/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Free solutions & answers for Yet Another Introduction to Analysis Chapter 1 - (Page 2) [step by step] | 91Ó°ÊÓ

91Ó°ÊÓ

Problem 13

Let \(x\) and \(y\) be real numbers with \(y>x+1 .\) By considering the largest member of the set $$ \\{n \in \mathbb{Z}: n

Problem 14

We shall now try to establish that between any two different numbers there is a rational number. To try to see first an informal verification of this fact imagine for a moment that the two numbers are more than \(1 / 101\) apart. Then surely at least one of the fractions \(\cdots \frac{-3}{101}, \frac{-2}{101}, \frac{-1}{101}, 0, \frac{1}{101}, \frac{2}{101}, \frac{3}{101}, \ldots\) illustrated below would lie between the two numbers? Formally suppose that \(x\) and \(y\) are two real numbers with \(x1 /(y-x)\). (How do we know that such an integer exists?) Use the previous exercise to show that there exists an integer \(M\) with \(N x

Problem 15

Prove by induction that for each positive integer \(n\) (i) \(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\cdots+\frac{1}{n(n+1)}=1-\frac{1}{n+1}\) (ii) \(1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1)\)

Problem 16

Prove by induction that for each positive integer \(n\) (i) \((1+x)^{n} \geqslant 1+n x\) where \(x\) is any number with \(x \geqslant-1\). (Think carefully about where in your solution you need the fact that \(x \geqslant-1\).) This result is known as Bernoulli's inequality and is credited to Jacob Bernoulli, one of a great Swiss mathematical family of the seventeenth and eighteenth centuries. (ii) \((1+x)^{n}=\left(\begin{array}{l}n \\\ 0\end{array}\right)+\left(\begin{array}{l}n \\ 1\end{array}\right) x+\left(\begin{array}{l}n \\ 2\end{array}\right) x^{2}+\cdots+\left(\begin{array}{l}n \\ r\end{array}\right) x^{\prime}+\cdots+\left(\begin{array}{l}n \\ n\end{array}\right) x^{n}\) where \(x\) is any number and \(\left(\begin{array}{l}n \\ r\end{array}\right)\) denotes the 'binomial coefficient' and equals \(n ! /(r !(n-r) !)\). (You will need the fact that $$ \left(\begin{array}{l} k-1 \\ r-1 \end{array}\right)+\left(\begin{array}{c} k-1 \\ r \end{array}\right)=\left(\begin{array}{l} k \\ r \end{array}\right) $$ which you can easily verify for yourselves.) This result is known as the binomial theorem.

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