Preparation For Arithmetic

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The Arithmetic preparation includes addition, subtraction, multiplication and division problems. The properties of integers are explained with the help of arithmetic operations. The arithmetic problems include real numbers, integers, complex numbers, rational numbers etc. The arithmetic preparation helps to solve basic algebraic expression problems, linear algebra problems etc. In mathematics arithmetic functions plays a important role for solving fundamental problems. The following are some of the example problems deals with preparation for arithmetic.

Example problems for arithmetic preparation:

Addition problems:

Question 1: Solve 5x + 4y +2x +8y

Solution: The given problem is

= 5x + 4y +2x +8y

= 5x + 2x + 4y + 8y

= 7x + 12y

Question 2: Solve 40 + 5 + 8 + 3

Solution: The given problem is
= 45 + 8 + 3
= 53 + 3
= 56

Question 3: Solve 4x + 2x + 3x

Solution: The given problem is
= 6x + 3x

= 9x
= 9x
Subtraction problems:

Question 1: Solve 40 - 5 - 8 – 2

Solution: The given problem is
...
... = 40 - 5 - 8 – 2
= 35 -8 - 2
= 27 – 2

= 25

Question 2: Solve 7x - 4y -2x -5y

Solution: The given problem is

= 7x - 4y -2x - 5y

= 7x - 2x - 4y - 5y

= 5x - 9y

Question 3: Solve 15x - 8x - 3x

Solution: The given problem is
=15x - 8x -3x

= 7x - 3x

= 4x
Multiplication problems:

Question1:

Multiply 4( 2x + 4x) + 4( 5y)

Solution:

Given problem is

= 4(2x + 4x) + 4(5y)

= 4(2x) + 4(4x) + 4(5y)

= 8x + 16x + 20y

= 24x + 20y

Question 2:

Multiply 380 by 101

Solution:

380 × 101 means 101 times 380

Or 100 times 380 + one time 380

Or 38000 + 380

Or 38380

Therefore 380 × 101 = 38380
Division problems:

Question 1: solve (2/4) + (2/8)

Solution:
= (2/4) + (2/8)
= (2+2) / 8
= 4/8

= 1/2

Question 2: solve (3/10) – (5/20)

Solution:
= (6/20) – (5/20)
= (6 – 5) / 20
= 1/20

Question 3: Simplify, 2x + 20+ (– x)
= 2x – x + 20
= x + 20

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Practice problems for arithmetic preparation:

Practice problems are given below for test preparation,

1) Solve the math expression

32 + (2 * 3) – 205

Answer: - 190

2) Solve the math expression

(–3)2 + 202 – 10 – 1

Answer: 200

3) Solve the math expression

3x + 2y – 20 + 4y – 2x

Answer: x + 6y – 2

Airthmetic

Everyone knows that set of integers can be broken up into the following two classes:

the even numbers (..., 6, 4, 2, 0, 2, 4, 6,...); and
the odd numbers (..., 5, 3, 1, 1, 3, 5,...).
There are certain generalizations we can make about the arithmetic of numbers based on which of these two classes they come from. For example, we know that the sum of two even numbers is even. The sum of an even number and an odd number is odd. The sum of two odd numbers is even. The product of two even numbers is even, etc.

Modular arithmetic lets us state these results quite precisely, and it also provides a convenient language for similar but slightly more complex statements. In the above example, our modulus is the number 2. The modulus can be thought of as the number of classes that we have broken the integers up into. It is also the difference between any two "consecutive" numbers in a given class.

Now we represent each of our two classes by a single symbol. We let the symbol "0" mean "the class of all even numbers" and the symbol "1" mean "the class of all odd numbers". There is no great reason why we have chosen the symbols 0 and 1; we could have chosen 2 and 1, or 32 and 177, but 0 and 1 are the conventional choices.

The statement "the sum of two even numbers is even" can be expressed by the following:

0 + 0 = 0 mod 2.
Here, the "=" symbol is not equality but congruence, and the "mod 2" just signifies that our modulus is 2. The above statement is read "Zero plus zero is congruent to zero, modulo two." The statement "the sum of an even number and an odd number is odd" is represented by

0 + 1 = 1 mod 2.
Those examples are natural enough. But how do we write "the sum of two odd numbers is even"? It is the (at first strange looking) expression

1 + 1 = 0 mod 2.
Here the symbols "=" and "mod 2" are suddenly very important! We have analogous statements for multiplication:

0 × 0 = 0 mod 2,
0 × 1 = 0 mod 2,
1 × 1 = 1 mod 2.
In a sense, we have created a number system with addition and multiplication but in which the only numbers that exist are 0 and 1. You may ask what use this has. Well, our number system is the system of integers modulo 2, and because of the previous six properties, any arithmetic done in the integers translates to arithmetic done in the integers modulo 2. This means that if we take any equality involving addition and multiplication of integers, say

12 × 43 + 65 × 78 = 5586,
then reducing each integer modulo 2 (i.e. replacing each integer by its class "representative" 0 or 1), then we will obtain a valid congruence. The above example reduces to

0 × 1 + 1 × 0 = 0 mod 2,
or 0 + 0 = 0 mod 2.

More useful applications of reduction modulo 2 are found in solving equations. Suppose we want to know which integers might solve the equation

3a 3 = 12.
Of course, we could solve for a, but if we didn't need to know what a is exactly and only cared about, say, whether it was even or odd, we could do the following. Reducing modulo 2 gives the congruence

1a + 1 = 0 mod 2,
or

a = 1 = 1 mod 2,
so any integer a satisfying the equation 3a 3 = 12 must be odd.

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