Question

prove that root 7 is rational number

Answer 1

Answer:

Lets assume that √7 is rational number. ie √7=p/q.

suppose p/q have common factor then

we divide by the common factor to get √7 = a/b were a and b are co-prime number.

that is a and b have no common factor.

√7 =a/b co- prime number

√7= a/b

a=√7b

squaring

a²=7b² .......1

a² is divisible by 7

a=7c

substituting values in 1

(7c)²=7b²

49c²=7b²

7c²=b²

b²=7c²

b² is divisible by 7

that is a and b have atleast one common factor 7. This is contridite to the fact that a and b have no common factor.This is happen because of our wrong assumption.

√7 is irrational

Step-by-step explanation:

MARK BRAINLIEST!

Answer 2

To prove that √7 is not a rational number we must do the following steps exactly to get the full designated marks.

Step-by-step explanation:

Let's assume that √7 is a rational number.

$\sqrt{7}$=$\frac{p}{q\\\\}$            (where p and q are positive integers, p≠0; p and q are co-prime)

Squaring both sides

$(\sqrt{7} )^{2}$= $(\frac{p}{q})^{2}$

 7=$\frac{p}{q}^{2}$

 $p^{2}$=7$q^{2}$-------------------------------------EQUATION 1

⇒ $p^{2}$ is divisible by 7                

⇒ $p^{}$ is divisible by 7

( If p divides $a^{2}$  then p divides $a^{}$ also, where p is a prime number)

Let $p^{}$=7m, m is a positive integer.

Substitute $p^{}$ in equation 1

$(7m)^{2}$=7$q^{2}$

49$\\ m$=7$q^{2}$

7$\\ m$=$q^{2}$

⇒$q^{2}$ is divisible by 7

⇒$\\ q$ is divisible by 7

( If p divides $a^{2}$  then p divides $a^{}$ also, where p is a prime number)

$\\ q$=7$\\ n$ , $\\ n$ is a positive integer.-------------------------EQUATION 2

From Equation 1 & 2 we get

$\frac{p}{q\\\\}$ =$\frac{7m}{7n}$

∴ WE SEE THAT  $p^{}$ AND $\\ q$ HAVE THE LOWEST COMMON PRIME FACTOR, 3. HENCE, WE CAN SAY THAT  $\sqrt{7}$ IS NOT A RATIONAL NUMBER.

REASONING:

THE DEFINITION OF A RATIONAL NUMBER IS THAT THEY HAVE FACTORS OTHER THAN THEMSELVES AND 1. BUT, HERE IN THE END WE SEE THAT THEY ARE CO-PRIME. SO, WE CAN SAY THAT $\sqrt{7}$ NOT A RATIONAL NUMBER.