Question

Provethat√7isrationalnumber.​

Answer 1

Step-by-step explanation:

let √7 be rational

√7=p/q

square both side

7=p^2/q^2

7q^2=p^2. .equation (1)

q^2=p^2/7

here p^2 is divided by 7

now let p=(7r)

put in equation 1

7q^2=(7r)^2

7q^2=49r^2

q^2=7r^2

q^2/7=r^2

here q^2 is divided by 7

hence it can not be possible that both p^2 and q^2 divided by 7

hence proved that √7 is

irrational

Answer 2

Correct Question

Prove that√7 is an irrational number

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

Step by step explanation

1. In this type of Questions first we have to assume that the given root value is rational.
2. Now , equate with two co-prime numbers.

[co-prime numbers are those numbers which having one factor].

3.After equating we know that √7 have alteast one factor it may be 2 -3 also so,here we have to prove our assumption is wrong and we can also prove that √7 is irrational.

=>Let us assume that √7 is rational number.

=>√7=a/b ( a and b are some co-prime integers)

a=b√7-----(1)

Squaring both sides

we obtained

=>a²=7b²-----(2)

Here,a² is divisible by 7, if a² is divisible then a is also divisible by 7.

We can also write as :a=7p ----(3)

Now , substitute a's value from equation 3 in 2.

=>(7p)²=7b²

=>49p²=7b²

=>b²=7p²

Here,b² is divisible by 7 it means b is also divisible by 7.

∴a and b have atleast one common factor i.e.,7

But this contradicts said that a and b are co-primes.Hence,our assumption is wrong

∴√7 is an irrational number.