The degree of a polynomial is the highest degree of its terms when the polynomial is expressed in its canonical form consisting of a linear combination of monomials.
The degree of a term is the sum of the exponents of the variables that appear in it.
(i)The degree of the polynomial 5x3 + 4x2 + 7x is 3.
(ii) The degree of the polynomial 4 – y2 is 2
(iii) The degree of the polynomial 5t – √7 is 1
(iv) The degree of the polynomial 3 = 3x0 is 0.
A real number is a value that represents a quantity along a number line which can be positive numbers, negative numbers, fractions or zero. Real numbers are represented by R.
Real numbers can be categorized into two classes as:
Real numbers can be thought of as points on an infinitely long line called the numbers line or real line, where the points corresponding to integers are equally spaced. Any real number can be determined by a possibly infinite decimal representation, where each consecutive digit is measured in units one tenth the size of the previous one.
Rational Numbers and Irrational Numbers together forms real numbers.
A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers.
A number ‘r’ is called a rational number, if it can be written in the form p/q ,where p and q are integers and q ≠ 0. The collection of rational numbers is denoted by Q.
Every whole number is a rational number, because any whole number can be written as a fraction. For example, 10 can be written as 10/1, 101 can be written as 101/1, and 9876 can be written as 9876/1.
The set of rational numbers includes all positive numbers, negative numbers and zero that can be written as a ratio (fraction) of one number over another.
Whole numbers, integers, fractions, terminating decimals and repeating decimals are all rational numbers.
If a number ends with any of the digits 0, 2, 4, 6, or 8, you immediately say the number is divisible by 2. Why? We write any such number a as a = 10k + r, where r is the remainder when divided by 10. Hence r is one of the numbers 0, 2, 4, 6, 8. We know see that 10 is divisible by 2 and r is also divisible by 2. We conclude that 2 divides a.
Here are few more divisible tests with example and statements:
Consider the numbers 2, 23, 234, 2345, 23456, 234567. We observe that among these 6 numbers, only 234 and 234567 are divisible by 3. Here, we cannot think of the number formed by the last 2 digits or for those matter even three digits. Note that 3 divides 234, but it does not divide 34. Similarly, 3 divides 456 but it does not divide 23456.
STATEMENT: An integer a is divisible by 3 if and only if the sum of digits of a is divisible by 3. An integer b is divisible by 9 if and only if the sum of digits of b is divisible by 9.
Example 1: Check whether the number 12345321 is divisible by 3. Is it divisible by 9?
Solution:
The sum of digits is 1+2+3+4+5+3+2+1 = 21. Hence the number is divisible by 3, but not by 9. In fact 12345321 = (9 x 1371702) + 3.
Example 2: Is 444445 divisible by 3?
Solution:
The sum of digits is 25, which is not divisible by 3. Hence 444445 is not divisible by 3. Here the remainder is 1.
If a number is divisible by 4, it has to be divisible by 2. Hence the digit in the units place must be one of 0, 2, 4, 6, and 8. But look at the following numbers: 10, 22, 34, 46, and 58. We see that last digit in each of these numbers is as required, yet none of them is divisible by 4. Thus, we can conclude that it is not possible to decide the divisibility on just reading the last digit. Perhaps, the last two digits may help.
If a number has two digits, we may decide the divisibility by actually dividing it by 4. All we need is to remember the multiplication table for 4. Suppose the given number is large, say it has more than 2 digits. Consider the numbers, for example, 112 and 122. We see that 112 is divisible by 4. But 112 = 100 + 22; here 100 is divisible by 4 but 22 is not. Hence 122 is not divisible by 4.
We invoke the following fundamental principle on divisibility:
STATEMENT: If a and b are integers which are divisible by an integer m ≠ 0, then m divides a+b, a-b and ab.
Now, let us see how does this help us to decide the divisibility of a large number by 4. Suppose we have number a with more than 2 digits. Divide this number by 100 to get a quotient q and remainder r; a = 100q + r, where 0 ≤ r < 100. Since 4 divides 100, you will immediately see that a is divisible by a 4 if and only if r is divisible by 4. But r is the number formed by the last two digits of a. Thus we may arrive at the following test:
STATEMENT : A number (having more than 2 digit ) is divisible by 4 if and only if the 2 digit number formed by the last two digits of a is divisible by 4.
Example 1: Check whether 12456 is divisible by 4.
Solution:
Here, the number formed by the last two digits is 56. This is divisible by 4 and hence so is 12456.
Example 5: Is the number 12345678 divisible by 4?
Solution:
The number formed by the last 2 digits is 78, which is not divisible by 4. Hence the given number is not divisible by 4.
STATEMENT:
Given number n in decimal form, put alternatively – and + signs between the digits and compute them sum. The number is divisible by 11 if and only if this sum is divisible by 11. Thus a number is divisible by 11 and only if the difference between the sum of the digits in odd places and the sum of digits in even places is divisible by 11.
Example 1: Is the number 23456 divisible by 11?
Solution:
Observe that 2-3+4-5+6 = 4 and hence not divisible by 11. The test indicates that 23456 is not divisible by 11.
A palindrome is a number which leads the same from left to right or right to left. Thus a palindrome is a number n such that by reversing the digits of n, you get back n. For example, 232 is a 3 digit palindrome; 5445 is a 4 digit palindrome.
Example 2: Find all 3 digit palindromes which are divisible by 11.
Solution:
A 3 digit palindrome must be of the form aba, where a≠0 and b are digits. This divisible by 11 if and only if a-b+a = 2a-b is divisible by 11.
This is possible only if 2a-b = 0, 11 or -11. Since a≤1 and b≤9, we see that 2a-b ≥ 2(1)-(9) = 2-9 = -7 > -11. Hence, 2a-b = -11 is not possible. Suppose 2a-b = 0. Then 2a = b. Thus, a = 1,b = 2 ; a = 2, b = 4; a=3, b = 6; and a = 4, b = 8 are possible.
We get the numbers, 121, 242, 363, 484.
For a = 6, b = 1, we see that 2a-b = 12-1 = 11 and hence divisible by for which 2a-b is divisible by 11. We get four more numbers 616, 737, 858 and 979.
Thus required numbers are 121, 242, 363, 484, 616, 737, 858 979.
Example 3: Prove that 12456 is divisible by 36 without actually dividing it.
Solution:
First notice that 36 = 4 x 9. So it is enough if we prove 12456 is divisible by 4 and 9 both. The last 2 digits of the given number 12456 is 56 which is divisible by 4 and it is left to show that, the given number 12456 is divisible by 9. Let us find the sum of digits of the given number 12456, i.e., 1+2+4+5+6 = 18, which is divisible by 9. Thus, the number 12456 is divisible by 36.
