Real Numbers Class 10 Maths Notes

Examples

Use Euclid's algorithm to find the 65 and 117.

Solution :
Step:1 Since 117 > 65 we apply the division lemma to 117 and 65 to get ,
$117 = 65 \times 1 + 52$
Step:2 Since 52 > 0 , we apply the division lemma to 65 and 52 to get
$65 = 52 \times 1 + 13$
Step:3 Since 13 > 0 , we apply the division lemma to 52 and 13 to get
$52 = 13 \times 4 + 0$
The remainder has now become zero, so our procedure stops. Since the divisor at this Step is 13, the HCF of 117 and 52 is 13.

Prime Number help
PRIME NUMBERS to 100 ={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97}

Rules to find Prime Number
Rule Number 1:
Since even numbers are always divisible by 2, no even number (other than 2) can be a prime number because they'll always have more than two factors. For example, the factor pairs for 6 are: 1,2,3,6 So 6 has four factors and therefore is not a prime number.

Rule Number 2:
Numbers that end in 5 or 0 are always divisible by 5. So no numbers divisible by 5 (other than 5) can be prime numbers because they'll always have more than two factors. They'll have 1 and themselves, and they'll also have 5 and some other factor.

Rule Number 3:
0 and 1 are not prime numbers. It may seem at first glance that they are, but remember that a prime number has only 1 and itself as factors. But 0 has an infinite number of factors because 0 * 1 = 0, and 0 * 2 = 0, and 0 * 3 = 0, and so on. So 0 is far from prime.

In the case of 1, it doesn't have two distinct factors because it's only factor is itself: 1 * 1 = 1. If you change one of those factors to any other integer, you no longer get a product of 1. Since it has only one factor, it's not prime.

Rule Number 4:
Add all of the digits composing the number. If the sum of the digits is divisible by 3, then so is the original number. Example: If the number is 111, add 1 + 1 + 1 = 3. The sum of the digits (3) is divisible by 3, so 111 is also divisible by 3.

Rule Number 5:
Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule repeatedly as necessary. Example: 826. Twice 6 is 12. So, take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also

Rule Number 6:
If you are checking if a number is a prime number,then you don't need to check all the prime number below it. you need to check run for small primes(2,3,5,7) and you're not done factoring, then keep trying bigger and bigger primes (11, 13, 17, 19, 23, etc.) until you find something that works or until you reach primes whose squares are bigger than what you're dividing into. if you don't find then it is a prime Number Reason: If your prime doesn't divide in, then the only potential divisors are bigger primes. Since the square of your prime is bigger than the number, then a bigger prime must have as its remainder a smaller number than your prime. The only smaller number left, since all the smaller primes have been eliminated, is 1. So the number left must be prime, and you're done.
Lets take it with an Example
The number given is 1017
We can observer that 1031 lies between square of 30 and 33. So we can just keep trying till 33
a. 1031 is not divisible by 2
b. 1031 is not divisible by 3
c. 1031 is not divisible by 5
d. 1031 is not divisible by 7
e. 1031 is not divisible by 11
f. 1031 is not divisible by 13
g. 1031 is not divisible by 17
h. 1031 is not divisible by 19
i. 1031 is not divisible by 23
h. 1031 is not divisible by 29
j. 1031 is not divisible by 31
The next prime is 37. So we are done factoring. We can say 1031 is a Prime Number
List of Prime Number till 1000
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101
103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197
199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311
313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431
433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557
569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673
677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811
821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941
947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039

Examples
4,6,8,9,10.....

Since 4 is divided by 2,6 is divided by 2 and 3

1.Prove that $\sqrt {2}$ is irrational number
Solution: Since we cannot clearly express in p/q form,it is difficult to say,So lets us assume this is rational number
then
$\sqrt {2}=\frac {p}{q}$

where p and q are co primes.
or
$q\sqrt {2}=p$
squaring both sides
$2q^2=p^2$

So 2 divides p²,from theorem we know that,
2 will divide p also. p=2c

$2q^2=4c^2$
or
$q^2=2c^2$
So q divided by 2 also

So both p and q are divided by 2 which is contradiction from we assumed
So $\sqrt {2}$ is irrational number
2. Prove that $\sqrt {3}$ is irrational number
Solution: Since we cannot clearly express in p/q form,it is difficult to say,So lets us assume this is rational number
then
$\sqrt {3}=\frac {p}{q}$

where p and q are co primes.
or
$q\sqrt {3}=p$
squaring both sides
$3q^2=p^2$

So 3 divides p²,from theorem we know that,
2 will divide p also. p=3c

$3q^2=9c^2$
or
$q^2=3c^2$
So q divided by 3 also

So both p and q are divided by 3 which is contradiction from we assumed
So $\sqrt {3}$ is irrational number