Number Theory: Proving a and b in R, a*b = 0 → a = 0 or b = 0

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Homework Help Overview

The discussion revolves around proving a property in number theory regarding real numbers, specifically that if the product of two real numbers \(a\) and \(b\) is zero, then at least one of them must be zero. The participants explore various mathematical concepts, including continuous functions, ring homomorphisms, and properties of ideals.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the continuity of functions and the conditions under which a function has an inverse. There is exploration of constructing specific functions that satisfy the conditions of the problem, as well as considerations of polynomial functions versus general continuous functions.

Discussion Status

The discussion is active, with participants questioning assumptions and definitions. Some guidance has been offered regarding the construction of functions, and there is an ongoing exploration of the properties of the functions involved. Multiple interpretations of the problem are being considered.

Contextual Notes

There is mention of the need for clarity on the definitions of functions and their continuity, as well as the implications of using polynomials versus other types of continuous functions. Participants are also navigating the requirements for ring homomorphisms and ideals in the context of the problem.

Firepanda
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For i) I said for a in [0,1], then the group of units are = {f in R | f(a) =/= 0}

i.e a continuous function f on [0,1] would have a continuous function g on [0,1] such that f.g=1

but the function would have to be g = 1/f, but this wouldn't be continuous if f(a) = 0

for ii) I have to show for a,b in R then if ab=0 then either a = 0 or b = 0

but I havn't gotton any further here since I'm having a hard time finding functions a and b where the product is 0

for iii) I believe I have to construct a surjective ring homomorphism ( f: R -> F ) and show that the kernal is equal to the Ideal.

But I'm not sure how what field F to choose for this

Or I have to show that the quotient group R/I is a field.

Thanks for any help!
 
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The statement of (i) doesn't have any 'a' in it. You aren't thinking very abstractly. Yes, the inverse of f would be 1/f. What conditions on f do you need to ensure that exists and is continuous?
 
Dick said:
The statement of (i) doesn't have any 'a' in it. You aren't thinking very abstractly. Yes, the inverse of f would be 1/f. What conditions on f do you need to ensure that exists and is continuous?

the condition that f is a polynomial?
 
Firepanda said:
the condition that f is a polynomial?

Is that just a random guess?
 
Dick said:
Is that just a random guess?

No it wasn't, I looked up a more precise definition of the conditions for continuity and it stated that all polynomials are continuous.

Otherwise I'm not sure what you mean sorry :/
 
Firepanda said:
No it wasn't, I looked up a more precise definition of the conditions for continuity and it stated that all polynomials are continuous.

Otherwise I'm not sure what you mean sorry :/

Ok. Maybe I'm just being overly fussy about your original answer because you didn't specify what 'a' is supposed to be. Is it some point in [0,1], or is it ANY point in [0,1]?
 
Dick said:
Ok. Maybe I'm just being overly fussy about your original answer because you didn't specify what 'a' is supposed to be. Is it some point in [0,1], or is it ANY point in [0,1]?

Any point.
 
Firepanda said:
Any point.

Ok. Be careful when you get to part (iii). There 'a' means 'some point'. So f is a nonzero continuous function on [0,1]. So 1/f exists. If you can say why 1/f is continuous then you've show f is a unit.
 
For part iii) I've said

that I is a proper ideal since the constant function f(x)=1 is not an element of the ideal.

So R/I is isomorphic to the image of the functions, and the Im(f) = the real numbers

Since the real numbers are a field then I is maximal. Is this ok?
 
  • #10
It's not very clear. What exactly is the ring homomorphism from R (the ring) to F (the reals) that I is the kernel of?
 
  • #11
Dick said:
It's not very clear. What exactly is the ring homomorphism from R (the ring) to F (the reals) that I is the kernel of?

phi : R -> R

f -> f(a)=0

Pretty sure that's a ring homomorphism
 
  • #12
"f(a)=0" is a truth-value (well, a truth-value-valued equation) not a real number...
 
  • #13
Hurkyl said:
"f(a)=0" is a truth-value (well, a truth-value-valued equation) not a real number...

hmm so what could I change it to, just 0? Or do I need to change my co-domain?
 
  • #14
Firepanda said:
hmm so what could I change it to, just 0? Or do I need to change my co-domain?

You just aren't expressing yourself very well. What is the real number phi(f)?
 
  • #15
Dick said:
You just aren't expressing yourself very well. What is the real number phi(f)?

I'm having a hard time getting my head around the ring of functions, usually it's something simpler.

Perhaps it's just

phi : R -> R

f -> f(a)
 
  • #16
Yes, it is. Is I the kernel? You said before you were only 'pretty sure' that it was a homomorphism. Can you check that it satisfies the properties of a homomorphism, so you can say you are 'really sure'?
 
  • #17
For part ii) I can only think of the fact that

'Theorem: Let R be an integral domain and let R[x] be the ring of polynomials in powers of x whose coefficients are elements of R. Then R[x] is an integral domain if and only if R is.'

These polynomials R[x] are functions that are most likely continuous on [0,1], so if I can choose an R that isn't an integral domain then by the theorem the continuous functions on [0,1] are not an integral domain?

Just a thoguht as I'm having trouble with this..
 
  • #18
The problem is not about polynomials, it's about continuous functions, which don't have to be polynomials. Surely you can think of two nonzero continuous functions on [0,1] whose product is zero.
 
  • #19
You make it sound trivial but nothings jumping out at me
 
  • #20
Firepanda said:
You make it sound trivial but nothings jumping out at me

How about defining the functions piecewise?
 
  • #21
hmm if they're piecewise then how to make sure they both stay continuous on [0,1] so that the product is zero?

is it still continuous if the endpoints match up?

so f(x) = x for x<0.5

and g(x) = x for x>=0.5

is sufficient?
 
  • #22
You haven't defined what f(x) is for x>=0.5. If you mean f(x)=0 for x>=0.5, then f(x) isn't continuous. Making sure the pieces match up so it's continuous is YOUR job.
 
  • #23
Dick said:
You haven't defined what f(x) is for x>=0.5. If you mean f(x)=0 for x>=0.5, then f(x) isn't continuous. Making sure the pieces match up so it's continuous is YOUR job.

how about for some small epsilon e > 0

f(x) = e for x<0.5 and f(x) = 0 for x>=0.5, would that be continuous?
 
  • #24
Firepanda said:
how about for some small epsilon e > 0

f(x) = e for x<0.5 and f(x) = 0 for x>=0.5, would that be continuous?

What do you think? How small would e have to be to make it continuous? This isn't supposed to be anything horribly exotic. Just pick a nonzero function on [0,.5] such that f(.5)=0.
 
  • #25
Dick said:
What do you think? How small would e have to be to make it continuous? This isn't supposed to be anything horribly exotic. Just pick a nonzero function on [0,.5] such that f(.5)=0.

oh so something like f(x) = -x+0.5 for x<0.5, f(x)=0 for x>=0.5

g(x) = x - 0.5 for x>0.5, g(x) = 0 for x<=0.5
 
  • #26
Firepanda said:
oh so something like f(x) = -x+0.5 for x<0.5, f(x)=0 for x>=0.5

g(x) = x - 0.5 for x>0.5, g(x) = 0 for x<=0.5

Right. That's wasn't so hard, was it?
 
  • #27
Dick said:
Right. That's wasn't so hard, was it?

ye sorry, thanks for the help
 

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