prove that 3 root 2×root 8 is a rational number
Answers
Answer:
Answer
Let
8
be a rational number
Then
8
=
n
m
where m, n are integers and m, n are expenses and n
=0
⇒m=
8
n
Squaring both sides we get
m
2
=8n
2
⇒
8
m
2
=n
2
……………(iv)
⇒8 divides m
2
i.e., 8 divides m
Then m can be written as
m=8k for some integer k.
Substituting value of m in (iv) we get
⇒
3
(8k)
2
=n
2
⇒8k
2
=n
2
⇒k
2
=
8
n
2
⇒8 divides n
2
i.e., 8 divides n
Thus we get that 8 is a common factor of m and n but m and n are co-primes which is a contradiction to our assumption.
Hence
8
is an irrational number.
Now consider
3
+
8
to be an rational number
Then
3
+
8
=
b
a
where a, b are integers, co-primes and b
=0
⇒
3
=
b
a
−
8
Squaring both sides we get
⇒3=
b
2
a
2
+(
8
)
2
−2
b
a
8
⇒3=
b
2
a
2
+8−
b
2a
8
⇒3−
b
2
a
2
−8=−
b
2a
8
⇒
b
2
3b
2
−a
2
−8b
2
=
b
−2a
8
⇒
8
=
−2ab
(3b
2
−a
2
−8b
2
)
⇒
8
=
2ab
a
2
+8b
2
−3b
2
⇒
8
=
2ab
a
2
+5b
2
Where a
2
+5b
2
is an integer and 2ab is also an integer, but
8
is an irrational number, which is a contradiction.
Hence
3
+
8
is an irrational number.