Question

If xth position in a pattren of numbers is given by the expression '8x-5' then find its 5th term.​
Explain with all the steps

Answer 1

Answer:

ssume the contrary, namely that √2 + √3 + √5 = r, where r is a rational number.

Square the equality √2 + √3 = r − √5 to obtain 5 + 2

√6 = r2 + 5 − 2r

√5. It follows

that 2√6 + 2r

√5 is itself rational. Squaring again, we find that 24 + 20r2 + 8r

√30

is rational, and hence √30 is rational, too. Pythagoras’ method for proving that √2 is

irrational can now be applied to show that this is not true. Write √30 = m

n in lowest

terms; then transform this into m2 = 30n2. It follows that m is divisible by 2 and because

2( m

2 )2 = 15n2 it follows that n is divisible by 2 as well. So the fraction was not in lowest

terms, a contradiction. We conclude that the initial assumption was false, and therefore

√2 + √3 + √5 is irrational.

2. Assume that such numbers do exist, and let us look at their prime factorizations. For

primes p greater than 7, at most one of the numbers can be divisible by p, and the partition

cannot exist. Thus the prime factors of the given numbers can be only 2, 3, 5, and 7.

We now look at repeated prime factors. Because the difference between two numbers

divisible by 4 is at least 4, at most three of the nine numbers are divisible by 4. Also, at

most one is divisible by 9, at most one by 25, and at most one by 49. Eliminating these at

most 3 + 1 + 1 + 1 = 6 numbers, we are left with at least three numbers among the nine

that do not contain repeated prime factors. They are among the divisors of 2 · 3 · 5 · 7,

and so among the numbers

2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210.

Because the difference between the largest and the smallest of these three numbers

is at most 9, none of them can be greater than 21. We have to look at the sequence

1, 2, 3,..., 29. Any subsequence of consecutive integers of length 9 that has a term

greater than 10 contains a prime number greater than or equal to 11, which is impossible.

And from 1, 2,..., 10 we cannot select nine consecutive numbers with the required

property. This contradicts our assumption, and the problem is solved.