Determine whether the set is a vector space

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Discussion Overview

The discussion revolves around determining whether the set of all fifth-degree polynomials forms a vector space under standard operations. Participants explore the implications of various axioms related to vector spaces and identify specific axioms that fail in this context.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant states that axioms 1, 4, 5, and 6 fail for the set of fifth-degree polynomials, seeking clarification on why this is the case.
  • Another participant questions whether the failure of axioms 1 and 6 implies the failure of axiom 7, expressing difficulty in finding counterexamples for axioms 4, 5, and 6.
  • A third participant explains that axioms 4, 5, and 6 fail because the zero polynomial is not a fifth-degree polynomial, thus there is no zero element to satisfy these axioms.
  • One participant notes that while the set of all polynomials of degree less than or equal to 5 forms a vector space, the requirement for precisely fifth-degree polynomials excludes the zero polynomial.
  • Another participant questions whether the zero vector must be a fifth-degree polynomial, suggesting that the zero vector could simply be 0, and presents an example to illustrate their point.
  • A later reply critiques the imprecision in the formulation of the axioms, emphasizing that the zero element must be a polynomial and cannot simply be the number 0.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the zero vector in relation to the axioms of vector spaces. There is no consensus on the implications of the axioms' failures, and the discussion remains unresolved regarding the interpretation of the zero vector in this context.

Contextual Notes

Participants highlight the importance of precise definitions in the axioms and the implications of including or excluding the zero polynomial from the set of fifth-degree polynomials.

eyehategod
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15. Determine wheter the set is a vector space.
The set of all fifth-degree polynomials with the standard operations.
AXIOMS
1.u+v is in V
2.u+v=v+u
3.u+(v+w)=(u+v)+w
4.u+0=u
5.u+(-u)=0
6. cu is in V
7.c(u+v)=cu+cv
8.(c+d)u=cu+cd
9.c(du)=(cd)u
10.1(u)=u

the back of my book says that axioms 1,4,5, and 6 fail. I don't know why 4,5,and 6 fail. Can anyone help me?
 
Last edited:
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hm..if it fails 1 and 6 shouldn't it also fail 7? o.o...Sorry I can't see a counterexample that 1/4/5/6 fails.

say you have
u=ax^5+bx^4y+cx^3y^2+dx^2y^3+exy^4+fy^5
v=gx^5+hx^4y+ix^3y^2+jx^2y^3+kxy^4+ly^5
where a,b,...,l are arbitrary constants where at least 1 of a,b,...,f and g,h,...,l doesn't equal 0.

does:
f(u+v)=f(u)+f(v)?
f(u+0)=f(u)?
f(u+(-u))=0?
f(cu)=cf(u)?

you can write it them out and I think they do hold, or can you give 1 counter-example that proves it doesn't?
 
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Oh boy is that too complicated. 4,5,6 simply fail because 0 is not a 5th degree polynomial, thus there is no 0 to add to u in 4, no 0 for u+(-u) to equal, and 0*u is not a 5th degree poly
 
Notice that the set of all polynomials with degree less than or equal to 5 is a vector space. But here, you must have polynomials of precisely degree 5. The "0" polynomial is not in that set.
 
to matts response:
im still getting use to this abstract way of thinking and am not sure if this is stupid or not. For axiom 4 to pass, does the zero vector have to be a 5th degree polynomial? can't the zero vector just be 0? b/c let's say (x^5+x)+(0)=(x^5+x). Whats wrong with the way I am approaching this?
 
eyehategod said:
to matts response:
im still getting use to this abstract way of thinking and am not sure if this is stupid or not. For axiom 4 to pass, does the zero vector have to be a 5th degree polynomial? can't the zero vector just be 0? b/c let's say (x^5+x)+(0)=(x^5+x). Whats wrong with the way I am approaching this?

Well, first of all, you wrote the axioms in such an unprecise manner, that some of them represent nothing. For example, axiom 4 sould be something like: "There exists 0 in V, such that for every x from V, 0 + x = x + 0 = x", etc, which means that merely by reading the axioms you can get the answers to your questions.

You can't add a polynomial and a number. Of course 0 has to be a polynomial. And since it is not a polynomial of degree 4 (or whatever you defined the space to be, I don't remember right now), it fails to be in the vector space at all.
 

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