Convince a classmate that division is not associative.

[okunyg]

Homework Statement

How would you convince a classmate that division is not associative?
By "associative" the book aims at the Associative properties of multiplication and addition.

Is this equation correct?
a/(b/c) = (a/b)/c

Homework Equations

a + (b + c) = (a + b) + c
a(bc) = a(bc)

a/(b/c) = (a/b)/c (?)

The Attempt at a Solution

The parenthesis' are top priority, and the arithmetic in it should be done first.
- This means that a/(b/c) yields a/<new number>.
- This means that (a/b)/c yields <new number>/c.
-- This means a/<new number> does _not_ equal <new number>/c.

Or can it?

I would like to solve it algebraically, but I don't know how. Is it possible to solve it with algebra? I mean, does it take more advanced mathematics or can I simply use basic algebra as a tool for solving this?

This is not homework actually, I'm trying to learn math by myself with the book Algebra and Trigonometry (Wesley 2007).

Thanks.

 

[honestrosewater]

Well, the associative property claims to hold for all values of a, b, and c. So you could just find one set of values for which it doesn't hold, yes?

 

[okunyg]

Yes, that is correct. But are answers like that accepted? What if I would like to prove it without trial and error (if that's the correct expression)?

 

[honestrosewater]

Yes, counterexamples are perfectly acceptable.

You could try to prove from the axioms

There exists some a, b, and c such that a/(b/c) != (a/b)/c​

where "!=" means "not equal", sure. Counterexamples are one way to do that.

By the bye, proving things most ways involves a bit of trial and error.

 

[turdferguson]

Its called indirect proof or proof by contradiction, and its a perfectly legitamate form of proof. Assume to the contrary (division is associative) and work until you reach something that can't be true