Question

Prove that √7 is not a rational number.​

Answer 1

Answer:

there u go

Step-by-step explanation:

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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.