How to prove that 3 is divisor of two numbers and their sums?

Question

Original question: Prove that if 3 is a divisor of two numbers, it is a divisor of their sum and their difference. Generalize this and prove that if d is a divisor of two numbers b1 and b2, then d is a divisor of b1 + b2 and of b1 -b2. [Question from Rational and Irrational numbers by Ivan Niven]

I am new to proofs and I have limited methods to approach the problem.

Consider that if num₁ and num₂ which are exactly divisible by 3, then

num₁ = 3 q₁ [q1 is a quotient] and

num₂ = 3 q₂

Consider, if 3 divides exactly num₁ + num₂

Then, (num₁ + num₂)$\div$3

= (3 q₁ + 3 q₂)$\div$3

= 3 (q₁ + q₂)$\div$3

= q₁ + q₂

= Q (Where Q = q₁ + q₂ which in fact is integer)

So, can we address the first half of the question with this proof?

Answer 1

I would change a few things in the second half of the proof. There are two things to take into account:

• Avoid using $/3$, $\div 3$ or $\frac{}{3}$. Your proof should use only integer arithmetics, no fractions.
• Add a few words to make sure that logic links between statements are more explicit; for instance, you can say "Now, let's use this to prove that 3 divides exactly $\operatorname{num}_1 + \operatorname{num}_2$" but not "Now, 3 divides exactly $\operatorname{num}_1 + \operatorname{num}_2$".

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Let $\operatorname{num}_1$ and $\operatorname{num}_2$ be two integers divisible by $3$. We want to prove that their sum $\operatorname{num}_1 + \operatorname{num}_2$ is also divisible by $3$; i.e., we want to prove that there exists an integer $Q$ such that $\operatorname{num}_1 + \operatorname{num}_2 = 3Q$.

By assumption, $\operatorname{num}_1$ and $\operatorname{num}_2$ are divisible by $3$, so there exists two integers $q_1$ and $q_2$ such that:

\begin{align} \operatorname{num}_1 &= 3 q_1; \\ \operatorname{num}_2 &= 3 q_2. \end{align}

Consider now the sum $\operatorname{num}_1 + \operatorname{num}_2$:

\begin{align} \operatorname{num}_1 + \operatorname{num}_2 &= 3 q_1 + 3 q_2 \\ \operatorname{num}_1 + \operatorname{num}_2 & = 3(q_1 + q_2) \end{align}

Let $Q = q_1 + q_2$. We have proven that $\operatorname{num}_1 + \operatorname{num}_2 = 3 Q$. Hence, $\operatorname{num}_1 + \operatorname{num}_2$ satisfies the definition of divisibility by $3$.

Answer 2

Yes.

The definition of divisibility is usually as follows: $a$ divides $b$ if there exists an integer $k$ such that $b = ak$.

And you are essentially showing that $k = q_1 + q_2$ for $b = \textbf{num}_1 + \textbf{num}_2$ and $a = 3$. The proof can be generalized to arbitrary divisor and difference quite easily.