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Addition

In simple way, addition means to add two or more numbers.

For example:-If anyone has one ball and his/her mother gives him/her another ball, then the total number of balls will be two. In mathematical way, 1+1=2.

Additive Identity

The additive identity of a number x(say) is the number that when added to x provides a result of x,i.e., the additive identity of every number is zero.

For example:- The additive identity of 3 is 0.

Additive Inverse

The additive inverse of a number is the number that when added to the first number provides a result of zero. 

For example:- The additive inverse of a number a(say) is going to be -a because on adding these two numbers we get a zero.                      Now try putting values for a. Let a be 3 then its additive inverse will be -3.

Note:- The additive inverse of a positive number is negative and that of a negative number is positive.

Mean, Mode and Median

Mean, Mode and Median are the three most common kind of averages that you will see in your classes till near about 9th class and later there are some more bizarre averages for you in statistics.

Mean:- It is just the ordinary average that you may be familiar with but its better to revise, so, Mean is the average in which you just add all the given values and divide their sum by the number of values.

For example:- The mean of 6,12,3 is 7,i.e., 6+12+3(sum of all values) divided by 3(The number of values).

Mode:- The mode is the value that appears most often in a group of values. For simplicity you may write the digits in ascending order or descending order and find out which number appears most often. 

For example:- The mode of 2,4,5,2,5,3,1,4,2 is 2. As (I prefer ascending order) 1,2,2,2,3,4,4,5,5.Clearly we can observe that 2 is coming most often that is 3 times.

Median:- It is the middle value of a set of values. For finding the median we have to arrange the values in ascending or descending order and then the value counted to be in the middle from both the sides(left and right) is called the Median.

For example:- The median of 2,4,8,1,3 is 3. As I stated above I prefer ascending order, so, 1,2,3,4,8. Clearly 3 is the middle value from both the sides.

In this case the number of values were odd but what would happen if the number of values are even because we will have two middle numbers. In that case we take the average of the two middle numbers,i.e.,add them and divide by two.

For example:- The median of 2,4,1,3 is 2.5. As on arranging we get 1,2,3,4 in this the two middle terms that appear are 2 and 3. So when we add 2 and 3 and divide by 2,i.e.,5 divided by 2, we get 2.5.

Midrange

Mid range is a bit different concept than a Range. It is the arithmetic mean or simply just the mean of  the highest and the smallest values of a set of values.

For example:- The mid range of 1,2,3,4,12,19 is 10. As the mean of 19(the highest value) and 1 (the smallest value) is 10,i.e., 19+1(Sum of highest and smallest values) divided by 2(the number of values).

Multiplicative Identity

Multiplicative identity of a number x(say) is the number which on multiplying with x provides a result of x, i.e., the multiplicative identity of every number is 1.

For example:- The multiplicative identity of 5 is 1.

Multiplicative Inverse

Multiplicative inverse of a number x(say) is the number which on multiplying with x provides a result of 1, i.e., 1/x(this is for fraction).

For example:- The multiplicative inverse of 5 is 1/5 or 0.2.

Polynomial Identities

The most important ten polynomial identities for students till class 9th are the following which are related to binomials.                      i. (a+b)²= a² + 2ab + b² 

Proof of identities:- (a+b)² means (a+b) is multiplied by itself . When we multiply (a+b) by (a+b), then we first multiply (a+b) by a and then by b and then we add them up. Like this:- i)  (a+b)×(a+b)= a×(a+b)+b×(a+b)   Then we get    ii) (a+b)×(a+b)or (a+b)²= a² + ab +ba(or you can write ab)+b²   This will give you the final result, i.e.,  iii)  (a+b)² = a² + 2ab + b²
Proof with the help of example:- (6+8) ² = 6² + 2×(6)×(8) + 8²   (because on observing here a is 6 and b is 8.) (it means implies that)  (14)²=36 + 96 + 64=196 
                                                                             Or by the other simple way
                                                                                    (14)²= 14×14 =196 
ii. (a-b)²= a² - 2ab + b²
Proof of identities :- (a-b)² means (a-b) is multiplied by itself . When we multiply (a-b) by (a-b), then we first multiply (a-b) by a and then by -b and then we add them up. Like this:- i)  (a-b)×(a-b)= a×(a-b)+(-b)×(a-b)   Then we get    ii) (a-b)×(a-b)or (a-b)²= a² - ab -ba(or you can write ab)+b²    This will give you the final result, i.e.,  iii) (a-b)² = a² - 2ab + b²
Proof with the help of example:- (10-8) ² = 10²-+ 2×(10)×(8) + 8²  (because on observing here a is 10 and b is 8.) (it means implies that)  (2)²=100- 160 + 64=4 
                                                                               Or by the other simple way
                                                                                         (2)²= 2×2 =4 
iii. (a+b)×(a-b)= a² - b²
Proof of identities :-  When we multiply (a+b) by (a-b), then we first multiply (a+b) by a and then by -b and then we add them up. Like this:- i)  (a+b)×(a-b)= a×(a+b)+(-b)×(a+b)   Then we get    ii) (a+b)×(a-b)= a² +ab -ba(or you can write ab)-b²    This will give you the final result, i.e.,  iii) (a-b)(a+b) = a² - b²  
Proof with the help of example:- (10+8)(10-8) = 10² - 8² (because on observing here a is 10 and b is 8.)  (it means implies that)  (18)(2)=100-64=36 
                                                                                 Or by the other simple way
                                                                                      (10+8)(10-8)= 18×2 =36 
iv. (x+a)×(x+b)= x² + (a+b)×x + ab
Proof of identities :-  When we multiply (x+a) by (x+b), then we first multiply (x+a) by x and then by b and then we add them up. Like this:- i)  (x+a)×(x+b)= x×(x+a)+(b)×(x+a)   Then we get    ii) (x+a)×(x+b)=x² +xa(or you can write ax)+ bx + ab  On simplifying xa +bx we get x in common so it becomes (a+b)× x  On putting this in the equation you will get the final result, i.e.,  iii) (x+a)(x+b) = x² + (a+b)×x + ab
Proof with the help of example:- (2+3)×(2+1) = 2² + (3+1)×2  + 3×1 (because on observing here x is 2, a is 3 and b is 1.)  (it means implies that)  (5)×(3) =4 + 8 + 3=15 
                                                                                  Or by the other simple way
                                                                                        (2+3)×(2+1) = 5×3 =15 
v. (x+y+z)²= x² + y² + z² + 2xy + 2yz + 2zx
Proof of identities:- When we multiply (x+y+z) by itself, then we first multiply (x+y+z) by x, then by y and then finally by z and add them up.Like this:- i) (x+y+z)×(x+y+z)= x×(x+y+z)+ y×(x+y+z)+ z×(x+y+z).      This will give us   ii) (x+y+z)×(x+y+z)= x² +xy+xz (or you can write zx) + yx(or you can write xy) + y² + yz+ zx + zy(or you can write yz) + z²      On adding all of these we will get the final result,i.e.,  iii) (x+y+z)²= x² + y² + z² + 2xy + 2yz + 2zx.
Proof with the help of an example:- (1+2+3)²= 1² + 2² +3² + 2×1×2 + 2×2×3 + 2×3×1 (because on observing here x is 1,y is 2 and z is 3.)  (6)²= 1 +4 +9+4+12+6 = 36
                                                                                     Or by the other simple way       
                                                                                        (1+2+3)² = 6² =6×6 =36  
vi. (x+y)³= x³ + y³ + 3xy(x+y)
Proof of identities:- First of all you may be thinking why no sign is between 3xy and (x+y), this means multiplication when we don't write any sign when including the brackets. When we multiply (x+y) by itself thrice or you can say three times, then first we have to find the square of it and then multiply it by (x+y). Like this:- i) (x+y)² × (x+y)= (x² + 2xy +y²)( As by the first identity )(x+y) Now, we will first multiply (x² + 2xy +y²) by x and then by y and then we add them up. Like this:- ii) x(x² + 2xy +y²)+ y(x² + 2xy +y²)   This will give us:- iii) (x+y)³ = x³+2x²y + xy²+x²y + 2xy² +y³   iv)  (x+y)³= x³+3x²y + 3xy² +y³ On simplifying it we get v)  x³ + y³ + 3xy(x+y).  
Proof with the help of an example:- (1+2)³ = 1³+2³ + 3(1)(2)(1+2)  ⇒ 3³= 1+8 + 18 =27
                                                                                     Or by the other simple way.
                                                                                         (1+2+3)³= 3³= 3×3×3= 27
vii. (x-y)³= x³ - y³ - 3xy(x-y)
Proof of identities:- When we multiply (x-y) by itself thrice or you can say three times, then first we have to find the square of it and then multiply it by (x-y). Like this:- i) (x-y)² × (x-y)= (x² - 2xy +y²)( As by the first identity )(x-y) Now, we will first multiply (x² - 2xy +y²) by x and then by  -y and then we add them up. Like this:- ii) x(x² + 2xy +y²)+ y(x² + 2xy +y²)  This will give us:- iii) (x+y)³ = x³-2x²y + xy²-x²y + 2xy² -y³   iv)  (x+y)³= x³-3x²y + 3xy² +y³ On simplifying it we get v)  x³ - y³ - 3xy(x-y).
Proof with the help of an example:- (3-2)³ = 3³-2³ - 3(3)(2)(3-2) ⇒ 1³= 27 - 8 - 18=1
                                                                         Or by the other simple way.
                                                                              (3-2)³= 1³= 1×1×1= 1
viii

Pythagorean Theorem




                          Figure 1


The Great Pythagoras






        Figure 2

Pythagorean theorem is an interesting theorem given by the Greek philosopher and mathematician, Pythagoras whom you can see in the above picture. He found that for any right triangle (90°), if you construct a square on each of the sides the square of the longest side is equal to the sum of the squares of other two sides. It can be stated in a short equation as follows For any right triangle ABC (as in the adjacent figure) A²+ B² = C².

Definition of Pythagorean theorem:- After the above discussion we conclude a simple definition of Pythagorean theorem that the square of hypotenuse is equal to the sum of squares of other two sides.             

Hypotenuse:- The longest side of a right triangle is called its hypotenuse.In the adjacent figure C is the hypotenuse.

Proof with the help of an example:- In Figure 2 above A=7, B=a, C=10. So A²+ B² = C²

7² + a² = 10²,  This will give us- 49+a²= 100     On subtracting 49 from both sides we get  a²=100-49  or a²=51  a= √51   a=7.14 approx.

Range

It is also a common term. Even it is the one of the easiest terms. It is the difference of the highest value and the smallest value in a set.

For example:- The range of 1,2,12,23 is 22. As 23(The highest value) - 1(The lowest value)=22