HW2Notes

# HW2Notes - Notes on Homework #2 Exercise 1.4.3 We are given...

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Notes on Homework #2 Exercise 1.4.3 We are given real numbers a < b and we have to show that there exists an irrational number t with a < t < b . Let a 0 = a = 2 and b 0 = b = 2. Since a 0 < b 0 we can apply Theorem 1.4.3 to justify saying that there is a rational number r with a 0 < r < b 0 , in other words a - 2 < r < b - 2. Adding 2 to both sides we have that a < r + 2 < b. Now t = r + 2 is the sum of rational and irrational numbers, so by Exercise 1.4.2 t is irrational. Thus there does exist an irrational number t with a < t < b . Exercise 2.2.1 b) First, observe that ± ± ± ± 3 n + 1 2 n + 5 - 3 2 ± ± ± ± = ± ± ± ± 2(3 n + 1) - 3(2 n + 5) 2(2 n + 5) ± ± ± ± = ± ± ± ± - 13 4 n + 10 ± ± ± ± = +13 4 n + 10 . (The absolute value is important — without it you get an expression which is negative and so will always be < ± .) From here we see that the n we want are those which are n > (13 ± - 1 - 10) / 4. So we want to use an N with N > (13 ± - 1 - 10) / 4. Here is what the ﬁnished solution should look like: Consider any ± > 0. By the Archimedian Property there exists a natural number
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## This note was uploaded on 11/07/2009 for the course MATH 3224 at Virginia Tech.

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