a|c$ then $a|(sb+tc)$ for any $s, t \in \mathbb{Z}$"? I can do the proof for any integers $s$ and $t$, but if any and all mean the same thing, then I can do this proof. If not, how can I do this?

Edward Jiang
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JCMcRae
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3 Answers3

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try writing both numbers as a combination of integers and remainders b=qa+r c=q'a+r sb+tc=sqa+sr+tq'a+tr=a(sq+sq')+r(s+t) since a divides b, c, and 0 <= r <=a , we conclude r must be 0. from here, forall s,t in Z, a divides any linear combination of b and c

FisherDisinformation
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In response to your question about any and all, they generally mean the same thing. The symbol $\forall$ can be read "for all" or "for every" or "for any" or "for each". These all mean the same thing.

This is how you could write the proof, with meticulous rigor.

Let $a,b,c \in \mathbb{Z}$ with $a \mid b$ and $a \mid c$.

Let $s,t \in \mathbb{Z}$ be arbitrary.

Then by definition, $\exists\, k_1, k_2 \in \mathbb{Z}$ such that $$b = ak_1$$

$$c = ak_2$$

Then $$sb = s(ak_1)$$

$$tc = t(ak_2)$$

By commutativity and associativity,

$$sb = a(sk_1)$$

$$tc = a(tk_2)$$

Adding these equations, we get

$$sb + tc = a(sk_1) + a(tk_2)$$

By distributivity,

$$sb + tc = a(sk_1 + tk_2)$$

Since $\mathbb{Z}$ is closed under addition and multiplication, $$(sk_1 + tk_2) \in \mathbb{Z}$$

Thus, by definition, $$a \mid (sb+tc)$$

Zubin Mukerjee
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Assume that $a|b$ and $a|c$, i.e., $\frac ba$ and $\frac ca$ are integers. For any $s,t\in\mathbb Z$ we have $$\frac{sb+tc}a=s\left(\frac ba\right)+t\left(\frac ca\right)\in\mathbb Z$$ (since sums and products of integers are integers), so $a|(sb+tc)$.

bof
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    A (minor) weakness with this approach is that $a | b$ allows for $a = 0 $ which would force $b=0$ and $ c=0$, However, in this case the fraction would be undefined. To fix this you just need to add a condition $a \neq 0$. Pedagogicaly this proof is useful as it is intuitive. – john Sep 25 '19 at 16:23