I need help in the right direction. I am showing youwhat I am doing in blue text

Question

I need help in the right direction. I am showing youwhat I am doing in blue text.

Problem:
W = {p P(Real Universe) : p(xy) = p(x)p(y) for allx,y Real Universe}

Determine whether the above subset of the given vector space is asubspace of that vector space. Give details.

My answer:
I'm checking whether the given subset isclosed under addition.

This is what I'm doing:

let p,q Real Universe
let x,y be elements of the Real Universe

(p+q)(xy) = (p+q)(x) * (p+q)(y)
p(xy) = p(x)p(y)
q(xy) = q(x)q(y)

I believe p(xy) + q(xy) = p(x)p(y) + q(x)q(y)
however my tutor told me that:

p(xy) + q(xy) = [p(x) + q(x)] * [p(y) + q(y)] = p(x)p(y) + p(x)q(y)+ p(y)q(x) + q(x)q(y)

how does that happen? I do not get that at all. Why isn't p(xy) +q(xy) = p(x)p(y) + q(x)q(y)?

Then my tutor told me to plug in arbitrary things into it to showthat it is not a subspace.
Could you go over how I should go about and do that? Do not give methe answer please.

Thank you so much!

Explanation / Answer

Whatever your tutor said is right In the definition , we have p(xy) = p(x)p(y) , where x,yReal universe In this definition , the distribution first takes place on theentries of the real universe and not on the function Therefore, in (p+q)(xy), first we distribute the real entriesx,y, we get [(p+q)(x)][(p+q)(y)] and then we use the distributiveproerty in addition Hence (p+q)(xy) = [(p+q)(x)][(p+q)(y)]                          = [p(x)+q(x)] [p(y)+q(y)]                          = p(x)p(y)+p(x)q(y)+q(x)p(y)+q(x)q(y)

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