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Discrete Math Problems

There are basically two types of data used in mathematics - discrete and continuous. The branch of mathematics that deals with discrete variables is known as discrete math. It is the study of objects which are not continuous. In other words, in contrast with continuous math which deals with smooth and continuous variables, discrete math is all about non-smooth and distinct points. The points or objects studied in discrete mathematics are not connected to each other.

The main areas of study of continuous maths are calculus and analysis etc. The concepts of discrete mathematics are used in computer programming and software development also. Discrete math has been proved very useful in higher and college-level mathematics. Though, it is studied in high school and middle school level too.

The fields of study of discrete mathematics are:

  • Number Theory
  • Set Theory
  • Functions
  • Vectors
  • Matrices
  • Boolean Algebra
  • Probability
  • Sequences
  • Discrete calculus
  • Discretization of continuous math
  • Topology
  • Game Theory
  • Graph Theory
  • Linear programming
  • Operational Research (OR), etc.

Let us have a look at simple discrete math problems.

Sequence Problem: Let us consider a sequence $<a_{n} >$ = 1, 4, 9, 16, 25, ..... Determine its $n^{th}$ term.
Solution: $<a_{n} >$ = 1, 4, 9, 16, 25, .....
$a_{1}= 1=1^{2}$
$a_{2}= 4=2^{2}$
$a_{3}= 9=3^{2}$
$a_{4}= 16=4^{2}$
$a_{5}= 25=5^{2}$
....... and so on
Hence $a_{n}=n^{2}$

Set Theory Problem: Two sets are given such that -
A = $\left \{ x:x\leq 5,x\in N \right \}$ and
B = $\left \{ x:x\leq 11,\ x\ is\ an\ odd\ number,\ x\in N \right \}$
Find $A\cup B$ and $A\cap B$.
Solution: A = $\left \{ x:x\leq 5,x\in N \right \}$
Therefore, A = {1, 2, 3, 4, 5}
B = $\left \{ x:x\leq 11,\ x\ is\ an\ odd\ number,\ x\in N \right \}$
Therefore, B = {1, 3, 5, 7, 9, 11}
$A\cup B$ = {1, 2, 3, 4, 5, 7, 9, 11}
$A\cap B$ = {1, 3, 5}

Matrix Problem: A matrix M is given such that
M=$\begin{bmatrix}
1 &2 \\
-2 &3
\end{bmatrix}$
Determine $M^{2}$.
Solution: $M^{2}=M*M$
$M^{2}$=$\begin{bmatrix}
1 &2 \\
-2 &3
\end{bmatrix}*\begin{bmatrix}
1 &2 \\
-2 &3
\end{bmatrix}$
= $\begin{bmatrix}
1*1+2*(-2) \ & 1*2+2*3\\
 (-2)*1+3*(-2)\ &(-2)*2+3*3
\end{bmatrix}$
= $\begin{bmatrix}
1-4 \ & 2+6\\
-2-6\ &-4+9
\end{bmatrix}$
= $\begin{bmatrix}
-3 \ & 8\\
-8\ &-5
\end{bmatrix}$

Boolean Algebra Problem: Draw the truth table of $!(A\wedge B)$.
Solution: Truth table of $!(A\wedge B)$ is given below:

A
B
$(A\wedge B)$ $!(A\wedge B)$
T
T
T
F
T
F F
T
F T
F
T
F
F
F
T

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