prove that:√3 is Irrational!Neat answer please!. Created by Noah11. Math

prove that:√3 is Irrational!Neat answer please!...

MasterYaduvanshi $\color{indigo}{Thanks \: for \: asking \: this \: question}$

$\color{blue}{Bonjour}$

$\color{red}{ Required Answer}$

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.

$\color{pink}{Proved}$

$\color{green}{Sujeet \: kumar}$

$\color{red}{Have \: an \: awesome \: day \: dear \: friend}$

$\color{blue}{Hope \: its \: help \: you}$

DaIncredible Great :o

MasterYaduvanshi Thanks brother. !!!!!

DaIncredible :p

MasterYaduvanshi :-)

Prakhar2908 Gr8

MasterYaduvanshi Thanks jee

Prakhar2908 : )

MasterYaduvanshi :-)

MasterYaduvanshi Thanks!!

vikram991 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

nitya1984 hi

DaIncredible No chats ✌

Noah11 @nitya1984,don't tend to start a conversation here

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