Question

show that under root 3 is an irrational number

Answer 1

Solution :
Let us assume on the contrary that root √3 is a rational no.There exists positive integers a and b such that
√3 = a/b

Now √3 = a/b
Squaring both sides
3 =a^2 /b^2
A^2 =3b^2
By theorem
a = 3c (( i eq.))
a^2 = 9c^2
So 3b^2 = 9 c^`2
B^2 = 3c^2
3 /b^2
3/b. (( ii eq.))
By theorem

..From i and ii.Eq that we observe a and b as a common factor

.So our supposition is wrong

.Hence √3 is an irrational no.

....Thanks✔️✔️✔️✔️✔️✔️✔️

Answer 2

Hey !

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We have to prove √3 is irrational.

Let us assume in Contradiction that √3 is rational number.

Hence , √3 can be written in the form of a/b where a and b are co-primes ( no common factor other than 1 )


√3 = a / b

√3 b = a

squaring on both sides.

( √3b )² = a²

3b² = a²

a²/3 = b²

Hence , 3 divides a²

By theorem : If p is prime number , and p divides a² , then p divides a , where a is a positive number.

So , 3 divides a also.........(1)

Hence , we can say a /3 = c where c is some integer.

So , a = 3c

Now we know that

3b² = a²

putting a = 3c

3b² = ( 3c )²

3b² = 9c²

b²= 3c²

b²/3 = c²

Hence 3 divides b²

So, 3 divides b also........(2)

By (1) and (2)

3 divides both a & b

Hence , 3 is a factor of a and b .

So., a & b have a factor 3

Therefore , a & b are not co-primes.

Here , our assumption is wrong.

By contradiction...

√3 is irrational number.


# Hope it helps #