Question

What does the notation
p:[0,1]-> Real
mean?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Yes, any real number in that interval.

Answer 5

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...

Answer 6

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)

\[(p+(q+r))(x) = p(x)+(q(x)+r(x)) = p(x)+q(x)+r(x)\]\[ = (p(x)+q(x))+r(x) = ((p+q)+r)(x)\]