SOLVED MATHEMATICS for STUDENTS

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Height and distance. Mensuration, GEOMETRY, TRIGONOMETRY
 FOR CLASS  IX/ X
CBSC/ICSC EXAMS.

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 Use Euclid's Division Algorithm to find the HCF of 255 and 867 (CBSE Sept 2012)


            Since  867> 255 taking  867 as dividend and 255 as divisor


                                    867= 255 X 3 +102


Now 102 is the remainder, which is not zero
 We have    255= 102 X 2 +51

Now 51 is the remainder and is not zero, so once again we have to apply Euclid's division algorithm

We have     102= 51 X 2+0
Now the remainder has become zero and 51 is the divisor.

Hence HCF of  867 and 255 is 51 (ANS)


Euclid Division Lemma 

Let there are two positive integers  p and q
There will exist two unique positive integers a and b which can be expressed as 

q= p*a + b 



where b is in between 0 and q, which is mathematically expressed 

We have taken two positive integers  p and q ,supposing p

         p | q |a
               x          let x =a*p
______________
let remainder =b      b= q-x





Let us take a real integer example, let two digits are   17 and  3

let us divide  17 by 3

              3  | 17 | 5
                     15

                    _____
                       2
This can be expressed as  17= 3*5 + 2

In the above example there is a remainder which is 2

Now let us take another example  of  15 and 3

         3| 15 | 5
              15
           _____        the equation becomes  15 = 3*5 +0
               0
                             So any two integer p and q , where supposing p< q 

can be expressed in a equation     q= a* p + b 






Now an important sum 

Prove that  squre root of any integer is irrational

Let us assume it is  square root of 3 

To solve a mathematical problem sometimes we are to help of inversion

So let us assume that  square root of 3 is rational

So  square root of 3 can be expressed in its simplest and lowest form of a/b ,

b is not equal to zero. a/b has no common factor other than  1

Square root of 3 = a/b 
then squaring  3 = a2/b2
                     or a2/b2= 3
                     or a2= 3 b2 

now  as 3b2 is divided by 3  so a2 is also divisible by 3 

as  a2 is divisible by 3 , a is also divisible by 3 ( 3 is a prime no.)

Now  let us assume that   a = 3c  ( for solving such mathematics it is general to assume to get to the solution)

Now assumption   a = 3c
                       or  a2= 9c2
                        or, 3b2= 9c2
                         or,  b2= 3 c2

Now with the same logic as above  3c2  can be divided by 3 so b2 can also be divided by 3 and if b2 is divided by 3 , b is also divided by 3 . ( 3 is also a prime no)

Now initially we have assumed that a/b is not divisible by any no other than 1 , now it is appearing that a and b are both can be divided by 3 .

Both logic are contrary to each other. So square root of 3 is irrational.