Prove that 3
5 – 3
7
is an irrational number.
Answers
Step-by-step explanation:
let it be a rational number
5 - \frac{3}{7} \sqrt{3} = \frac{a}{b}5−
7
3
3
=
b
a
= > - \frac{3}{7} \sqrt{3 } = \frac{a}{b} - 5=>−
7
3
3
=
b
a
−5
= > \frac{3}{7} \sqrt{3} = \frac{5b - a}{b}=>
7
3
3
=
b
5b−a
\sqrt{3} = \frac{7(5b - a)}{3b}
3
=
3b
7(5b−a)
it seems that √3 is rational
so our assumption get wrong because √3 is irrational number so 5-3/7√3 is irrational
ok bro
Answer:
We have to prove that 3+7is 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 = ba⇒
7 = ba−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.