MHB Showing that P_3 is a subspace to P_n

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To show that P_3 is a subspace of P_n, it is essential to define P_n as the set of all polynomials of degree less than or equal to n. The proof requires demonstrating three conditions: that the zero polynomial is in P_3, that the sum of any two polynomials in P_3 is also in P_3, and that any scalar multiple of a polynomial in P_3 remains in P_3. It is important to note that the assertion does not hold for n=0, 1, or 2. The discussion concludes with the original poster indicating they have resolved their query.
cbarker1
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Dear everyone,

I need to show that the $${P}_{3}$$ is a Subspace to $${p}_{n}$$. how to start the proof?
 
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Could you clarify what $P_3$ and $P_n $ are? Also, what is the condition on $n$?
 
Cbarker1 said:
I need to show that the ${P}_{3}$ is a Subspace to ${p}_{n}$. how to start the proof?

Perhaps you mean the vector space $P_n=\{p\in\mathbb{R}[x]:\deg p\le n\}.$ In such case prove the 3 conditions for $P_3$ to be a subspace :
$\begin{aligned}&(1)\quad0\in P_3.\\
&(2)\quad p,q\in P_3\Rightarrow p+q\in P_3.\\
&(3)\quad \lambda\in\mathbb{R},p\in P_3\Rightarrow\lambda p\in P_3.
\end{aligned}$
 
Euge said:
Could you clarify what $P_3$ and $P_n $ are? Also, what is the condition on $n$?

$${P}_{n}$$ is the set of all polynomials of degree less than or equal to n. So $${P}_{3}$$ is the set of all polynomials of degree less than three.
 
Fernando Revilla said:
Perhaps you mean the vector space $P_n=\{p\in\mathbb{R}[x]:\deg p\le n\}.$ In such case prove the 3 conditions for $P_3$ to be a subspace :
$\begin{aligned}&(1)\quad0\in P_3.\\
&(2)\quad p,q\in P_3\Rightarrow p+q\in P_3.\\
&(3)\quad \lambda\in\mathbb{R},p\in P_3\Rightarrow\lambda p\in P_3.
\end{aligned}$
What should be the first words that I should use in this verification? I am not good with proof-writing.
 
Cbarker1 said:
What should be the first words that I should use in this verification? I am not good with proof-writing.
Do you know what "0\in P_3" means?

(What you want to prove is not true for n= 0, 1, or 2!)
 
I solved it.
 

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