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D)S (−1)a = −a. a/b ad ad = = . c/d bc bc c ad + bc a . (f)S If b = 0 and d = 0, then + = b d bd (e) If b, d = 0, then 2. Let r, s ∈ Q. Assuming that Z is closed under addition and multiplication, prove that r ± s, rs, r/s ∈ Q, the last provided that s = 0. S If r = 0 ∈ Q and x ∈ I, prove that r ± x, rx, r/x ∈ I. 4. Let n ∈ N. Prove the following identities without using mathematical induction: n (a) ⇓ x − y = (x − y) 3 n xn−j y j−1 . n j=1 n (b) xn + y n = (x + y) (−1)j−1 xn−j y j−1 if n is odd.

Let {an } and {bn } be sequences with an → a and bn → b. If an ≤ bn for infinitely many n, then a ≤ b. Proof. Suppose b < a. Then b < (a + b)/2 < a, hence we may choose indices N1 and N2 such that bn < (a + b)/2 for all n ≥ N1 and an > (a + b)/2 for all n ≥ N2 . But then bn < an for all n ≥ max{N1 , N2 }, contradicting the hypothesis. Note that, as a consequence of the preceding theorem, a convergent sequence in a closed interval I must have its limit in I. 5 Theorem (Squeeze principle). Let {an }, {bn }, and {cn } be sequences in R such that an ≤ bn ≤ cn for all n.

Let {an } and {bn } be sequences in R. The following limit properties hold in R in the sense that if the expression on the right side of the equation exists in R, then the limit on the left side exists and equality holds. (a) limn (san + tbn ) = s limn an + t limn bn , s, t ∈ R. (b) limn an bn = limn an limn bn . (c) limn an /bn = limn an / limn bn , if limn bn = 0. (d) limn |an | = | limn an |. √ √ (e) limn an = limn an if an ≥ 0 for all n. Proof. Let an → a, bn → b. We prove the theorem first for the case a, b ∈ R.

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