Question

prove that root 2 is irrational number​

Answer 1

Answer:

Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.

We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.

From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something.

If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:

2 = (2k)2/b2

2 = 4k2/b2

2*b2 = 4k2

b2 = 2k2

assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational.

hope it will help you.......

Answer 2

Answer:

Hey mate, Good evening ❤

#Here's ur answer...☆☆☆

Step-by-step explanation:

$\boxed{ \orange{ \huge{answer...}}}$

$\bold{ \red{to \: prove \: }}= > \bold{ \sqrt{2} \: is \: irrational...}$

$\bold{ \green{let \: us \: assume \: that \: \sqrt{2} \: is \: rational.. }}\\ \bold{hence \: \sqrt{2} \: can \: be \: written \: as \: \frac{p}{q}} \\ \bold{where \: p \: and \: q \: are \: coprimes...} \\ = > \sqrt{2} = \frac{p}{q} \\ = > 2 = \frac{p {}^{2} }{ {q}^{2} } \\ = > {p}^{2} = 2 {q}^{2} \\ \red{ \bold{p \: is \: a \: factor \: of \: 2 }}\\ then \: we \: can \: write \: p = 2m \\ = > {p}^{2} = {2m}^{2} \\ = > {4m}^{2} = 2 {q}^{2} \\ = > {q}^{2} = 2 {m}^{2} \\ \bold{ \red{ so \: q \: is \: a \: factor \: of \: 2}} \\ \bold{but \: we \: had \: assume \: p \: and \: q \: are} \\ \bold{ \: coprimes(no \: factor \: other \: than \: 1)} \\ \bold{ \green{ hence \: our \: assumption \: is \: wrong}} \\ { \underline{ \pink{ \bold{\sqrt{2} \: is \: irrational...}}}}$

➡️Hope this helps you..✌✌

➡️Pls mark as brainliest..❇❇❇

《♤》♤《♤》♤《♤》