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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

*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.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,258

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.

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

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

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.

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

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)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,258

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)*

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

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

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

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

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)*

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**Al-Allo****Member**- Registered: 2012-08-23
- Posts: 293

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)*

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,258

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

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

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