Methods of Proof
1.Assume the contrary, namely that√
2 +
√
3 +
√
5 =r, whereris a rational number.
Square the equality√
2 +
√
3 =r−√
5 to obtain 5+ 2√
6 =r^2 + 5 − 2 r√
- It follows
that 2
√
6 + 2 r√
5 is itself rational. Squaring again, we find that 24+ 20 r^2 + 8 r√
30
is rational, and hence√
30 is rational, too. Pythagoras’ method for proving that√
2is
irrational can now be applied to show that this is not true. Write√
30 =mn in lowest
terms; then transform this intom^2 = 30 n^2. It follows thatmis divisible by 2 and because
2 (m 2 )^2 = 15 n^2 it follows thatnis 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
primespgreater than 7, at most one of the numbers can be divisible byp, 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 numbers2 , 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.