Hence show that 5 – √3 is an irrational number.

Prove that √3 is a rational number. Hence show that 5 – √3 is an irrational number.

Solution:

If √3 is a rational number
Consider √3 = p/q where p and q are integers
q ˃ 0 and p and q have no common factor
By squaring both sides
3 = p²/q²
So we get
P² = 3q²
We know that
3q² is divisible by 3
p² is divisible by 3
p is divisible by 3
Consider p = 3 where k is an integer
By squaring on both sides
P² = 9k²
9k² is divisible by 3
p² is divisible by 3
3q² is divisible by 3
q² is divisible by 3
q is divisible by 3
Here p and q are divisible by 3
So our supposition is wrong
Therefore, √3 is an irrational number.
In 5 – √3
5 is a rational number
√3 is an irrational number (proved)
We know that
Difference of a rational number and irrational number is also an irrational number
So 5 – √3 is an irrational number.
Therefore, it is proved.

More Solutions:

• Find the value of p and q where p and q are rational numbers.
• Taking √2 = 1.414, √3 = 1.732, upto three places of decimal:
• If a = 2 + √3, find 1/a, (a – 1/a):
• Solve: If x = 1 – √2, find 1/x, (x – 1/x)4:
• Solve: If x = 5 – 2√6, find 1/x, (x2 – 1/x2):
• If p = (2-√5)/(2+√5) and q = (2+√5)/(2-√5):
• Find the value of x2 + 5xy + y2.
• Choose the correct statement:
• Between two rational numbers: