Are linear functions always continuous? I'm trying to clarify a definition for middle school mathematics students (grade 8) for "linear function." 
I have always viewed the NCTM (National Council of Teachers of Mathematics) definition of a linear functions as accurate: 
"a relationship whose graph is a straight line." 
A physicist and mathematics teacher is stating this definition is incorrect and says that linear functions can be discrete, as in an arithmetic sequence. 
Because my area of expertise is not higher-level maths, but I want to be accurate, I am asking for confirmation that a linear function is always continuous.
 A: The problem  here is that the NCTM's definition is not the same definition that is used by working mathematicians and physicists.  This does not mean that it is wrong; it is simply different, in deference to the level of the students.  The NCTM's definition is, perhaps, better stated as

Definition 1:  A function $f:\mathbb{R}\to\mathbb{R}$ is said to be linear if its graph is a straight line.

This is actually a perfectly reasonable definition.  Assuming the postulates of Euclidean geometry, two points determine a line.  So suppose that $(x_1,y_1)$ and $(x_2,y_2)$ are any two points on the graph of $f$, and further suppose that $(x,y)$ is any other arbitrary point on that line, as in the figure below (and interactive Desmos demo).

A line parallel to the $x$-axis through $(x_1,y_1)$ and a line parallel to the $y$-axis through $(x_2,y_2)$ determine a right triangle.  Dropping a second line parallel to the $y$-axis through $(x,y)$ determines a second right triangle which is similar to the original triangle.  By properties of similar triangles, we have
\begin{align}
\frac{y_2 - y_1}{x_2 - x_1} = \frac{y - y_1}{x - x_1}
 &\implies y-y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x-x_1) \\
 &\implies y = \underbrace{\left(\frac{y_2 - y_1}{x_2 - x_1}\right)}_{=m} x + \underbrace{\left(y_1 - \left(\frac{y_2 - y_1}{x_2 - x_1}\right)x_1\right)}_{=b},
\end{align}
which gives the "usual" slope-intercept equation for a line.
So this definition is perfectly reasonable from a mathematical point of view, and is relatively simple for middle schoolers to understand.  Pedagogically, it makes a lot of sense.
The difficulty is that linearity is often defined differently in higher mathematics.  Specifically,

Definition 2: A function $f : X \to Y$ between vector spaces $X$ and $Y$ is linear if
$$ f(ax + by) = af(x) + bf(y), $$
for all scalars $a$ and $b$, and all vectors $x$ and $y$.  Such a function may also be called a linear map.

Note that this is different from the middle school definition of a linear function in two very important ways:  first, the definition applies to functions with arbitrary vector spaces for domain and codomain, rather than the real numbers; and second, a linear function must be zero at zero.  A function of the form
$$ f(x) = ax + b $$
is linear if and only if $b=0$ (though if $b\ne 0$, then $f$ is a translation of a linear function, and may be called an affine map).
It is worth noting that such a function is generally continuous,[1] though the intuition which we give to middle schoolers (i.e. a function is continuous if its graph can be drawn without lifting the pen) fails.  For example, it is impossible to draw the graph of a linear function (or any function, for that matter) from $\mathbb{Q}$ to $\mathbb{Q}$ without lifting one's pen.  This middle school definition of continuity doesn't generalize well—one really needs to introduce the notion of a topology (or, at least, epsilons and deltas) to properly describe continuity.  But this is entirely inappropriate for middle school. ;)

[1]  It should be noted that, in more general settings, a linear function needn't actually be continuous.  If $V$ is an infinite dimensional vector space over some field, then it may be possible to find linear functions (in the sense of Definition 2) which are not continuous.  For example, the space of all smooth, real-valued functions on $\mathbb{R}$ is an infinite dimensional vector space over $\mathbb{R}$.  The derivative operator (i.e. the function which takes a smooth function $f$ as input, and outputs $f'$) is linear, but not continuous.
A: ==== Answer 1:  In which I don't actually explain what anything means but I explain why this isn't a problem =====
THe 8th grade definition of linear as "in a straight line" is fine.  But the 8th grade definition of continuous as "can be drawn without lifting your pencil" is not. 
The linear function of $x\mapsto 2x+3; x\in \mathbb Z$ is a set of a bunch of separatee points all in a line.  These points: $(0,3), (1, 5), (2,7), etc.$  all lay on a straight line.
Intuitively the OP is probably assuming that this is not continuous.  it jumps from point $(0,3)$ to $(1,5)$ and doesn't pass through any points in between.  BUT, according to the mathematical definition,  it IS continuous.  At every point where the function exists you can draw that part of function without lifting your pencil.  The stuff in between doesn't count because.. it is in between.
A function has a domain.  And when we analyze if a function is continuous (or if the function is anything) we are only concerned with input values within the domain.  If a value is not in the domain it doesn't matter because it is not input so it just doesn't count.  It might as well not exist.  
If $f: \mathbb Z \to \mathbb R$ is $f(x) = 2x + 3$ it is only defined for $x\in \mathbb Z$.  We are no more concerned about what $f(1.5)$ is, than we are concerned about what $f(\text{babar, the elephant})$ is.  $1.5$ is not an integer so $f(1.5)$ doesn't exist.  It is as non-existant as $f(blue)$ or $f(hungry)$.
So it doesn't matter if we have to "lift our pencil" to get from $(0, 3)$ to $(1,5)$ because all the $x$es between $0$ and $1$ do not exist as far as we are concerned.  The just don't exist.
So... if we were to use the 8th grade definition of "continuous" it would be "we can draw the function without lifting our pencil at any point in our universe".  If however our universe has hyperspace portals where $1$ comes immediately after $0$, then jumping from $0$ to $1$ does not count as "lifting a pencil". 
===== Answer 2: where I explain what a few things mean ====
The two problems with the definition of a linear function meaning "graph is a straight line" are 1) who don't really have a definition of what a "straight line" and 2) this assumes we are measuring everything on geometric plane and not on any other type of "universe".
I think the best definition of "linear function" would be: a function where the difference in outputs is always proportional to the difference of the corresponding inputs.  That is if your inputs are difference of $a$ and your outputs are a difference of $b$ then i) that will be true no matter which points you take for input and ii) if your inputs are difference of $k*a$ then your outputs will be a difference of $k*b$.
For example.  $f(x) =x^2$ is not linear because $f(0)=0$ and $f(1)=0$ have a difference of $1$ but $f(99) = 9801$ and $f(100) = 10000$ have a difference of $199$.  And $f(1)=1$ and $f(2)=4$ have a difference of $3$ but $f(1)$ and $f(3)$ (where the inputs are twice as far apart) does not have a difference of $2*3$ (the difference of the outputs are not twice as far apart).
What this definition means is that $\frac {f(b) - f(a)}{b-a}$ will always be some constant proportional factor we can call $m$ and .....
.... tl;dr.....
A linear function is in the form $f(x) = mx + b$ and this always makes a graph where $m$ is a constant slope (all lines have a constant slope; that's what makes them lines--- and functions that aren't lines will have points where sometimes the graph is steeper or less steep than others).  $b$ is not that important for the point I am trying to make.  It's a $y$-intercept of a vertical offset.
[I'd be remiss if I didn't mention that "linear function" has a diffeerent meaning in linear algebra.  In linear algebra al linear function is one where $f(x+y) = f(x) + f(y)$.  That basically means algebraically that $f(x) =mx $ where $m = f(1)$... I'll let you play with that.  It's the same idea but we are not allowed to have any veritical offsets.]
And that's all I really want to say about linear functions.
The graph is going to depend on the domain.  If the domain is all of $\mathbb R$ the graph will look like a solid straight line.  If the domain is $\mathbb Z$ the graph is going to look like a dotted straight line.  And if the domain is $\mathbb Q$ the graph is going to look like a misty straight line.  (Every part of it will have holes and points infinitely close together.)
Okay, so what about "continuous"?
The lifting the pencil off you paper definition doesn't cut it, because your paper is the domain.  If your domain has rips then your paper has rips and lifting you pencil off the paper to get for one part of your domain to the other to get over the rips is perfectly fair.  It's just that on your paper you can't have any breaks.  Breaks where your paper breaks is acceptable.
So continuous means:  well, if you hone in closely on the inputs then you will also be honing in on the outputs.
So a linear function $f(x)= 2x + 3$ is continuous.  If you hone in really tight on the $x$, so you hone in on $x=1$ by looking at $x=0.99$ and $x=1.01$ and you look at the outputs of $4.98$ and $5.02$ we see that we have honed in closely to $f(x) = 5$.
Now if our domain is ... spotty ... like $\mathbb Z$.  If we try to hone in that tight.... well there is only one point to consider.  That is $x = 1$.  We don't consider $x=0$ or $x=2$ because that's not tight enough to matter.  There is only one point that is tight enough and that is $x=1$ exactly.  And we have $f(1) = 5$ exactly.  So by our definition of "continuous".  $f(x) = 2x + 3$ wwith $x \in \mathbb Z$ is still continuous.
[Actually the definition  for "continuous at $a$" is for any $\epsilon > 0$ (for example $\epsilon = 0.02$ but $\epsilon$ can be as small as we like) we can find a $\delta > 0$ (for example $\delta = 0.01$ but how small the $\delta$ is will depend on the size of $\epsilon$, where it is true that if $a- \delta < x < a+\delta$ then $f(a) - \epsilon < f(x) < f(a) + \delta$.  (so for example:  If $1-0.01 < x < 1 + 0.01$ then $5-0.02 < 2x + 3 < 5+0.02$.)]
[This is true if our domain is all reals.  But it is also true if our domain is $\mathbb Z$.  If $1- 0.01 < x < 1+0.01$ and $x \in \mathbb Z$ that means that $x = 1$ *exactly.  Which means $f(x) = f(1) = 5$ exactly.  So it is certainly true that $5-0.02 < f(x) < 5+0.02$.]
A: In the strict sense, a function is linear in a vector space when the following properties hold:
$$f(u+v)=f(u)+f(v),\\f(\lambda u)=\lambda f(u).$$
Hence its expression is of the form
$$f(u)=au$$ (notice that there is no independent term, the graph is a line through the origin). If the scalars are taken from $\mathbb R$, the function may not be discrete (because $\lambda u$ need not be integer).
In a wider sense, the expression can be
$$f(u)=au+b,$$ and this is technically an affine function (general line).
One also speaks of linear growth (corresponding to affine), and this both applies to functions in a discrete or continuous domain.
A: Yes; a linear function ($f(x) = ax + b$, where $a,b$ are real and $a \neq 0$) is a polynomial and all polynomials are continuous over $\mathbb{R}$.
