NUMBER SYSTEM

NUMBER SYSTEM

 THE DECIMAL NUMBER SYSTEM:

The Decimal number system is a number system of base that are equal to 10, this implies that we have 10 basic counting numbers called Arabic numerals, symbols used to represent number : 0, 1, 2, 3,…….,9 , which are used for counting.  Then others are just like a combination of two more basic numerals.

To represent more than nine units, we must either develop additional symbols or use those we have in combination. When used in combination, the value of the symbol depends on its position in the position in the combination of symbols (i.e its place in the basic numerals). We refer to this as positional notation and are categorized into tens, hundreds, thousands, and so on while the basics are called the units.

The units symbol occupies the first position to the left of the decimal point is represented as 10⁰. The second position is represented as 10¹, and so forth. To determine what the actual number is in each position, take the number that appears in the position, and multiply it by 10^X, where x is the power representation. This is expressed mathematically of the first nine positions as.

10⁰    10¹    10²     10³    10³              10⁴   

units  tens    hundreds     thousands    ten thousands

am sure you got what and where we are going now let’s play around with some examples.

EXAMPLE 1: find the combination of this symbols 89

Solution

8×10 ^{1 +}  9×10⁰

8×10+ 9 ×1= 80+9 =89

EXAMPLE 2: find the combination of this symbols 975

Solution

 9× 10² + 7 ×10¹ + 5×10⁰

9×100 +7×10 +5×1 = 900+70+5 =975

Same goes for all other higher combination. Now let’s consider the negative powers of the decimals.

 The positions to the right of the decimal point carry a positional notation and corresponding weight as well. The exponents to the right of the decimal point are negative and increase in integer steps starting with-1. This is expressed mathematically for each of the first four positions as;

10^-1              10^-2              10^-3              10^-4

Tenth          hundredth   thousandth  ten thousandth

Let’s also play around with some few examples.

EXAMPLE 1: find the combination of this symbols 45.34

Solution

Now 4×10¹ + 5×10⁰ + 3×10^-1 +4×10^-2

4×10 + 5×1 +  +    =  40+5+0.3+0.4 = 45.35 and the same procedure goes for all other combinations.

The Binary Number System:

The binary number system is a number system of base equal to 2, which means that there are two symbols used to represent numbers : 0 and 1. Since there are only two symbols, we can represent two numbers , 0 and 1, with individual symbols. The position of the 1 or 0 in a binary number system indicates its weight or value within the number. We then combine the 1 with 0 and with itself to obtain additional numbers.

TABLE OF BINARY

	Decimal Number		Binary NO
	0		0000
	1		0001
	2		0010
	3		0011
	4		0100
	5		0101
	6		0110
	7		0111
	8		1000
	9		1001
	10		1010

Binary-to-Decimal Conversion:

Since we are programmed to count in the decimal number system, it is only natural that we think in terms of the decimal equivalent value when we see a binary number. The conversion process is straight forward and is done as follows: Multiply binary digit (1 or 0) in each position by the weight of the position and add the results. The following examples explain the process.

Example 1: Convert the following binary number to their decimal equivalent. (a) 1101 (b) 1001

Solution:

(a)  1101 = (1 ×2³) + (1 ×2²) + (0 ×2¹) + (1 ×2^o) = 8 + 4 + 0 + 1 = 13

(b) 1001 = (1 ×2³) + (0 ×2²) + (0× 2¹)(1 ×2⁰)

= 8 + 0 + 0 + 1 = 9

Example 2: Convert the following binary numbers to their decimal equivalent. (a) 0.011 (b) 0.111

Solution:

(a)  0.011 = (0 × 2^-1) + (1 × 2^-2) + (1 ×2^-3)

= 0 +  +

= 0.25 + 0.125 = 0.375

(b) 0.111 = (1 × 2^-1) + (1× 2^-2) + (1 ×2^-3)

=  +  +  

= 0.5 + 0.25 + 0.125 = 0.875

You can continue in the same vein If you encounter any problem just fall back to us.

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FURTHER PROBLEMS ON STANDARD FORMS

FURTHER PROBLEMS ON STANDARD FORMS

PROBLEM. FIND THE VALUE OF:

(a) 7.9×10⁻² −5.4×10⁻²

(b) 8.3×10³ +5.415×10³ and

(c) 9.293×10² +1.3×10³

expressing the answers in standard form. Numbers having the same exponent can be added or subtracted by adding or subtracting the mantissa and keeping the exponent the same. Thus:

(a) 7.9×10⁻² −5.4×10⁻²

=(7.9−5.4)×10⁻² =2.5×10⁻²

(b) 8.3×10³ +5.415×10³

=(8.3+5.415)×10³ =13.715×10³

=1.3715×10⁴ in standard form

(c) Since only numbers having the same exponents can be added by straight addition of the mantissa, the Numbers are converted to this form before adding.

Thus:

9.293 × 10² + 1.3 × 10³

= 9.293 × 10² + 13 × 10²

= (9.293 + 13) × 10²

= 22.293 × 10² = 2.2293×10³ in standard form.

Alternatively, the numbers can be expressed as decimal fractions, giving:

9.293 × 102 + 1.3 × 103

= 929.3 + 1300 = 2229.3

= 2.2293×10³  in standard form as obtained previously. This method is often the ‘safest ‘way of doing this type of problem.

PROBLEM. Evaluate

(a) (3.75×10³)(6×10⁴)

and (b) 3.5×10⁵ /7×10²

expressing answers in standard form

(a) (3.75×10³)(6× 10⁴)=(3.75 × 6)(10³⁺⁴)

=22.50×10⁷

=2.25×10⁸

(b)

3.5×10⁵/7×10²

= 3.5/7×10⁵⁻²

=0.5×10³ =5×10²

NOW PRACTICES THIS

In Problems 1 to 4, find values of the expressions given, stating the answers in standard form

1. (a) 3.7×10² +9.81×10²

(b) 1.431×10⁻¹ +7.3×10⁻¹

[(a) 1.351×10³ (b) 8.731×10⁻¹]

2. (a) 4.831×10² +1.24×10³

(b) 3.24×10⁻³ −1.11×10⁻⁴

[(a) 1.7231×10³ (b) 3.129×10⁻³]

3. (a) (4.5×10⁻²)(3×10³)

(b) 2×(5.5×10⁴)

[(a) 1.35×10² (b) 1.1×10⁵]

4. (a)6 × 10⁻³/3 × 10⁻⁵

 (b)(2.4 × 10³)(3 × 10⁻²)(4.8 × 10⁴)

[(a) 2×10² (b) 1.5×10⁻³]

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NUMERIC PROCESS 2 (STANDARD FORMS)

NUMERIC PROCESS 2 (STANDARD FORMS)

A number written with one digit to the left of the decimal point and multiplied by 10 raised to some power is said to be written in Standard Form. Thus: 5837 is written as 5.837×10^3 in standard form, and 0.0415 is written as 4.15×10⁻² in standard form. When a number is written in standard form, the first factor is called the Mantissa and the second factor is called the EXPONENT. Thus the number 5.8 × 10³ has a mantissa of 5.8 and an exponent of 10³.

(i) Numbers having the same exponent can be added or subtracted in standard form by adding or subtracting the mantissa and keeping the exponent

the same.

 Thus:

2.3 × 10⁴ + 3.7 × 10⁴

= (2.3 + 3.7) × 10⁴ = 6.0 × 10⁴

and 5.9 × 10⁻² − 4.6 × 10⁻²

= (5.9 − 4.6) × 10⁻² = 1.3 × 10⁻²

When the numbers have different exponents, one way of adding or subtracting the numbers is to express one of the numbers in non-standard form, so that both numbers have the same exponent.

Thus:

2.3 × 10⁴ + 3.7 × 10³

= 2.3 × 10⁴ + 0.37 × 10⁴

= (2.3 + 0.37) × 104 = 2.67 × 104

Alternatively,

2.3 × 10⁴ + 3.7 × 10³

= 23 000 + 3700 = 26 700

= 2.67 × 10⁴

(ii) The laws of indices are used when multiplying or dividing numbers given in standard form. For example,

(2.5 × 10³) × (5 × 10²)

= (2.5 × 10⁵) × (10³⁺²)

= 12.5 × 10⁵ or 1.25 × 10⁶

Similarly,

6 × 10⁴

1.5 × 10²

=

 6

1.5× (10⁴⁻²) = 4 × 10²

2.5 Worked problems on standard

Problem  

Express in standard form:

(a) 38.71 (b) 3746 (c) 0.0124

For a number to be in standard form, it is expressed with only one digit to the left of the decimal point. Thus:

(a) 38.71 must be divided by 10 to achieve one digit to the left of the decimal point and it must also be multiplied by 10 to maintain the equality, i.e.

38.71 =

38.71    ×10 = 3.871×10 in standard form

  10       

(b) 3746=

3746

1000    ×1000=3.746×10³ in standard form

(c) 0.0124=0.0124× 100

100

1.24

100     =1.24×10⁻² in standard form

Problem 15. Express the following numbers, which are in standard form, as decimal numbers:

(a) 1.725×10−2 (b) 5.491×104 (c) 9.84×100

(a) 1.725×10⁻² =

1.725

100         =0.01725

(b) 5.491×10⁴ =5.491×10 000= 54 910

(c) 9.84×10⁰ =9.84×1=9.84 (since 10⁰ =1)

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