Question

show that root 5 + root 3 is an irrational number

​

Answer 1

Answer:

√5+√3=a/b

√5=a/b-√3

(√5)2=(a/b-√3)2

5=a2/b2-2a√3/b+3

2=a2/b2-2√3a/b

2√3a/b=a2-2b2/b2

√3=a2-2b2/2ab

√3 is rational a contradiction

hence √3+√5 is irrational

Answer 2

$\huge\bf{\textsf{\underline{\underline {Answer:-}}}}$

$\large\sf{Let\:\sqrt{5}+\sqrt{3}\:is\:a\:rational}$

$\large\sf{number\:,\:say\:'r'.}$

$\large\sf{Then,}$

$\large\sf{\implies\:\sqrt{5} + \sqrt{3} = r}$

$\large\sf{\implies\:\sqrt{5} =r-\sqrt{3} }$

$\large\sf{On\:squaring\:both\:sides, we\:get-}$

$\large\sf{\implies {( \sqrt{5}) }^{2} = {(r - \sqrt{3} )}^{2} }$

$\large\sf{\implies 5 = {r}^{2} + 3 - 2 \times r \times \sqrt{3} }$

$\large\sf{\implies 5 = {r}^{2} + 3 - 2 \sqrt{3} \: r }$

$\large\sf{\implies 2 \sqrt{3} \: r = {r}^{2} + 3 - 5 }$

$\large\sf{ \implies \sqrt{3} = \frac{ {r}^{2} + 3 - 5}{2r} </p><p> }$

$\large\sf{Since,\:r\:is\:rational.\:(According \:to \:our\:assumption) }$

$\large\sf{Therefore, \frac{ {r}^{2} + 3 - 5}{2r} \:is\:also\:rational. }$

$\large\sf{\implies\:\sqrt{3}\:is\:also\:rational. }$

$\large\sf{But,\:it\:is\:an\:irrational\:number. }$

$\large\sf{Therefore,\:our\:assumption\:is\:wrong. }$

$\large\sf{Hence, \sqrt{5}+\sqrt{3}\:is\:an\:irrational }$

$\large\sf{number.}$

$\large\sf{Hence,\:proved. }$