# Well-Ordering and Division Theorem

32. Show that there is no rational number b/a whose square is 2, as follows: if b^2 = 2a^2, then b is even, so b = 2c, so, substituting and cancelling 2, 2c^2 = a^2. Use that argument and well-ordering to show that there can be no natural number a > 0 with b^2 = 2a^2 for some natural number b.

33. Let m be the least common multiple of a and b, and let c be a common multiple of a and b. Show that m divides c. Hint: use the division theorem on m and c, and show that the remainder r is a common multiple of a and b, hence r = 0.

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Let's us review the concept of well ordering principle: if a subset of integers is bounded below, then it has a minimal element. With this in mind, let us do these problems:

32. Suppose otherwise, let b/a = âˆš2, then b2 = 2a^2. The given argument tell us

you can always find b = 2c such that a^2 = 2c^2. The trick is to notice that a/c = âˆš2, and a < b.

Let us use well ordering principle: choose b/a = âˆš2 such that b is the smallest (please think about why we can apply the well ordering principle: think about the following set {x âˆˆ N : x/y =âˆš2}, is it a subset of integers? is it bounded below?). Then apply the given argument, we find another c such that a/c = âˆš2 where a < b. This contradicts our assumption that b is the smallest.

33. Division algorithm says: given x, p, where p 6= 0, there exists a, b such that x = a Â· p + b, and 0 â‰¤ b < p. Here, x, p, a, b are all integers. Now given m, c apply division algorithm to get c = x Â· m + r where 0 â‰¤ r < m.

Suppose m does not divide c, then r 6= 0. Rewrite it as r = c âˆ’ x Â· m, where x is an integer. Now a, b both divides c, m. Think about why it follows that a, b also divides r.

Now m is the least common multiple of a, b, r is another common multiple of a, b with 0 < r < m. So we get a smaller common multiple which gives us a contradiction (Recall the least common multiple is the smallest positive natural number that can be divided by both a, b).

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