MHB Proving Ideal Property of f(x)=0 for Every Rational x in $\mathcal{F}(\mathcal{R})$

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The discussion centers on proving that specific sets are ideals of the ring of real-valued functions, $\mathcal{F}(\mathcal{R})$. The first set includes functions that equal zero for every rational input, while the second set consists of functions that equal zero at zero. A key point raised is the confusion over the multiplicative operation in the ring, questioning whether it is standard multiplication or function composition. It is clarified that the usual operation is pointwise multiplication, which supports the validity of the proposed ideals. Participants are encouraged to proceed with the proof using pointwise multiplication and report any further issues.
Kiwi1
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I am asked:

Prove that each of the following is an ideal of $\mathcal{F}(\mathcal{R})$:
a. The set of all f such that f(x)=0 for every rational x
b. The set of all f such that f(0)=0

My question is how do I know what the multiplicative operation is within the ring? Is multiplication the standard multiplication on real numbers or is it composition of functions?

I would expect it to be composition of functions but then if I choose g(x)=1 then I get g(f(x))=g(0) for any real x and this is not generally zero. So it must be the multiplication on reals.

Am I just supposed to know that?
 
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Hi Kiwi,

I don't know where this exercises come from, but it must be specified what the product is.

It is usual to consider $\mathcal{F}(\mathbb{R})$ the ring of real valued functions with pointwise product, and with this product your statements are true (they aren't with composition).

Try to prove it with pointwise multiplication and let us know if you encounter any other problem. ;)
 
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