# Prove or Disprove: if a | bc, then a|b or a|c

## Homework Statement

Prove or disprove that if a|bc, where a, b, and c are positive integers, then a|b or a|c.

Division! LOL...

## The Attempt at a Solution

Suppose that a|b and a|c are both NOT true. Then... what???

I really, really, really super suck at proofs.

Last edited:

matt grime
Homework Helper
Stop thinking about writing proofs. Just think about the statement. Write down a few examples, try to see what is happening.

Let's try a = 42, b = 3, c = 7.

$$\frac{a}{bc}\,=\,\frac{42}{21}\,=\,2\,\longrightarrow\,a|bc$$

$$\frac{a}{b}\,=\,\frac{42}{3}\,=\,14\,\longrightarrow\,a|b$$

$$\frac{a}{c}\,=\,\frac{42}{7}\,=\,6\,\longrightarrow\,a|c$$

I don't think that there is a way to choose a, b and c such that a|b or a|c are not true. But how do I "show" such?

matt grime
Homework Helper
a is supposed to divide bc. How does 42 divide 21?

Let's try a = 42, b = 3, c = 7.

$$\frac{bc}{a}\,=\,\frac{21}{42}\,=\,0.5\,\longrightarrow\,a|bc\,not\,true$$

$$\frac{b}{a}\,=\,\frac{3}{42}\,=\,0.071\,\longrightarrow\,a|b\,not\,true$$

$$\frac{c}{a}\,=\,\frac{7}{42}\,=\,0.167\,\longrightarrow\,a|c\,not\,true$$

Let's try different numbers...

a = 7, b = 3, c = 14, bc = 42

$$\frac{bc}{a}\,=\,\frac{42}{7}\,=\,6\,\longrightarrow\,a|bc\,is\,true$$

$$\frac{b}{a}\,=\,\frac{3}{7}\,=\,0.428\,\longrightarrow\,a|b\,is\,NOT\,true$$

$$\frac{c}{a}\,=\,\frac{14}{7}\,=\,2\,\longrightarrow\,a|c\,is\,true$$

This only "shows" for these exact values of a, b, and c. How do I show for all a, b, and c?

matt grime
Homework Helper
So from one example you think it is true for all examples? try some more. if you just thnk about prime factorization, rather thatn actually putting numbers into a calculator, it becomes easy.

Prime factorization?

Dividing by two until the number is no longer divisible by two to get a prime at the end?

I don't see how that applies to this proof though.

matt grime
Homework Helper
Sigh.
Every number divides itself. What if a number is not prime?

A number is not prime when it can be divided by a number other than itself or one.

matt grime
Homework Helper
So 6, say, divides 6. 6 isn't prime, so how can you write 6 as the product of 2 smaller numbers?

3 * 2 = 6, yes, it is not prime.

How does this help with the proof though? I just don't understand how I am supposed to use prime numbers to prove anything about divisibility.

matt grime
Homework Helper
So 6 divides 2*3. Now, what does the question ask?

Prove or disprove:

if a divides bc then a divides b or b divides c.

Now do you see? And do you see why primes are important?

I still don't get it!

a = 6, b = 2, c = 3, bc = 6

$$\frac{bc}{a}\,=\,\frac{6}{6}\,=\,1$$ <----- claim holds

$$\frac{b}{a}\,=\,\frac{2}{6}\,=\,\frac{1}{3}$$ <----- claim does NOT hold

$$\frac{c}{a}\,=\,\frac{3}{6}\,=\,\frac{1}{2}$$ <----- claim does NOT hold

I guess using this case, the claim is disproved?

Dick