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A one dimensional vector is only with one coordinate such as [1] , and a two dimensional vector is one with two coordinates such as [1,2]. It is possible to represent them in a two dimension, the former in the number line and the latter in a two dimensional space. Yet, why can't they be added? Why can't the second component of the former(one dimensional vector) be taken as zero?
(My personal opinion is that the addition means that the corresponding elements are added, and then when there is no corresponding component, it is equal to 0. So it can be added?)

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  • $\begingroup$ Vector addition is just a definition. You can define whatever you want, as along as it satisfies the vector space axioms. $\endgroup$ – user251257 Feb 8 '18 at 15:21
  • $\begingroup$ Addition is an operation $+ : X \times X \to X$ that takes two elements $a$ and $b$ of some vector space (for example $\mathbb{R}$ or $\mathbb{R}^2$) and gives you another element of the same vector space, denoted by $a+b$. To define an addition as you want, you would have to make some strange choice, like ignoring the second component (but why not the first?). So it can be done, but it would not be an interesting operation. $\endgroup$ – 57Jimmy Feb 8 '18 at 15:22
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Well, firstly the property of closure is valuable when dealing with operations. On some level we want to know that when we define operations on some set that we don't pop out of the universe we're considering somehow. Say I'm looking at moves on a Rubix Cube. Certainly I can allow weird moves like an 1/8th turn on some face which would disallow certain other moves until another 1/8th turn on the same face occurs. While we can in fact make such moves they aren't terrible valuable in understanding the structure of the cube or how to solve it so it's typical to disallow them as being "outside" the universe of meaningful moves we consider. We want to know that any move can be followed by any other move to get a "normal" sort of position on the cube.

This is the same for vector spaces. Now of course we can define the operation as you've described and it can work but there is no value in doing so. We can maintain the property of closure simply by insisting we explicitly include the $0$ component rather than assuming it's there tacitly. Nothing is lost by taking this approach but we do gain the additional valuable property of closure, ensuring the vectors stay in the same vector space when we add them or take scalar multiples. For example in a programming setting if you get a vector returned to you as a value from some function and you are expecting a vector with two components and ask the vector for the value of it's second component it will either throw an error or give you zero, and in the second case it still has two components and in the first you have to write special code to deal with a special case. Organizing your ideas this way can cause unexpected bugs so it makes sense to value the closure property in this setting.

I would also caution you from talking about the dimension of a vector! Dimension is a property of a vector space and has nothing to do with the number of components used to represent a vector. The space spanned by $[2,1]$ is still one dimensional even though it is being embedded in $\mathbb{R}^2$. A vector in and of itself tells you nothing about the dimension of the underlying space.

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Usually, an operation such as addition is defined for elements within the same set; i.e. for elements of the same kind. For vectors with two components, the classical addition is defined as: $$(a,b)+(c,d) = (a+c,b+d)$$ which comes down to, as you say, adding the corresponding components.

With this definition, there's no way to add $x$ and $(a,b)$; that's undefined.

You could define that yourself or alternatively, you could identify every one-dimensional element $x$ with the two-dimensional vector $(x,0)$; notice that this identification $x \leftrightarrow(x,0)$ is a one-to-one correspondence. You then get what you seem to want, without having to define an addition between different objects.

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If you have a 1-dimensional vector $(1)$, then you are free to assume that this is the same as $(1,0)$. But someone else might think different and will associate it with $(0,1)$ or even $(1,17)$. The last one might seem strange to you, but there is not the right way to place a 1-dimensional number line in the 2-dimensional plane, but infinitely many equally justified ways.

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You can write $(1)+(1,2)$ and interprete this as $(1,0)+(1,2)=(2,2)$, but you have to mention it because it is not consensus and noone knows that you are doing it that way.

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