Question

prove that root 5 is irrational
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Answer 1

Answer:

Let us assume the contrast i.e

Let $\sqrt{5}$be a rational number.

$\sqrt{5}$= ab where a and b both are co prime number.

$b \sqrt{5 } = a$

Squaring both sides we have,

$(b \sqrt{5 }) {}^{2} = a {}^{2}$

$5b {}^{2} = a {}^{2} ...equation \: 1$

So that

$a {}^{2} \: is \: \: divisible \: by \: 5 \:$

then a is also divisible by 5

Now let a = 5c

$then \: 5b {}^{2} = (5c) {}^{2}$

$5b {}^{2} = 25c {}^{2} \\ b {}^{2} = 25c {}^{2} \div 5 \\ b {}^{2} = 5c {}^{2} ... \: equation \: 2$

$so \: b {}^{2} is \: divisible \: by \: 5 \:$

then b is also divisible by 5

Therefore, a and b have atleast 5 as a common factor and its contradict that a and b are co - prime.

It happened due to incorrect assumption.

$so \: \sqrt{5} \: is \: an \: irrational \: number$

Step-by-step explanation:

Hence, proved.

Answer 2

Answer:

Step-by-step explanation:

There is the answer in image