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I am trying to prove whether a specific form of polynomials form a subspace of $P_3$. I know that any set $W$ forms a subspace of a vector space $V$ if the set $W$ is closed under addition and scalar multiplication (is this right? please clarify).

Based on this, are polynomials of form $a_0 + a_1x + a_2x^2 + a_3x^3$ where $a_0,a_1,a_2,a_3$ are rationals, a subspace of $P_3$ ? I am not sure how any polynomials of this form actually violates the closure properties.

Thanks in advance

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  • $\begingroup$ "closed under addition and scalar multiplication" AND nonempty. $\endgroup$
    – Gerry Myerson
    Oct 16, 2017 at 4:10
  • $\begingroup$ What is "P3"? Is it the space of all real polynomials of degree $\leq 3$? $\endgroup$
    – Hayden
    Oct 16, 2017 at 4:11
  • $\begingroup$ Do you really mean $a_2x^2a_3x^3$, as you have written? or do you mean $a_2x^2+a_3x^3$? $\endgroup$
    – Gerry Myerson
    Oct 16, 2017 at 4:12
  • $\begingroup$ @Hayden Yes. Polynomials of degree <= 3 $\endgroup$
    – Abrar Hossain
    Oct 16, 2017 at 4:13
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    $\begingroup$ @GerryMyerson I have edited the post to reflect a2x^2 + a3x^3 $\endgroup$
    – Abrar Hossain
    Oct 16, 2017 at 4:14

2 Answers 2

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The product of an irrational and a rational is irrational. .. this appears to present a problem, and your subset would appear not to be closed under scalar multiplication. ..

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  • $\begingroup$ I know from real analysis that product of rational and irrational is irrational. So, correct me if I am wrong, then for a scalar k=sqrt(2), the a(sub i) * k is irrational. Then, it must be true that the scalar multiples of k in this case are not in P3. $\endgroup$
    – Abrar Hossain
    Oct 16, 2017 at 4:33
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    $\begingroup$ @AbrarHossain Close. They are in $P_3$ but not in your proposed subspace. .. $\endgroup$
    – user403337
    Oct 16, 2017 at 4:44
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Hint: Are all scalar multiples of $1$ ($= 1+0x+0x^2+0x^3$) in the set $$ \{ a_0+a_1x+a_2x^2+a_3x^3 \mid a_0,a_1,a_2,a_3 \in \mathbb{Q}\} $$ ?

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  • $\begingroup$ According to your hint, the answer should be no. All scalar multiples of 1 of the form (1 + 0x + 0x^2 + 0x^3) are not in the set P3. I know this is the right answer but, I would be glad if you expand on the hint and, derive a contradiction (if possible) by assuming that all multiples of 1 are indeed in the set of P3 for polynomials of the given form. Thanks. $\endgroup$
    – Abrar Hossain
    Oct 16, 2017 at 4:26
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    $\begingroup$ e.g. $\pi \cdot 1 = \pi$ $\endgroup$
    – Hayden
    Oct 16, 2017 at 4:34

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