My mind has gone fallow, and I can't quite understand factoring

[roger]

I don't understand what the difference is between bijection and surjection?
I see that f:R->R f(x)=x^2 is epimorphic but I'm not sure about bijection.

Thanks in advance.
Roger

 

[matt grime]

A bijection is a surjection that is also an injection (i.e. look at the definition). Is squaring injective (as a function from R to R)? Obviously not. But then it is not surjective either (since the square of a real number is a positive real number), so you are mistaken in claiming that f above is a surjection (and you should avoid using epimorphism, since it is unnecessarily complicated at this stage).

 

[JonF]

You should look up and go over the definition of bijection and surjection. Or state what you think they mean in your own words. If you really understood them the difference between the two should be evident.

 

[roger]

iff a,b<p where p=prime, a^(p-2)+a^(p-3)b+a^(p-4)b^2...b^(p-2)=0modP?

I know that (a+b)^p-(a^p+b^p)=0modp but I don't know if that's relevant.


Thanks in advance.
Roger

 

[matt grime]

Have you considered checking a couple of counter examples, with say, p=2 or p=3, or b=0?

 

[roger]

Have you considered checking a couple of counter examples, with say, p=2 or p=3, or b=0?

Why would p=2 or 3 be counterexamples?

 

[roger]

it works iff a=/=b and a,b=/=0

 

[matt grime]

Sorry, but this is rubbbish: you have not even clearly stated a proposition. You cannot begin a sentence with if and only if. It makes no sense. And the best guess as to what you are conjecturing is trivially refuted with p=2 or p=3 without thinking very hard. When p=3 it reduces to a+b. So you're honestly attempting to claim that a+b is always 0 mod 3? Nonsense. Just think about it for more than one second and stop wasting people's time.

 

[roger]

When p=3 it reduces to a+b. So you're honestly attempting to claim that a+b is always 0 mod 3? Nonsense. Just think about it for more than one second and stop wasting people's time.

When p=3, given that a,b<p, a=/=b, and a,b=/=0 then a+b=0mod3.
Is that incorrect?

 

[matt grime]

a=2 b=-1? That any good for you? State your question clearly. Are a and b supposed to be integers? residues mod p, what? Since you've used inequalities it can't be a statement about residues since they are not ordered (in any nice way).

 

[roger]

a and b must be positive integers. could you try to explain the proof of it?

 

[roger]

can anybody help me pleae?

 

[matt grime]

It's a geometric progression question, roger. You're supposed to recognise how to simplify

x^n+yx^{n-1}+\ldots y^n

by thinking of geometric progressions. If you don't see why it is a geometric progression, try dividing by something.

 

[roger]

cheers matt.

But why must a and b be <p for it to work?

 

[roger]

a^3+b^5+c^7=d^11 iff a,b,c and d are natural numbers.

Is there a way to simplify this in order to find a,b,c and d?
and is there any significance in the powers being prime?

 

[matt grime]

cheers matt.

But why must a and b be <p for it to work?

What makes you think they must be? Try finding some counter examples and try to spot a pattern. I already gave you many counter examples, and they all had the same theme to them.

 

[matt grime]

a^3+b^5+c^7=d^11 iff a,b,c and d are natural numbers.

This makes no sense, again. You need to write things in sentences, and ones that make sense at that. We keep telling you this.

Are you asking to find the integer solutions to this?

 

[roger]

This makes no sense, again. You need to write things in sentences, and ones that make sense at that. We keep telling you this.

Are you asking to find the integer solutions to this?

I'm asking firstly whether it can be simplified to find a,b,c, and d such that the equality is true, and secondly whether there is any significance in the powers being all prime?

But the 'counter examples you gave were based on any value of a and b whereas afterwards I stated that they must be positive integers.

 

[HallsofIvy]

a^3+b^5+c^7=d^11 iff a,b,c and d are natural numbers.

Is there a way to simplify this in order to find a,b,c and d?
and is there any significance in the powers being prime?

Your first statement is "a^3+ b^5+ c^7= d^11 iff a, b, c, and d are natural numbers". Do you understand that that asserts that a^3+ b^6+ c^7= d^11 for all natural numbers a, b, c, d? It certainly is NOT true since you can take a= b= c= d= 1 and get a false statement.

What you want to say is "Find natural numbers a, b, c, and d such that a^3+ b^5+ c^7= d^11 (or prove that no such numbers exist)".

 

[matt grime]

But the 'counter examples you gave were based on any value of a and b whereas afterwards I stated that they must be positive integers.

I give up. You're on your own.

 

[roger]

can someone else help me please?
The counter examples were all false since a and b must be positive integers.

 

[roger]

What you want to say is "Find natural numbers a, b, c, and d such that a^3+ b^5+ c^7= d^11 (or prove that no such numbers exist)".

I understand, so can it be simplified? and is there any significance in the powers being all prime?

 

[roger]

p=2 or p=3 aren't counter examples since when p=2, it reduces to 1+1 so 2=0mod2 and when p=3, 2+1=0mod3.

 

[matt grime]

can someone else help me please?
The counter examples were all false since a and b must be positive integers.

Seems I can't help myself. The point is that the counter examples all broke your initially unstated rules that 0<a<b<p. You asked why must a,b<p for this to work (which is a bad sentence by the way - use proper sentences). The answer is that they need not be, but that this condition, with the positivity condition will force it to work. Besides, you're just reducing everything mod p and so you only care about a,b mod p anyway.

The assertion will still be true for some values of a,b,p breaking that condition, but the triples where it fails all have something in common. Just look at them.

It will work for, say, a=-1,b=1 p=3. It won't work for a=0=b,p anything. It didn't work for a=-1,b=2,p=3, but it will for a=-1,b=2,p=5. Final hint: what is 2-(-1)?

 

[roger]

What if the conditions are that a and b must be positive integers in addition to being<p?

 

[matt grime]

wtf? I thought you'd proved it for 0<a<b<p.

 

[roger]

I have, but I just wanted to know why a and b had to be less than p. I couldn't see why it need to be.

 

[matt grime]

For hopefully the last time THEY DO NOT HAVE TO BE IN THE RANGE 0<a<b<p, but if they are not in that range they may or may not satisfy the condition. This is at least the 3rd time I've told you this.

It's a simple standard observation: A implies B, does not in mean that not(A) implies not(B).

 

[roger]

So my question is why is it that outside the range, they do not always work, but if they're inside the range, it will always work?

 

[matt grime]

But I've explained this to you as well. Look at the 'counter examples' outside the range. E.g. post 54. And don't think about posting another question on this until you've gone away and thought about why

a=-1,b=2,p=3 (and the fact that 2 - (-1)=3) is important. Look at your proof. Don't you divide by something at some point? Something that might be zero outside the specified range but isn't inside the specified range?