Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

*If I turn out to have a wrong answer, please no hints or showing an valid proof. I want to do it on my own !

http://postimg.org/image/buax1y1sd/

ad/bd-bc/bd=+-1/bd is neighbor fraction

Now, reduce the common numbers :

a/b-c/d=+-1/bd

We must now prove that the left hand side has irreductible fractions. Lets see what would happen if these fraction were reductible.

let a=z*y , b=z*l , c=p*m d=p*n

z*y/z*l-p*m/p*n equality to be determined +-1/(z*l)*(p*n)

Reduce:

*ln (y/l-m/n) equality to be determined (+-1/(z*l)*(p*n))*ln

yn-lm equality to be determined +-1/z*p

yn-lm not= +-1/z*p

We have considered the initial fractions to be reductible and have arrived at a false result. An integer cannot be equal to (+-1/z*p)

So, the initial fractions must be irreductible.

*Of course, I'm considering the variables to represents integers.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,137

hi Al-allo

I couldn't follow this because of the way you had set out your proof. So, I've converted the fractions, taken out the * and written L instead of l.

I have added some words in red that make it clearer to me what you are doing.

In a proof by contradiction you may put equals even though you are hoping to show that the two sides are not equal. So you may replace "equality to be determined" by "=".

Instead of the word "let" you could put "suppose".

At the line of ************* I think you need to add one more step.

After my word But you need to justify why these expressions cannot be equal.

Bob

Now, reduce the common numbers :

We must now prove that the left hand side has irreductible fractions. Lets see what would happen if these fraction were reductible.

let

Reduce and multiply by Ln:

***************

But

We have considered the initial fractions to be reductible and have arrived at a false result. An integer cannot be equal to (+-1/z*p)

So, the initial fractions must be irreductible.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

The reason why the left hand side cannot be equal to the right hand side is because we have considered our variables (a=zy,etc.) to be integer whose factorization were also integers.(We are dealing with fractions, this is the reason why we need them to be integers.) So, when we arive at this result : yn-lm(not equal to) +-1/zp, the left hand side is a substraction of integers. But the right hand side is a fraction with 1/zp(zp also being integers). So, the result shall be inferior to 1. The only exception where it won't be inferior is the case were the left-right hand side shall be equal to +-1.

For the *****, I'm not sure what you're referring to... I shall add more later on.

Offline

**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

By the way, I forgot to treat the case where there might exist one fraction which is factorizable and the other not.

So :

a/b-c/d=+-1/bd

xi/xj-c/d = +-1/((xj)*d)

(i/j-c/d=+-1/xj*d) * jd

id-jc not =l +-1/x

except for 1 and -1

The same is done for the second fractions on the left side.

*Last edited by Al-Allo (2014-01-08 04:06:48)*

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,137

Explanation in post 3 is what I was seeking.

The ************* is missing a simplification of fractions to show where the LN has gone. It's fairly obvious but I would have put it in.

Bob

*Last edited by bob bundy (2014-01-08 04:15:57)*

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Oh I really need to put it in the numerator and show it ? I didn't know that !

Offline

**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

By the way, am I right by saying that the exception (the equality with 1 or -1) is telling us that such factorization exists?

because ovisously we could have 1*1/1*1-0*x/1*1=1/1*1*1*1

Which is one example of such a factorization that exists.

*Last edited by Al-Allo (2014-01-08 05:31:29)*

Offline

**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 324

Ok, yeah, I think that's it :

We have shown that such factorization exists which can make the equation equal :

1*1/1*1-0*x/1*1=+1/1*1*1*1

which results in 1-0=1

The other one is :

0*x/1*1-1*1/1*1=-1/1*1*1*1

0-1=-1

I don't know if there exists other ways of 1 or -1, but anyway, this is out of the context of the inital problem.

*Last edited by Al-Allo (2014-01-08 05:48:27)*

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,137

Al-Allo wrote:

Oh I really need to put it in the numerator and show it ? I didn't know that !

How many steps to put in is not a fixed thing. You have to demonstrate that your proof is rigorous. If there's marks in it, I would 'err' on the side of caution and put in too many rather than too few. No one can complain at that.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

Pages: **1**