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Determine if the given set is a subspace of $\mathbb{P}_{n}$ for an appropriate value of $n .$ Justify your answers.All polynomials of degree at most $3,$ with integers as coefficients.

Therefore, the set of "All polynomials of degree at most 3 with integers as coefficients"is not a subspace of $P_{n}$ .

Calculus 3

Chapter 4

Vector Spaces

Section 1

Vector Spaces and Subspaces

Vectors

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Hello. For this question, we need to determine if this set which is the set of all, Paul, No meals that are third degree and have into your coefficients. That's what this means is a which I've denoted V just for convenience. Um, is it a subspace of PN, which is the set of all polynomial of degree and where N is greater than or equal to three? This is a pretty standard question in, uh, when your algebra I have some well known set that I know is a vector space, and then I have some subset of it that I'm interested in. For whatever reason, maybe this is ah numeric six question or something on. I want to figure out if that is also a vector space. So I'm going to check the three sub space conditions to see if it is. And then depending on that, it's easier or nice sort of work with than normal. So there are three conditions that the set V would have to meet to be considered a vector space and in of itself, given that is the subset of a vector space. The first is that the zero vector has to be in there. The second is that if I have two vectors within this set that immediately gives me that the addition of those two vectors is within the set and that if I have a new, arbitrary constant, actually, let me call that Alfa within the field, um, than an any vector within my set in this also immediately implies that whoops that the scaler multiplication is within the subset is well, so for the first condition, that is immediately true, because I can just pick C zero c one C two and C three all equal to 00 is an integer, um and, uh, this is immediately done. Zero function is in the set. The second condition is also true. I'll give myself a little check mark for this because addition of polynomial of the same degree boils down to addition of the coefficients and Z is a is a field. So if I add, um, if I feel with respect to edition So if I add two elements of this, uh, set together, I will end up back in the set. Um, So if I have C 11 and see 12 for example, representing the coefficients If you wanna be too. I add them together. I will again get an imager coefficient. But it is not the case that this third condition is met. And it's important on the real stickler here. It's important to realize Is this so when we're talking about checking, subspace is a vector. Spaces, we have to use the same field. We can't cheat and swap fields. This would be true. Um, if my field waas uh, integers Because if I multiply two images together, I get an energy back. But the field in this case and the definition of PN is ah, actually the more proper way to write that if I get Eyes PM, which is the set of objects and the field, which is the rial numbers and because this field is the real numbers, I can pick Alfa equal toe one half, for example, or if you if you want to get even even fancier Ah, Alfa equal to one over square root of two in this constant multiplied by any integer is no longer an integer Uh, eso if I take any element of V and I multiplied by this constant, I will not end up back and, v um, for the specific choice. So V is not a vector space. Uh, it's not a It's not a vector space, because it is not a subspace. It does not meet the three conditions. So the set of all polynomial is with integer coefficients are not, um, a subspace of PM and we're done.

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