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$1,2,3\cdots$

$0,1,2,3 \cdots$

$-3,-2,-1,0,1,2,3 \cdots$

Rational numbers can be expressed as $\dfrac{a}{b}$ where $a$ and $b$ are integers and $b \ne 0$

Examples: $\dfrac{11}{2}$, $\dfrac{4}{2}$, $0$, $\dfrac{-8}{11}$ etc.

All integers, fractions and terminating or recurring decimals are rational numbers.

Any number which is not a rational number is an irrational number. In other words, an irrational number is a number which cannot be expressed as $\dfrac{a}{b}$ where a and b are integers.

For instance, numbers whose decimals do not terminate and do not repeat cannot be written as a fraction and hence they are irrational numbers.

Example: $\pi,\sqrt{2},(3+\sqrt{5}),4\sqrt{3},\sqrt[3]{6}$ etc.

Please note that the value of $\pi=3.14159 26535$ $89793 23846 26433$ $83279 50288 41971$ $69399 37510 58209 74944$ $59230 78164 06286 20899$ $86280 34825 34211 70679...$

We cannot $\pi$ as a simple fraction (The fraction $22/7=3.14....$ is just an approximate value of $\pi$)

Real numbers include counting numbers, whole numbers, integers, rational numbers and irrational numbers.

Let $a$ be any rational number and $n$ be any positive integer such that $\sqrt[n]{a}$ is irrational. Then $\sqrt[n]{a}$ is a surd.

Example: $\sqrt{3} $, $\sqrt[6]{10} $, $4 \sqrt{3}$ etc

Please note that numbers like $\sqrt{9}$, $\sqrt[3]{27}$ etc are not surds because they are not irrational numbers

Every surd is an irrational number. But every irrational number is not a surd. (eg : $\pi$ , $e$ etc are not surds though they are irrational numbers.)

- Sum of any number of even numbers is always even.
- Sum of even number of odd numbers is always even.
- Sum of odd number of odd numbers is always odd.

- Difference of two even numbers is always even.
- Difference of two odd numbers is always even.

- Product of even numbers is always even.
- Product of odd numbers is always odd.
- If there is at least one even number multiplied by any number of odd numbers, the product is always even.

Sanjeet

2015-10-20 01:52:59

Show whether the bigger number is a multiple of the smaller number

5250, 25

5250, 25

Chirantan

2015-10-25 09:53:33

first 5250 is divisible by 5.

on dividing we get 1050.

in order to show that 5250 is div. by 25, we have to show that 5250 is divisible by two 5's one by one

i.e. 1050 is divisible by 5 and it is because the last digit is 0.

therefore 5250 is divisible by 25.

on dividing we get 1050.

in order to show that 5250 is div. by 25, we have to show that 5250 is divisible by two 5's one by one

i.e. 1050 is divisible by 5 and it is because the last digit is 0.

therefore 5250 is divisible by 25.

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Anar

2015-10-06 12:01:08

It is very good.This pre-algebra information very interesting and important for the students.

Anar

Anar

mamatha

2015-09-15 18:54:50

It is very very good explanation. But i did not understand the divisibility rule 11 example 4. There we need to find sum of odd numbered digits and sum of even numbered digits. Then, why there is 9+4 and 5+8. I hope u'll clear this. Thank u so much

Javed Khan

2015-09-16 11:10:50

For 9548,

digits in odd positions (i.e., first and third digits from left) are 9 and 4, sum = 9+4=13

digits in even positions (i.e., second and fourth digits from left) are 5 and 8, sum = 5+8=13

Actually we can count from any direction. for example, if you count from right,

digits in odd positions (i.e., first and third digits from right) are 8 and 5, sum = 8+5=13

digits in even positions (i.e., second and fourth digits from right) are 4 and 9, sum = 4+9=13

This is true for all numbers. please revert for any clarification

digits in odd positions (i.e., first and third digits from left) are 9 and 4, sum = 9+4=13

digits in even positions (i.e., second and fourth digits from left) are 5 and 8, sum = 5+8=13

Actually we can count from any direction. for example, if you count from right,

digits in odd positions (i.e., first and third digits from right) are 8 and 5, sum = 8+5=13

digits in even positions (i.e., second and fourth digits from right) are 4 and 9, sum = 4+9=13

This is true for all numbers. please revert for any clarification

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pooja

2015-09-07 06:23:34

It is very good explanation ,equation, important note and additional short note

not only for student but also for teacher, really I would like to thank you

not only for student but also for teacher, really I would like to thank you

vin

2015-04-04 06:06:28

We don't need any text book if we read this formulas and practice perfectly.

Jay

2015-03-02 18:20:32

600 = 2*2*2*2*3*3 (prime factorization)

Take any combination of numbers from 2,2,2,2,3,3 and their product will be a divisor of 144

Among the numbers 2,2,2,2,3,3 , there are four identical 2s and two identical 3s

Total number of combinations possible = 5*3-1 = 14

*(from the formula - Number of ways of selecting one or more than one objects out of S1 alike objects of one kind, S2 alike objects of the second kind ,S3 alike objects of the third kind and so on ... Sn alike objects of the nth kind is (S1 + 1) (S2 + 1)(S3 + 1)...(Sn + 1) - 1)*

Required number of divisors is 14.

But we have to take 1 also as a factor. So required number of divisors will be 15

Take any combination of numbers from 2,2,2,2,3,3 and their product will be a divisor of 144

Among the numbers 2,2,2,2,3,3 , there are four identical 2s and two identical 3s

Total number of combinations possible = 5*3-1 = 14

But we have to take 1 also as a factor. So required number of divisors will be 15

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