Question

If$\mathit{a}\mathbf{=}\mathit{b}\mathit{c}$ with$\mathit{a}\mathbf{\ne }\mathbf{0}$ and b and cnon units, show that a is not an associate of b.

Short answer

It has been proved that if $\mathit{a}\mathbf{=}\mathit{b}\mathit{c}$ with$\mathit{a}\mathbf{\ne }\mathbf{0}$ and b and cnon-units, then a is not an associate of b.

Step-by-step solution

Given that

Given that $\mathit{a}\mathbf{=}\mathit{b}\mathit{c}$with $\mathit{a}\mathbf{\ne }\mathbf{0}$.

Also, b and c are non-units.

Proof of statement by contradiction

Suppose that a is an associate of b.

Then,$\mathit{a}\mathbf{=}\mathit{b}\mathit{u}$ for some unit u.

But $\mathit{a}\mathbf{=}\mathit{b}\mathit{c}$.

So $\mathit{b}\mathit{c}\mathbf{=}\mathit{b}\mathit{u}$ .

This implies c must be a unit.

But it is given that c is not a unit.

Hence, the supposition was wrong.

Conclusion

Thus, it can be concluded that a is not an associate of b