Precalculus and Discrete Mathematics

Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. In contrast to real numbers that have the property of varying "smoothly", the objects studied in discrete mathematics – such as integers, graphs, and statements in logic. Precalculus an advanced form of secondary school algebra, is a foundational mathematical discipline. Precalculus also called Introduction to Analysis. In this article we shall discuss about precalculus and discrete mathematics problem.(Source: wikipedia)


I like to share this Precalculus Problem Solver with you all through my article.

Discrete mathematics problem

Example:

Prove that (C, +) is an infinite abelian group.

Solution:

(i) Closure axiom: Sum of two complex numbers is always a complex number.

i.e., z1, z2 ∈ C ⇒ z1 + z2 ∈ C

Closure axiom is true.

(ii) Associative axiom: Addition is always associative in C

i.e., (z1 + z2) + z3 = z1 + (z2 + z3) ∀ z1, z2, z3 ∈ C

Therefore Associative axiom is true.

(iii) Identity axiom:

The identity element o = o + io ∈ C and o + z = z + o = z ∀ z ∈ C

Therefore Identity axiom is true.

(iv) Inverse axiom: For every z ∈ C there exists a unique − z ∈ C such that

z + (− z) = − z + z = 0. Inverse is true. ∴ (C, +) is a group.

(v) Commutative property:

∀ z1, z2 ∈ C, z1 + z2 = z2 + z1

Therefore the commutative property is true. Hence (C, +) is an abelian group.

Since C is an infinite set (C, +) is an infinite abelian group.

Example 2:

Find the order of each element of the group (Z4, +4)

Solution: Z4 = {[0], [1], [2], [3]} is an abelian group under the addition modulo 4. The identity element is [0] and notes that [4] = [8] = [12] = [0]

Therefore 0([0]) = 1

0([1]) = 4 [we have to add [1] four times to get [4] or [0]]

0 ([2]) = 2 [we have to add [2] two times to get [4] or [0]]

0 ([3]) = 4 therefore we have to add [3] four times to get [12] or [0]

Precalculus example problem

Find the H.C.F. of the polynomials 2x3 + 2x2 + 2x + 2 and 6x3 + 12x2 + 6x + 12

Solution: Let f(x) = 2x3 + 2x2 + 2x + 2 = 2 (x3 + x2 + x + 1) and

g(x) = 6x3 + 12x2 + 6x + 12 = 6(x3 + 2x2 + x + 2)

x3 + x2 + x + 1 )      x3 + 2x2 + x + 2                 (   1
x3 + x2    + x + 1
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x2 + 1                                       ≠ 0
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Dividing x3 + x2 + x + 1 by x2 + 1, we have

x2 + 1  )           x3 + x 2+ x + 1                          (   x + 1

x3 + x
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x2 + 1

x2 + 1
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0

H.C.F. of the polynomials = 2 (x2 + 1) (since the G.C.D. of 2 and 6 is 2).

Algebra is widely used in day to day activities watch out for my forthcoming posts on how to add and simplify fractions and iit jee 2013 syllabus. I am sure they will be helpful.