Question

prove 1/3+√2 as rational or irrational number​

Answer 1

$\huge{\underline{\sf{\pink{ᏗᏁᏕᏇᏋᏒ:-}}}}$

Let us assume that 3+√2 is rational.

Therefore we can find the co-prime integers a and b where b ≠ 0 such that

$3 + \sqrt{2} = \frac{a}{b} \\ Therefore \: \frac{a}{b} - 3 = \sqrt{2 }......(i) \\ Since \: a \: and \: b \: are \: integers, \\ \frac{a - 3b}{b} = \frac{integer - 3 \times integer}{integer ≠ 0 } \\ = \: rational \: number$

So, from (i) ,√2 is a rational number which contradicts the fact that √2 is an irrational number.

Therefore our assumption is wrong.

$\bf\star\underline\red{Hence, \: 3+ √2 \: is \: irrational \: number. }$

$\bf\star\underline\purple{ Hope \: it \: works \: for \: you☺☺}$

$\bf\star\underline\orange{ ᏕᏖᏗᎩ \: ᏂᏗᎮᎮᎩ \: ᏗᏁᎴ \: ᏰᏝᏋᏕᏕᏋᎴ☺❤}$

Answer 2

Answer:

irrational number

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