Question

prove that root 7 is an irrational number pls answer fast​

Answer 1

Hello Mate♥️♥️♥️

Answer:-

√7

Step-by-step explanation:

Let √7 is an irretional number

√7/1=a/b

b√7=a

________S.B.S

=>2b²=a²

=>b²=a²/7........1st equation

#If a² is divisible by 7, then a Is also divisible by 7.

=>a=7c........2nd equation

#Put the value of equation 1

=>b²=(7c²)

=>7b²= 49c²

=>b²=7c²

=>7/b²

=>7/b

So,√7 is an irretional number.

$thanku$

Hope it help uh✌️♥️✌️♥️✌️

Answer 2

$\huge{\underline{\underline{\red{Answer}}}}$

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$\mathbb{\boxed{\pink{GIVEN\:TO\:PROVE-}}}$

• $\sqrt{7} is \: irrational$

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$\bf{\underline{\underline{\red{SOLUTION}}}}$

Let's assume that $\sqrt{7}$ is rational.

Then, $\sqrt{7}$ can be expressed as $\frac{p}{q}$ where q $\neq$0 and p, q has no common factor other than 1.

$\sqrt{7} = \frac{p}{q} \\ \implies \: 7 = \frac{ {p}^{2} }{ {q}^{2} } \\ \implies \: 7 {q}^{2} = {p}^{2} - - - - i$

This means 7 is a prime factor of $p^{2}$

means...

$\underline{\boxed{\green{7\:is\:a\:prime\:factor\:of\:p}}}$

Let

$\underline{\boxed{\pink{p=7b}}}$

Now putting p=7b in i we get,

$7 {q}^{2} = {p}^{2} \\ \implies \: 7 {q}^{2} = {(7b)}^{2} \\ \implies 7{q}^{2} = 49 {p}^{2} \\ \implies \: {q}^{2} = 7 {p}^{2}$

This means, 7 is a factor of $q^{7}$

Means, $\underline{\boxed{\green{7\:is\:a\:prime\:factor\:of\:q}}}$

Thus both p and q has common factor 7.

But this is a contradicts as we assumed p and q has no common factor other than 1.

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Thus our assumption is wrong, and,

$\large{\red{\underline{\underline{\sqrt{7}}{is\:irrational}}}}$