Prove that \sqrt{5} is an irrational number.

 

 
 
 
 
 

Answers (1)
S safeer

To prove \sqrt{5} is irrational, let's assume it is a rational number.

Hence, \sqrt{5} can be written is form of \frac{a}{b}

Hence, (a,b) are co-prime and b\neq 0

\Rightarrow \sqrt{5}=\frac{a}{b}

\Rightarrow a=\sqrt{5}b

Squaring both sides

\Rightarrow a^{2}=5b^{2} _____(a)

\Rightarrow b^{2}=\frac{a^{2}}{5}\Rightarrow this means a^{2} is divided by 5

By theorem = If P is a prime number of P divides a^{2} then P also divides a.

So we can say \Rightarrow \frac{a}{5}=k   [hence k is integer]

                          \Rightarrow a=5k _____(b)

Putting the value of (b) in (a)

\Rightarrow (5k)^{2}=5b^{2}

\Rightarrow 25\; k^{2}=5\; b^{2}

\Rightarrow k^{2}=\frac{b^{2}}{5}

Here we could also say that b^{2} is divided by 5. By the theorem stated above, b should also be divisible by 5

\Rightarrow hence a & b both are divisible by 5 but as we know a&b are co-prime (hence they can't have a co-factor =5)

\Rightarrow Hence by contradiction, \sqrt{5} is an irrational number.

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Buy Now
Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions