4. Represent √10 and √20 on the number line.
5. Prove that the following numbers are irrational.
(1) 2+ √3
(2)3-√5
(3) √2+√5
Answers
Step-by-step explanation:
5. Prove that the following numbers are irrational.
(1) 2+ √3
Solution -
Let's assume 2+ √3 is a rational number.
Rational number = if a and b are in the form of a/b where a and b are integers and b is is not equal to 0
2+ √3= a/b
√3=a/ b-2
Rational number =a/b =a/b-2
So, √3 is also rational number
But we know that √3 is an irrational number
So our contradiction is wrong
Hence 2+ √3 is an irrational number. (proved)
(2)3-√5
Solution -
Let's assume 3-√5 is a rational number.
Rational number = if a and b are in the form of a/b where a and b are integers and b is is not equal to 0
3-√5=a/b
-√5=a/b-3
Rational number =a/b =a/b-3
So, √5 is also rational number
But we know that √3 is an irrational number
So our contradiction is wrong
Hence 3- √5 is an irrational number. (proved)
(3) √2+√5
Solution -
Let us assume that √2+√5 is a rational number.
A rational number can be written in the form of p/q where p,q are integers and q≠0
√2+√5 = p/q
On squaring both sides we get,
(√2+√5)² = (p/q)²
√2²+√5²+2(√5)(√2) = p²/q²
2+5+2√10 = p²/q²
7+2√10 = p²/q²
2√10 = p²/q² – 7
√10 = (p²-7q²)/2q
p,q are integers then (p²-7q²)/2q is a rational number.
Then √10 is also a rational number.
But this contradicts the fact that √10 is an irrational number.
Our assumption is incorrect
√2+√5 is an irrational number.
Hence proved.