Prove that 3+√7 is an irrational number
Answers
Answer:
Here is ur answer
Step-by-step explanation:
Assume rt 7 is rational
rt 7 = a/b
in reduced form
Squarificate it
(rt7)^2 = (a/b)^2
7 = a^2 / b^2
a^2 = 7 b^2
But if a^2 is a multiple of 7 then it is a multiple of 49, and 'a' is a multiple of 7
a = 7c
(7c)^2 = 7 b^2
49c^2 = 7b^2
7c^2 = b^2
Same argument as before with 'a' being a multiple of 7, we see that b is a multiple of 7 now.
But that means a/b = 7c/ 7d = c/d is further reduced, thus contradicting our premise.
Conclusion...
rt7 is irrational.
3 + rt 7 is therefore also irrational
I hope it will help you a lot
ANSWER
We have to prove that 3+
7
is irrational.
Let us assume the opposite, that 3+
7
is rational.
Hence 3+
7
can be written in the form
b
a
where a and b are co-prime and b
=0
Hence 3+
7
=
b
a
⇒
7
=
b
a
−3
⇒
7
=
b
a−3b
where
7
is irrational and
b
a−3b
is rational.
Since,rational
= irrational.
This is a contradiction.
∴ Our assumption is incorrect.
Hence 3+
7
is irrational.
Hence proved.