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I'm given the equation $$ a+b+\mathbf x = \mathbf y $$ With the vectors $\mathbf x=(x_1,x_2,x_3)$, $\mathbf y=(y_1,y_2,y_3)$ and the two scalars $a$, $b$.

Is the following correct? As a vector equation: \begin{align} a+b+\mathbf x &= \mathbf y \tag 1\\ a + b + (x_1,x_2,x_3)&=(y_1,y_2,y_3) \tag 2 \\ (a+b+x_1,a+b+x_2,a+b+x_3)&=(y_1,y_2,y_3) \tag 3 \end{align} And the vector equation as three separate scalar equations: \begin{align} a+b + x_1 = y_1 \tag 4\\ a+b + x_2 = y_2 \tag 5\\ a+b + x_3 = y_3 \tag 6\\ \end{align}

Also, if it is illegal, does $a+b+\mathbf x = \mathbf y$ have any meaning or is it just nonsense?

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    $\begingroup$ Don't do it. It is illegal. $\endgroup$ – JCAA Jun 28 at 16:22
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    $\begingroup$ wait..that's illegal $\endgroup$ – sai-kartik Jun 28 at 16:29
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    $\begingroup$ Who gave you this equation? Adding a scalar and a vector is ill-defined, unless it has previously been explicitly stated what it means. This is definitely non-standard, and without such an explicit clarification, it is meaningless. $\endgroup$ – TonyK Jun 28 at 16:42
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    $\begingroup$ @Jam: I agree. There is nothing special about the vector $(1,1,1)$. $\endgroup$ – TonyK Jun 28 at 16:45
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    $\begingroup$ Regarding whether you can add two different objects in general, consider that addition is a function, i.e., a mapping from one specific domain set to a range set. In particular, for addition of $a+b$, both $a$ and $b$ need to be in the same set (or at least type-converted naturally like $1\mapsto 1+0i$). If they aren't in the same set, then $a+b$ is undefined. $\endgroup$ – Jam Jun 28 at 16:53
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Use notation that will be easily understood $$ (a+b)\left<1,1,1\right>+\bf{x}=\bf{y} $$

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    $\begingroup$ But only if forced at gunpoint. $\endgroup$ – TonyK Jun 28 at 16:43
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Others have already stated that the answer is no, but I want to reinforce why this is in a formal way. The problem here is that you are viewing scalar and vector addition as the same thing, when in fact they are not. Although we use the same symbol, "$+$", for both operations, they are not the same thing because they operate on different sets. Suppose we have a vector space $V$ over a field $K$ - then scalar multiplication is an operation between two scalars that outputs another scalar. Formally speaking, $+:F\times F\to F ~ ; ~ +:(a,b)\mapsto a+b$ given some $a,b \in F.$ Vector addition takes in two vectors and outputs another vector. But we happen to use the same symbol for it because of its similarities to scalar addition. So, $+:V\times V \to V ~ ; ~ +:(\mathbf{u},\mathbf{v})\mapsto \mathbf{u}+\mathbf{v}$ given some $\mathbf{u},\mathbf{v}\in V.$ But trying to do $a+\mathbf{u}$ isn't defined, since we don't have any addition operators that can add elements of different sets like $?:F\times V \to V ~;~ ?:(a,\mathbf{u})\mapsto a+\mathbf{u}$. It's about as sensible as trying to add apples and oranges, so to speak.

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    $\begingroup$ We do have operators that can act on elements of different sets. For instance, we can multiply a vector by a scalar. So there are nuances here that you have ignored. $\endgroup$ – TonyK Jun 28 at 19:44
  • $\begingroup$ Ok, I meant addition operators. I'll add that in. $\endgroup$ – K.defaoite Jun 28 at 20:01
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    $\begingroup$ I mostly like this answer but think "don't have any addition operators that can add elements of different sets" is a big overstatement, especially since the asker has implicitly defined one as $+:F\times V\to V; (a,\mathbf{u})\mapsto (a+u_1,a+u_2,a+u_3)$. Maybe "have in general a unique and natural addition operator" would fit better. $\endgroup$ – Jam Jun 28 at 20:23
  • $\begingroup$ Perhaps my wording is not the best. But I think the general idea is clear enough. $\endgroup$ – K.defaoite Jun 29 at 1:12

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