## anonymous 5 years ago What does the notation p:[0,1]-> Real mean?

1. anonymous

It means that p is a function from the closed interval [0,1] to the real numbers.

2. anonymous

It means that p maps values from 0 to 1 to Real values.

3. anonymous

Brilliant thanks! I'm guessing that means any number between 0 and 1 which isn't necessarily an integer? (sorry if that sounds obvious)

4. anonymous

Yes, any real number in that interval.

5. anonymous

So, uhm, the rest actual question asks you to show that the set $\Pi _{n}=\left\{ p:[0,1]\rightarrow \mathbb{R} | \sum_{n}^{j=0} a _{j}x ^{j}, a _{0},...,a_{n} \in \mathbb{R}\right\}$ equipped with pointwise addition and scalar multiplication is a vector space, by verifying the 8 axioms of vector spaces. Oh and $n \in \mathbb{N}$ So with the associativity of addition axiom, I would need to show: $p(x),q(x),r(x) \in \Pi_n, (p+(q+r))(x) = ((p+q)+r)(x)$ (Correct me if I'm wrong) Would I do this by say that by pointwise addition and because p,q,r(x) in Real, (in which associative addition holds) $(p+(q+r))(x) = p(x)+(q(x)+r(x)) = p(x)+q(x)+r(x) = (p(x)+q(x))+r(x) = ((p+q)+r)(x)$ I can't seem to get my head round this sort of maths. I never know how rigorous the proofs have to be either...

6. anonymous

This displays messily in chrome but shows the whole line, in Firefox the end of the line is missing so I'll post again... and I made a pigs ear of the sentence above it :") Would I do this by saying that, due to pointwise addition, and because p(x),q(x) and r(x) are real numbers (hence addition of p,q,r is associative) $(p+(q+r))(x) = p(x)+(q(x)+r(x)) = p(x)+q(x)+r(x)$$= (p(x)+q(x))+r(x) = ((p+q)+r)(x)$