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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I have an idea. Let this topic be the number theory QA topic.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

"e" means "element"**N** means Natural(>=1)**Z** means Integer**Sq** means Perfect square set:

x e **Sq** means "x is perfect square"**P** means prime number set.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

NumbTh1:

If a,b ∈ **N** and ab≡0(mod a+b) prove that a²b²/(a+b) has a rom x²y³, where x,y ∈ **N**

IPBLE: Increasing Performance By Lowering Expectations.

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

¡Ay chico, estás hablando Chino!

So, is (mod a+b) a modulus operation? And why bother if you multiply it by zero? And how can you define the product of two natural numbers to be zero?

What's a rom (besides a read-only memory)?

Or were you joking?

El que pega primero pega dos veces.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I've never seen rom either. Could we get a definition, krassi?

So, is (mod a+b) a modulus operation? And why bother if you multiply it by zero? And how can you define the product of two natural numbers to be zero?

modulo, to be exact. But it is the same idea.

k ≡ m (mod n) means that n divides k-m:

k-m = xn, x ∈ Z

So in this situation, a+b divides ab. Since a+b divides ab, a+b divides (ab)² or a²b².

Not knowing rom, I can't go any further than that.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,534

Rom is the capital of Ital

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

**slaps MathIsFun**

Everyone knows that Rom is the capital of rom.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

hey krassi-

I'm still waiting to hear what rom is...

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.

-Bertrand Russell

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Oh, a stupid muistake! not "rom"- it's "form". It means that

*Last edited by krassi_holmz (2006-02-11 03:47:00)*

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

for some k∈Z

Now let a + b = y^3 and k = x. Then:

Plugging this back in...:

So....

Which was to be shown.

*Last edited by Ricky (2006-02-11 05:48:32)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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"Now let a + b = y^3 and k = x"-

you cannot let a+b=y^3, because then y may not be integer.l

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Yea... I was hoping you wouldn't catch that.

Do you know the solution?

*Last edited by Ricky (2006-02-11 09:57:33)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I know it but i want you to get it.

IPBLE: Increasing Performance By Lowering Expectations.

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

Well, krassi...I think its safe to say that none of us came up with an answer and I've been really wanting to see the solution...

So can you post it, please?

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.

-Bertrand Russell

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**Ricky****Moderator**- Registered: 2005-12-04
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Give me till tuesday, I think I'm close, but I have a test in Advanced Calc and Modern Algebra that I got to get through first on monday.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

OK. I'll wait for some time.

The proof is not so hard.

IPBLE: Increasing Performance By Lowering Expectations.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Alright, I'm bout ready to give up. It seems like I'm missing some fact that being 0 mod gives, but I can't seem to find it. Being divisible by a+b doesn't seem to be enough.

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

In that case....give us the proof please, krassi!!!!

It's been bugging me since you posted this and I'm having the same problem...I've gone through all my notes to try to find some long lost number theory theorem that would make it all come together, but I can't....and I'm a bit obsessive so I need to see the proof...and I need to see it fast!

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.

-Bertrand Russell

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**damathamatician****Member**- Registered: 2006-02-28
- Posts: 10

what do the buttons do?

*Last edited by damathamatician (2006-02-28 10:58:15)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Let

d=GCD[a,b]

Then

a=di and b=dj and GCD[i,j]=1.

ab≡0(mod a+b) =>

(di)(dj)≡0(mod d(i+j))

dij≡0(mod (i+j)) ->>(i)

But GCD[i,j]=1

GCD[i,i+j]=1

GCD[j,i+j]=1

=>

GCD[ij,i+j]=1

but from (i) (i+j)|dij => (i+j)|d => exists k ∈ N : k(i+j)=d

Let go back to the main equation:

*Last edited by krassi_holmz (2006-02-28 19:43:21)*

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

NT#2:

Prove that for every a,b

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