Math, asked by Noah11, 1 year ago

prove that:

√3 is Irrational!

Neat answer please!

Answers

Answered by Anonymous
15
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Let to be √3 be Rational number

Let it's simplest form be a/b

then,

Here,

Note:-
a and b are Integer having no common factor other than 1.

Again,

√3=a/b

Now,

Squaring the both sides means (L.H.S and RHS)

we get,

(√3)²=(a/b)²

(3)=a²/b²

Here, using crossmultiplication...

3b²=a²

Now,.

3 Divide a²

Again,

3 Divide a

Note Reason:- 3 is a prime Number and 3 Divide by a²

Now,

Let to be a=3c

then,

(3b²)=9c²

Note:- 9 divide by 3

b²=3c²

Again,

3 Divide by b²

3 Divide by a

Now,

3 is a common factor of a and b...

Thus,

√3 is an irrational number..

Note :- Conclusion- √3 is an irrational number.

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Answered by vikram991
4
here is your answer OK dude


We can provide a contradictional proof for it..

Firstly let us assume

Assumption:let √3 be a rational ….then as every rational can be represented in the form p/q where q≠0

Let √3=p/q where p,q have no common factor.

Now squaring on both sides we get 3=p^2/q^2

i.e 3*q^2=p^2

Which means 3 divides p^2 which implies 3 divides p

Hence we can write p=3*k

This gives 3*q^2=9*k^2

q^2=3*k^2

Which means 3 divides q^2 which implies 3 divides q.

3 divides p and q which means 3 is common factor for p and q.

And this is contradiction for our assumption that p and q have no common factor…

Hence we can say our assumption that √3 is rational is wrong…

And therefore √3 is an irrational…

hope it help you

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