prove that root 3 is an irrational number
Answers
Answer:
Step-by-step explanation:
Let us assume on the contrary that \sqrt {3}
3
is a rational number. Then, there exist positive integers aa and bb such that
\sqrt { 3 } =\dfrac { a }{ b }
3
=
b
a
where, aa and bb, are co-prime i.e. their HCFHCF is 11
Now,
\sqrt { 3 } =\dfrac { a }{ b }
3
=
b
a
\Rightarrow \quad 3=\dfrac { { a }^{ 2 } }{ { b }^{ 2 } }⇒3=
b
2
a
2
\Rightarrow \quad 3{ b }^{ 2 }={ a }^{ 2 }⇒3b
2
=a
2
\Rightarrow \quad 3|{ a }^{ 2 }\quad \quad \left[ \because 3|3{ b }^{ 2 } \right]⇒3∣a
2
[∵3∣3b
2
]
\Rightarrow \quad 3|a\quad \quad ...\left( i \right)⇒3∣a...(i)
\Rightarrow \quad a=3c⇒a=3c for some integer cc
\Rightarrow \quad { a }^{ 2 }=9{ c }^{ 2 }⇒a
2
=9c
2
\Rightarrow \quad 3{ b }^{ 2 }={ 9c }^{ 2 }\quad \quad \left[ \because { a }^{ 2 }=3{ b }^{ 2 } \right]⇒3b
2
=9c
2
[∵a
2
=3b
2
]
\Rightarrow \quad { b }^{ 2 }={ 3c }^{ 2 }⇒b
2
=3c
2
\Rightarrow \quad 3|{ b }^{ 2 }\quad \quad \left[ \because 3|3{ c }^{ 2 } \right]⇒3∣b
2
[∵3∣3c
2
]
\Rightarrow \quad 3|b\quad \quad ...\left( ii \right)⇒3∣b...(ii)
From (i)(i) and (ii),(ii), we observe that aa and bb have at least 33 as a common factor. But, this contradicts the fact that aa and bb are co-prime. This means that our assumption is not correct.
Hence, \sqrt {3}
3
is an irrational number.