Poove that 2√3 +5√5 is
irrational number
a
Answers
Answer:
the word aw thanks for the na ki ham ko tho itna bhi nhi hai na ki
Answer:
Answer:
2√3+√5 is irrational.
Step-by-step explanation:
Let us suppose that 2√3+√5 is rational.
\begin{gathered}Let\: 2\sqrt{3}+\sqrt{5}=\frac{a}{b},\\where\:a,b\:are\: integers\\ \:and\:b≠0\end{gathered}
Let2
3
+
5
=
b
a
,
wherea,bareintegers
andb
=0
Therefore, \:2\sqrt{3}=\frac{a}{b}-\sqrt{5}.Therefore,2
3
=
b
a
−
5
.
Squaring on both sides, we get
12=\frac{a^{2}}{b^{2}}+5-2\times \frac{a}{b}\times \sqrt{5}12=
b
2
a
2
+5−2×
b
a
×
5
Rearranging
\implies \frac{2a}{b}\sqrt{5}=\frac{a^{2}}{b^{2}}+5-12⟹
b
2a
5
=
b
2
a
2
+5−12
\implies \frac{2a}{b}\sqrt{5}=\frac{a^{2}}{b^{2}}-7⟹
b
2a
5
=
b
2
a
2
−7
\implies \frac{2a}{b}\sqrt{5}=\frac{a^{2}-7b^{2}}{b^{2}}⟹
b
2a
5
=
b
2
a
2
−7b
2
\implies\sqrt{5}=\frac{a^{2}-7b^{2}}{b^{2}}\times \frac{b}{2a}⟹
5
=
b
2
a
2
−7b
2
×
2a
b
After cancellation, we get
\implies\sqrt{5}=\frac{a^{2}-7b^{2}}{2ab}⟹
5
=
2ab
a
2
−7b
2
Since , a ,b are integers , < /p > < p > \frac{a^{2}-7b^{2}}{2ab}</p><p>
2ab
a
2
−7b
2
is rational,and so √5 is rational.
This contradicts the fact that √5 is irrational.
Hence , 2√3+√5 is irrational