Given that √3 is an irrational number, show that (5+2√3) is an irrational number.
OR
An army contingent of 612 members is too much behind an army band of 48 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Let 5 + √3 be a rational number then it can be written in the
form ![]()
Where p and q are coprime integers.
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(p – 5q) and ‘q’ are integers, therefore RHS is a rational number
⇒ √3 is a rational number which is a contradiction.
∴ our assumption is wrong and 5 + 2√3 is an irrational number.
OR
Suppose, both groups are arranged in 'n' columns, for completely filling each column,
The maximum no of columns in which they can march is the highest common factor of their number of members. i.e. n = HCF(612, 48)
Now,
612 = 48 × 12 + 36
48 = 36 × 1 + 12
36 = 12 × 3 + 0
∴ n = HCF(612, 48) = 12
Couldn't generate an explanation.
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