Model problem-1: Prove that √2 is an irrational number-10th class mathematics-Proof of irrational number

Before going to learn the proof of irrational,

we know the following

Every rational number can be written as a fraction

let the simplest form of a rational number = p/q

Here p and q are co-primes and q≠0  

Co-prime numbers

Two numbers are called “co-prime numbers” If their H.C.F. is “1”

Co-primes have NO common factor other than “1”

Co-primes have only one common factor that is “1”

 

Theorem

a is a positive integer and p is a prime number.

“If a² is divided by p then a is also divided by p”

 

Division algorithm

Dividend(a) = Divisor(b) x Quotient(q)  +  Remainder(r)

                   a=bq + r

If remainder(r) is “0”

               a=bq

    we said that "b divides a"

In this case:

 “a” is the multiple of “b” and “q”

 “b” and “q” are called factors of  “a”

Contradiction Method

In this method, we take an assumption that “opposite to given statement”

After several steps, we get mathematical absurdity

Reason for this absurdity is our assumption

So our assumption is wrong

We conclude that given statement is correct.

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Prove that √2  is an irrational.

(We can prove this by contradiction method)

Proof:

Assume that √2 is not an irrational number

So √2   is a rational number

   ∴ √2  = p ⁄ q (Here p, q are co-primes and q≠0)

Squaring on both sides, we get

       (√2  )² = ( p ⁄ q )²

            2 = p²/q²

         2q² = p²

        p²  = 2q² (LHS and RHS are interchanged)

2 divides p² So 2 divides “p”……..…......(i)

That means p is multiple of 2

let take  p = 2k (k is an integer)

Substitute p = 2k in 2q² = p²

        2q² = (2k)²

        2q² = 4k²  

         q² = 2k²  (divided by 2)

    2 divides q², so 2 divides “q”………….(ii)

    From (i) and (ii), we get

     2 divides p and q

That means 2 is a factor of p and q

    But p and q are co-primes

(They have only “1” as common factor)

Above two statements are contradict each other

Because of our assumption

So our assumption “√2 is not an irrational” is wrong

Therefore √2  is an irrational

Hence proved

Practice the following problems

1. Prove that √3  is an irrational

2. Prove that √5  is an irrational

3. Prove that √7  is an irrational

Click the link for "Proof of  5√2+7√3  is an irrational number"

Click the link for "Introduction of Sets"

Click the link for "Euclid division algorithm&HCF"