Question

prove that √3 is an irrational number​

Answer 1

✍️Question:-

$\rule{300}{1}$

$\purple\leadsto$ Prove that $\sqrt{3}$ is an irrational number .

✍️How to solve?

• By assuming that the number given is rational and by correct methods of rational number solve the sum and last this contradicts will never be possible as we can't change some specific rules of maths
• ☀️So lets start.

✍️Solution:-

$\rule{300}{1}$

$\implies$ Let, us assume that the contrary , that $\sqrt{3}$ is a rational number.

Now,

$\tt \sqrt{3}=\dfrac{p}{q}$ (Here , p and q are co-prime )

$\tt p = \sqrt{3}q$

Now squaring this at both side we will get.

$\therefore \tt p^{2} = 3q^{2}$______(1)

$\red\cdot$Therefore, p² is divisible by 3 , it means p is also divisible by 3.

$\therefore$ p = 3c ( here c is some integer)

Now,

put p =3c in equation (1)

$\tt (3c)^{2} = 3q^{2} \\\\ \tt q^{2} = \dfrac{p}{3}c^{2} \\\\ \tt q^{2} = 3c^{2}$

$\red\cdot$ q² is divisible by 3 , it means q is also divisible by 3.

$\blue\implies$ p and q have at least 3 as common factor.

$\blue\implies$ This contradiction has arisen because of our incorrect assumption so $\sqrt{3}$ is irrational.

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⛄Hence we are done with the problem

Answer 2

$\huge{\underline{\mathbb{\red{A}\pink{n}\green{S}\blue{w}\purple{E}\orange{r}}}}$

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Let us assume that √3 is a rational number.

then, as we know a rational number should be in the form of p/q

where p and q are co- prime number.

So,

√3 = p/q { where p and q are co- prime}

√3q = p

$\small\underline\bold\red{Now, by \:squaring \:both\: equation\: we \:get}$

(√3q)² = p²

$\dashrightarrow$3q² = p² ........ ( i )

So,

$\dashrightarrow$if 3 is the factor of p²

$\dashrightarrow$then, 3 is also a factor of p ..... ( ii )

$\dashrightarrow$ Let p = 3m { where m is any integer }

squaring both sides

p² = (3m)²

p² = 9m²

putting the value of p² in equation ( i )

$\implies$3q² = p²

$\implies$3q² = 9m²

$\implies$q² = 3m²

So,

if 3 is factor of q²

$\implies$then, 3 is also factor of q

$\large\underline\bold\red{since}$

$\implies$3 is factor of p & q both

So, our assumption that p & q are co- prime is wrong

$\large\underline\bold\red{Hence,}$

$\implies$ √3 is an irrational number .

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