# Can we add a scalar and a vector? $a+b+\mathbf x = \mathbf y$?

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?

• Don't do it. It is illegal. – JCAA Jun 28 at 16:22
• wait..that's illegal – sai-kartik Jun 28 at 16:29
• 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. – TonyK Jun 28 at 16:42
• @Jam: I agree. There is nothing special about the vector $(1,1,1)$. – TonyK Jun 28 at 16:45
• 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. – Jam Jun 28 at 16:53

Use notation that will be easily understood $$(a+b)\left<1,1,1\right>+\bf{x}=\bf{y}$$
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.
• 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. – Jam Jun 28 at 20:23