direct proof or contraposition?

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June 24th 2007, 12:25 PM
Discrete

direct proof or contraposition?

I am asked to proof this conjecture:

If x is rational and x is not equal to 0, then 1/x is rational. How do I proof this? Should I use direct or contraposition proof? I just learned about this stuff and I don't know where to start at.
June 24th 2007, 12:40 PM
Plato

The answer depends upon the axiom set you are basing the proof upon.
What are you given about rational numbers?
June 24th 2007, 12:52 PM
Discrete

Quote:

Originally Posted by Plato

The answer depends upon the axiom set you are basing the proof upon.
What are you given about rational numbers?

What do you mean what I am given about rational numbers?
June 24th 2007, 01:02 PM
Plato

Quote:

Originally Posted by Discrete

What do you mean what I am given about rational numbers?

What is the set of axioms you have been given?
You can prove nothing without axioms.
June 24th 2007, 01:07 PM
Soroban

Hello, Discrete!

Quote:

$\text{Prove: If }x\text{ is rational and }x \neq 0\text{, then }\frac{1}{x}\text{ is rational.}$

You must have been given a defintion of a rational number.

Maybe somthing like:
. . $x$ is a rational number if it is of the form $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$ .

That's a good place to start . . .
June 24th 2007, 01:12 PM
Discrete

Quote:

Originally Posted by Soroban

Hello, Discrete!

You must have been given a defintion of a rational number.

Maybe somthing like:
. . $x$ is a rational number if it is of the form $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$ .

That's a good place to start . . .

yes soroban that's what I was actually given.

$x$ is a rational number if it is of the form $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$ .

but then I don't know the next step after that.
June 24th 2007, 01:15 PM
Jhevon

Quote:

Originally Posted by Discrete

yes soroban that's what I was actually given.

$x$ is a rational number if it is of the form $\frac{a}{b}$ where $a$ and $b$ are integers and $b \neq 0$ .

but then I don't know the next step after that.

Let $x$ be a rational number, then $x = \frac {a}{b}$ for $a,b \in \mathbb {Z}, b \neq 0$ . Since $x \neq 0$ it means that $a \neq 0$

So $\frac {1}{x} = \frac {1}{ \frac {a}{b}} = \frac {b}{a}$

Since $a,b \in \mathbb {Z}$ and $a \neq 0$ , the fraction above represents a rational number

QED
June 24th 2007, 01:20 PM
Plato

Quote:

Originally Posted by Jhevon

So $\frac {1}{x} = \frac {1}{ \frac {a}{b}} = \frac {b}{a}$

How do you know that? What axiom are you to use?
You are using properties of rational numbers that the student is expected to prove.
June 24th 2007, 01:43 PM
Ilaggoodly

not true, his argument is based off of the given assumption that x is a rational number, rewriting 1/x as b/a is nothing more than perforrming simple algebra, and then reusing the given definition of rationals
June 24th 2007, 01:52 PM
ThePerfectHacker

Quote:

Originally Posted by Ilaggoodly

not true, his argument is based off of the given assumption that x is a rational number, rewriting 1/x as b/a is nothing more than perforrming simple algebra, and then reusing the given definition of rationals

I think, Plato is asking how do you define 1/x to be.

The more elegant way to define a rational number is by using a "field of quotients of an integral domain"

In that case we can define Q=F(Z) where F(Z) represents the field of quotients of Z (integral domain). In that case it is valid to flip the fraction.
June 24th 2007, 01:53 PM
Plato

Quote:

Originally Posted by Ilaggoodly

rewriting 1/x as b/a is nothing more than perforrming simple algebra, and then reusing the given definition of rationals

Have you ever done a rigorous construction of the real numbers? Let’s say in a mathematical logic or set theory course.
June 24th 2007, 02:00 PM
Ilaggoodly

yes as a matter of fact i have taken a logic and set theory course, and im wel aware that you can make this problem MUCH more complicated than it is likely intended to be, i believe the way it has been shown by soroban and jhevon would satisfy the poster... in which case it was my mistake in saying that you were not correct, my apologies
June 24th 2007, 02:06 PM
Plato

Quote:

Originally Posted by Ilaggoodly

i believe the way it has been shown by soroban and jhevon would satisfy the poster...

But the point is: we don't know that!
The student should have given more of what sort of proof is expected.
June 24th 2007, 02:11 PM
Ilaggoodly

true enough, however considering some of his other posts...
S.O.S. Mathematics CyberBoard :: View topic - discrete math basic's
it would seem as if he/she is taking his or her first course in introduction to advanced mathematics...anyway, whatever
June 24th 2007, 02:25 PM
Jhevon

Yes, Plato is right. It is up to the poster to tell us how rigorous a proof is needed. My proof is not all that rigorous, it is simple algebra which assumes the reader knows the axioms behind it. If a more rigorous proof is needed, my post should only serve as an outline of what to do

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