Answers
Answer:
†
Answer:−
{\underline{\boxed{\sf{\purple{\mathtt{ Question :-}}}}}}
Question:−
\begin{gathered}Prove \: that \: \\ \\ 3+2√5 \: is \: irrational \: number.\end{gathered}
Provethat
3+2√5isirrationalnumber.
\small\underline\mathcal\pink{Requried \: Answer :-}
RequriedAnswer:−
Let 3+2√5 be rational.
3+2√2 = p/q where q≠ 0 & p and q are integers.
3 + \sqrt{5} = \frac{a}{b} \: \: \: \: \: \:3+
5
=
b
a
a and b are coprime integers
\begin{gathered}2 \sqrt{5} = \frac{a}{b} - 3 \\ 2 \sqrt{5} \: \frac{a - 3}{b} \\ \sqrt{5} = \frac{1}{2} ( \frac{a}{b} - 3)\end{gathered}
2
5
=
b
a
−3
2
5
b
a−3
5
=
2
1
(
b
a
−3)
\begin{gathered}Since \: a \: and \: b \: are \: integers \: \frac{1}{2} ( \frac{a}{b} - 3) \: \\ will \: also \: be \: rational. \\ therefore \: \sqrt{5} \: is \: rational \end{gathered}
Sinceaandbareintegers
2
1
(
b
a
−3)
willalsoberational.
therefore
5
isrational
This contradicts that the fact √5 is irrational.
Hence our assumption that 3+2√5 is irrational number is false
Therefore 3+2√5 is irrational number