prove that root 3 is irrational
Answers
Answer:
Given:
\sqrt {3}
3
To find:
\sqrt {3}
3
is an irrational number.
Solution:
\sqrt {3} = \frac {p}{q}
3
=
q
p
(Were p and q is a co-prime)
\sqrt {3 \times q} = p
3×q
=p
Squaring both the side in above equation
{ \sqrt {3 \times {q}^{2}} = p^{2}
3 \times q^{2} = p^{2}3×q
2
=p
2
if 3 is a factor of p^{2}p
2
Then, 3 will also be a factor of p
\Rightarrow Let\quad p = 3 \times m⇒Letp=3×m {where m is a integer}
Squaring both sides we get
p^{2} = {3 \times m}^{2}p
2
=3×m
2
p^{2} = 9 \times m^{2}p
2
=9×m
2
Substitute the value of p^{2}p
2
in the equation
3 \times q^{2} = p^{2}3×q
2
=p
2
3 \times q^{2} = 9 \times m^{2}3×q
2
=9×m
2
q^{2} = 3 \times m^{2}q
2
=3×m
2
If 3 is a factor of q^{2}q
2
Then, 3 will also be factor of q
Hence, 3 is a factor of p & q both
So, our assumption that p & q are co-prime is wrong.
So, \sqrt {3}
3
is an "irrational number". Hence proved.