Prove that 7 + 3√2 is an irrational number.
Answers
Answer:
SOLUTION :
Let us assume that ( 7 + 3√2 ) is a rational number
so,
\bf 7+ 3\sqrt{2} = \frac{p}{q} \:\:\:\:\:[in\:\: which\:\:'p'\:and \:'q'\:\:are\:\:integers \:\:and\:\: q\neq 0]7+3
2
=
q
p
[inwhich
′
p
′
and
′
q
′
areintegersandq
=0]
\bf \rightarrow 3\sqrt{2}=\frac{p}{q}-7→3
2
=
q
p
−7
\bf \rightarrow 3\sqrt{2}=\frac{p-7q}{q}→3
2
=
q
p−7q
\bf \rightarrow \sqrt{2}=\frac{p-7q}{q} \times \frac{1}{3}→
2
=
q
p−7q
×
3
1
\bf \rightarrow \sqrt{2}=\frac{p-7q}{3q}→
2
=
3q
p−7q
Since,
(p), (q), (7), (3) are integers and q ≠ 0 so,
\bf \frac{p-7q}{3q}
3q
p−7q
is a rational number
Therefore, √2 is also a rational number which is not possible
√2 is an irrational number
∴ This contradiction arise due to our wrong assumption.
Hence,
7+3√2 cannot be a rational number,
∵ it is an irrational number.
Hence Proved
Answer:
proved
Step-by-step explanation:
7+3√2
7+3;bibuihxdehibcfhueqbci