Relations and Functions

Algebra of Real Functions

Assume a case where Seema, whose monthly income is Rs. 15,000, spends Rs. 10,000. What will be her saving? Simple! Rs. 5,000. Saving = Income – Expenditure. Here, we see that the input and the output are the real numbers. We can say that real number input gives a real number output. Here, we will learn Real-valued functions and algebra of real functions. The above case is a representation of real mathematical functions and a case of subtraction in the algebra of real functions.

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Real-valued Mathematical Functions

In mathematics, a real-valued function is a function whose values are real numbers. It is a function that maps a real number to each member of its domain. Also, we can say that a real-valued function is a function whose outputs are real numbers i.e., f: R(stands for Real).

mathematical functions

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Algebra of Real Functions

In this section, we will get to know about addition, subtraction, multiplication, and division of real mathematical functions with another.

Addition of Two Real Functions

Let f and g be two real valued functions such that f: X→and g: X→where X ⊂ R. The addition of these two functions (f + g) : X→R  is defined by:

(f + g) (x) = f(x) + g(x), for all x ∈ X.

Subtraction of One Real Function from the Other

Let f: X→and g: X→R be two real functions where X ⊂ R. The subtraction of these two functions (f – g): X→R  is defined by:

(f – g) (x) = f(x) – g(x), for all x ∈ X.

Multiplication by a Scalar

Let f: X→be a real-valued function and γ be any scalar (real number). Then the product of a real function by a scalar γf: X→is given by:

(γf) (x) = γ f(x), for all x ∈ X.

Multiplication of Two Real Functions

The product of two real functions say, f and g such that f: X→R and g: X→R, is given by

(fg) (x) = f(x) g(x), for all x ∈ X.

Division of Two Real Functions

Let f and g be two real-valued functions such that f: X→and g: X→where X ⊂ R. The quotient of these two functions (f  ⁄ g): X→R  is defined by:

(f / g) (x) = f(x) / g(x), for all x ∈ X.

Note: It is also called pointwise multiplication.

Solved Example for You

Question 1: Let f(x) = xand g(x) = 3x + 1 and a scalar, γ= 6. Find

  1. (f + g) (x)
  2. (f – g) (x)
  3. (γf) (x)
  4. (γg) (x)
  5. (fg) (x)
  6. (f / g) (x)

Answer : We have,

  1. (f + g) (x) = f(x) + g(x) = x+ 3x + 1.
  2. (f – g) (x) = f(x) – g(x) = x– (3x + 1) = x– 3x – 1.
  3. (γf) (x) = γ f(x) = 6x
  4. (γg) (x) = γ g(x) = 6 (3x + 1) = 18x + 6.
  5. (fg) (x) = f(x) g(x) = x(3x +1) = 3x4 + x3.
  6. (f / g) (x) = f(x) / g(x) = x/ (3x +1), provided x ≠ – 1/3.

Question 2: What is meant by functions in algebra?

Answer: A function refers to an equation that consists of only one answer for y for every x. A function assigns only one output to each input that is associated with a specified type. It is common that a function is named as g(x) or f(x) but not y.

Question 3: Explain what makes a function a function?

Answer: A relation from a set X to a set Y is known as a function in case each element of X has a relation to exactly one element in Y. For example, consider an element x in X, so there shall be only one element in Y that x can have a relation to.

Question 4: Is it possible for an equation to be a function?

Answer: An equation shall be considered a function only when for every x’s value there is only one corresponding value for y.

Question 5: When will a function be well defined?

Answer: A function will be well defined when it provides the same result when a change takes place in the representation of the input without the change taking place in the value of the input.

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Al quba
Al quba

The example for onto function doesn’t qualify as a function in the first place. Does it??
It is a relation but not a function because a single element in the domain has been mapped to two elements in the co domain. Isn’t it??
Please tell me if I’m correct or not.
It is really confusing.

Eli Simeon
Eli Simeon

An onto function exists if and only the co-domain is equal to the range that is every element in set A (the domain) is mapped to every element in set B (the range/codomain) i.e without leaving out any element. Irrespective of whether it is a one to one mapping or not. Therefore it is a function. Put simply, take set A as a set of sons and set B as a set of fathers, a function requires that every son has one father (which is normal) yet every father can have more than one son(which is also normal) so if… Read more »

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