### Common Errors in Secondary Mathematics

Common Errors Committed  by the  Students  in Secondary Mathematics   Errors  that students often make in doing secondary mathematics  during their practice and during the examinations  and their remedial measures are well explained here stp by step.  Some Common Errors in Mathematics

# Maths Formulas Class X Ch -1 | Real Numbers

## NUMBER SYSTEM

Common topics for the students of IX and X standard
1. NATURAL NUMBERS :- Counting numbers are called natural numbers.
Eg:-  N =  {1, 2, 3, 4, …………}
2. WHOLE NUMBERS : Natural numbers along with zero are called whole numbers.   Eg:-  W = {0,1, 2, 3, 4,…….}  [N ⊂ W]
3. INTEGERS – All whole number and natural numbers with negative sign are called integers.  Eg :-  Z = {….-3, -2, -1, 0, 1, 2, 3…..}  [N ⊂ W ⊂ Z]
4. RATIONAL NUMBERS – The number which can be put in the form of P/q   (Where q ≠ 0) are called rational numbers.
Eg :-  Q = -5, 0, 7, 3/5, -7/8 etc.  [N ⊂ W ⊂ Z ⊂ Q]

5. FRACTIONS – Positive rational number are called fractions.
(OR)
Fraction is the ratio of two natural numbers. Eg:-  7/5, 1/2,  5/7 ….etc.

6. IRRATIONAL NUMBERS – The number which cannot be put in the form of  p/q are called irrational numbers.
For Example :   all are irrational numbers
7. REAL NUMBERS – All rational  and irrational numbers are called real numbers.
R =   S
8. PRIME NUMBERS:- The numbers which has only two factors one and itself are called prime numbers.  Eg:-   2, 3, 5, 7, 11, 13
"2" is the smallest prime number.
"2" is the only even prime number.
There are 10 prime number in a counting from 1 to 30
There are 25 prime numbers in a counting from 1 to 100

9. CO-PRIMES :- If H.C.F of two  numbers is 1 then numbers are called   co-prime numbers.       Eg:- (5, 7), (13, 27), (15, 16)…….
Note: HCF of any two consecutive Natural Numbers is always 1. So any two consecutive natural numbers are always a pair of coprime.

10. TWIN PRIMES :- Consecutive prime numbers which are differ by 2 are called twin primes.     Eg :-  (11, 13),  (17, 21), (29, 31)…….etc.

11. EVEN NUMBERS:- Natural numbers which are divisible by two are called even numbers. eg  2, 4, 6, 8, 10, ..........

12. ODD NUMBERS:- Natural numbers which are not divisible by 2 are called odd numbers.   Eg:- 1, 3, 5, 7, 9, ........

13. COMPOSITE NUMBERS:- The numbers which has more than two factors are called composite numbers. Eg:- 4, 6, 9, 10, ....
"4" is the smallest composite number.
"9" is the smallest odd composite number

Note:-  "1" is neither a prime number nor a composite number. It is a unit.

14. IMAGINARY NUMBER or Non-Real Number   :- Negative square root of a natural number is called Imaginary number Or  non-real numbers.
For Example  are irrational numbers.

## DIFFERENT TYPES OF DECIMALS

Common topics for the students of IX and X standard

Decimals are of three types

a. Terminating Decimal :-
If prime factors of denominator of a rational number is either the power of 2 or 5 or both (2m x 5n )then the decimal form is known as terminating decimal.
Example: 2.5, 3.6,  7.895

b. Non-Terminating But Repeating Decimal:- If prime factors of denominator of a rational number contain the factors other than power of 2 or 5 or both then the decimal form is called non terminating but repeating decimal.
Eg :-   3.6767….            14.367367…....

c. Non-Terminating Non-Repeating Decimal :–
Decimal expansion of the numbers like
is called non terminating no-repeating decimals
Eg:-  1.1010010001......,  2.5050050005......,  3.6060060006.......

12. RATIONAL DECIMALS :-
Terminating or non- terminating but repeating decimals are  called rational decimals.

13. IRRATIONAL DECIMALS:-
Non-terminating, Non-Repeating decimals are called Irrational decimals

SPECIFIC TOPICS FOR  X  STANDARD ONLY

EUCLID DIVISION LEMMA :(Deleted)
For given positive integers a and b there exist  unique integer q and r such that
a = bq + r ,     ................    where 0 ≤ r <b.
LEMMA :(Deleted)
It is the proven statement used for proving another statement.
ALGORITHM: (Deleted)
A series of well defined steps which gives a procedure for solving a type of problem.
FUNDAMENTAL THEOREM OF ARITHMETIC –
Every composite number can be expressed as a product of primes, and this factorization is unique irrespective of their order.
HCF – Product of common factors with smallest power.
LCM – Product of all factors with greatest power.
When two numbers are given then
HCF x LCM   =  Product of two numbers
This rule is not valid for more than two numbers.

***************************************************

## Q. 5 Check whether 6n can ends with digit 0 for any natural number n.

Solution: If 6n ends with 0 then its prime factorization must contain the power of 2 and 5. Prime factors of 6 are 2 x 3 and by the fundamental theorem of arithmetic this factorization is unique. This shows that prime factors of 6n does not contain 5.
Hence 6n never ends with zero for any value of n

## Question 1 Prove that √5 is an irrational number

Solution:

Let √5 is a rational number

Then there exist co-prime a and b such that

√5 = a/b

Cross multiply here we get

√5b = a

Now squaring on both side we get

5b2 = a2

b2 = a2/5  ……………..(1)

5 divides a2    5 divides a ………(2)

Let a = 5c ,  Squaring on both sides, we get

a2 = 25c2 ………………(3)

Putting  eqn.(3) in equation (1) we get

b= 25c2/5 ,

b2 = 5c2

b2/5  =  c2

5 divides b2  5 divides b…..(4)

From (2) and (4)

5 divides a and b both

a  and b are not co-prime

But it is given in the starting that a and b are co-prime numbers.

So there is a contradiction and our assumption is wrong

Hence  √5  is an irrational number

## Question: Prove that  6 - 2√5 is an irrational number

Solution:
Let 6 - 2√5 is a rational number.
Then there exist unique integers a and b such that

a, b, 2 and 6 all are integers

⇒
is a rational number.

But √5 is an irrational number.

So there is a contradiction and our assumption is wrong

Hence 6 - 2√5  is an irrational number.

## (Deleted)

Solution: For two given positive integers a and b there exist unique integers q and r such that
a = bq + r, where 0 ≤ r < b
Let b = 3
a = 3q + r
Possible values of r = 0, 1, 2
Possible values of  a = 3q + 0, 3q + 1, 3q + 2

Now taking  a = 3q

Squaring on both side we get

(a)2 = (3q)2

a2 = 9q2

a2 = 3(3q2 )

a2 = 3m1   … .......Where m1 = 3q2

Now taking  a = 3q +1

Squaring on both side we get

(a)2 = (3q + 1)2

a2 = 9q2 + 1 + 6q

a2 = 9q2 + 6q + 1

a2 = 3(3q2 + 2q) + 1

a2 = 3m2 +1       .........Where m2 = 3q2 + 2q

Now taking  a = 3q + 2

Squaring on both side we get

(a)2 = (3q + 2)2

a2 = 9q2 + 4 + 12q

a2 = 9q2 + 12q + 4

9q2 + 12q + 3 + 1

a2 = 3(3q2 + 4q + 1) + 1

a2 = 3m3 + 1        ............Where m3 = 3q2 + 4q + 1

Here we prove that the square of any positive integer is of the form 3m or 3m + 1

## Question 5:  Prove that the cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8 (Deleted)

Solution: For two given positive integers a and b there exist unique integers q and r such that
a = bq + r, where 0 ≤ r < b
Let b = 3
a = 3q + r
Possible values of r = 0, 1, 2
Possible values of  a = 3q + 0, 3q + 1, 3q + 2

Now taking  a = 3q

Cubing on both sides we get

(a)3 = (3q)3

a3 = 27q3

a3 = 9(3q3 )

a3 = 9m1   … .......Where m1 = 3q3

Now taking  a = 3q +1

Cubing on both side we get

(a)3 = (3q + 1)3

a3 = 27q3 + 1 + 27q2 + 9q

⇒  a3  = 27q3  + 27q2 + 9q +1

a3  = 9(3q3  + 3q2 + q) +1

a3 = 9m2 +1                ...........Where m2 = 3q3  + 3q2 + q

Now taking  a = 3q +2

(a)3 = (3q + 2)3

a3 = 27q3 + 8 + 54q2 + 36q

⇒  a3  = 27q3  + 54q2 + 36q +8

a3  = 9(3q3  + 6q2 + 4q) +8

a3 = 3m3 +8          ...........Where m3 = 3q3  + 6q2 + 4q

Here we prove that the cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8

1. Really it is very useful for we people.