Notes to a video lecture on http://www.unizor.com

Linear Functions

There are two approaches to linearity. One, more abstract, and another, traditionally characterized as a function of the first degree on an argument. We will briefly mention the former, but go into more details for the latter.
Abstract concept of linearity implies that a function y = f(x) can be called linear if the following equation is satisfied:
f(a·u+b·v) = a·f(u) + b·f(v).
Example of a such a function is a real function y = 2·x.
Indeed,
2·(a·u+b·v) = a·(2·u) + b·(2·v).
Here and now we will be talking about more "down to earth" concept of linear functions. Linear function is usually considered on a domain of all real numbers and taking values (having range) in a co-domain of the same set of all real numbers. In some cases it can be considered as a function of a complex argument with a complex range, but it's rare. We will only consider real linear functions.
The characteristic property of linear functions is their expression as an equality of a form
y = A·x + B
where x is an argument, y is the value of a function for this argument, A and B are given constants and A is not equal to 0.
Let's mention that in case B = 0 so defined linear function satisfies the abstract definition of linearity since, applying distributive and commutative laws,
A·(a·u+b·v) = a·(A·u) + b·(A·v).

The first observation we'd like to make is that this function can be applied to any real number. Therefore, the domain of this function is a set of ALL real numbers.
How about its range? Does it cover all real numbers? The answer is yes. Here is a proof.
For instance, we would like to know if some number R belongs to a range. Let's take a number x = (R − B)/A and substitute it into our linear function y = A·x + B as an argument. The result is:
y = A·(R − B)/A + B =
= (R − B) + B = R
Therefore, if an argument equals to
(R − B)/A then the value of the function is R. This is true for any real number R, which confirms that any real number can be a value of a linear function with some argument. So, the range of a linear function is a set of ALL real numbers.

Our second point is that a linear function defines a one-one correspondence between a set of all real numbers as arguments to this function and the same set of all real numbers as its range. Obviously, any real number as an argument is converted by a linear function into unique real number as a value of this function. But, to establish a one-to-one correspondence, this is not enough. We have to prove that different arguments are transformed into different values. Let's prove it.
Assume that two different arguments, u and v are transformed by a linear function y = A·x + B into the same value:
A·u + B = A·v + B
If from two equal values we subtract the same value B, the result should be the same:
(A·u + B) − B = (A·v + B) − B, which can be simplified using associativity of addition to
A·u + (B − B) = A·v + (B − B) or, performing subtraction in the parenthesis,
A·u + 0 = A·v + 0 and, considering that addition of 0 does not change the value,
A·u = A·v
If two equal values are divided by the same value A (not equal to 0), the result should be the same:
(A·u)/A = (A·v)/A, which can be simplified using commutative and associated laws of multiplication and division to
u·(A/A) = v·(A/A), which, using the fact that A/A = 1 and multiplication by 1 does not change the value, can be further simplified to
u = v.
We have assumed in the beginning that u and v are different arguments to a linear function that are transformed into the same value by this function, but came up at the end that u = v. This contradiction proves that different arguments are transformed into different values by a linear function. Combined with the fact that all real numbers represent the range of a linear function (our first observation), this concludes that any linear function defines a one-to-one correspondents of a set of real numbers to itself.

Our next point is related to a monotonic character of a linear function. More precisely, if a linear function y = A·x + B has positive coefficient A, it monotonically increases its value y as its argument x increases. If a coefficient A is negative, it monotonically decreases its value y as its argument x increases. Don't forget that coefficient A can be only positive or negative, not 0 (by definition of a linear function).
Let's prove this monotonic character of a linear function.
First, consider the case when A is positive. Start with two values of an argument, u and v and assume that u < v.
Smaller number, if multiplied by a positive number A, will result in a smaller number. Therefore, A·u < A·v.
Smaller number, if added to any number B, will result in a smaller number. Therefore, A·u + B < A·v + B.
This proves that for a positive coefficient A a linear function is monotonically increases as its argument increases.
Consider now a case of negative coefficient A.
Again, assume we have two values of argument, u and v such that u < v.
Smaller number, if multiplied by a negative number A, will result in a larger number. Therefore, A·u > A·v.
Larger number, if added to any number B, will result in a larger number. Therefore, A·u + B > A·v + B.
This proves that for a negative coefficient A a linear function is monotonically decreases as its argument increases.

Why do we avoid coefficient A being equal to 0? The reason is very simple.
The function y = A·x + B becomes, actually, y = B if A = 0. It means, for any argument x the value of a function is a constant B. Though, it does not contradict the definition of a function, this function is not considered as "linear". It does not establish one-to-one correspondence of a set of all real numbers to itself. It's not monotonic.
In addition, there is a well defined class of polynomial functions expressed as polynomials of its argument in a form
y = a₀xⁿ+a₁x^n-1+...+a_n-1x¹+a_n.
This class is classified by the highest degree the argument should be raised to, in this case it's n, the coefficient of which a₀ should not be 0. Thus linear functions are polynomial function of the first degree. In this classification function that takes a constant value can be considered as a polynomial function of the zero-th degree, quite different from the linear.