By Edwin Moise

Scholars can depend upon Moise's transparent and thorough presentation of easy geometry theorems. the writer assumes that scholars haven't any earlier wisdom of the topic and provides the fundamentals of geometry from the floor up. This finished procedure supplies teachers flexibility in educating. for instance, a complicated category may perhaps development swiftly via Chapters 1-7 and commit so much of its time to the cloth offered in Chapters eight, 10, 14, 19, and 20. equally, a much less complicated type may fit rigorously via Chapters 1-7, and forget a number of the more challenging chapters, comparable to 20 and 24.

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**Additional resources for Elementary Geometry from an Advanced Standpoint (3rd Edition) **

**Example text**

This is = alld = (a1) (1d) = ad. We define n-fold sums in exactly the same way and conclude by the same proof that n-fold sums satisfy the general associative law. That is, insertion of one pair of parentheses in an n-fold sum, a i + a2 + • • • + an, leaves the value of the sum unchanged. Finally we observe that we always have a(b, + b2+ • • • + bn) = ab1 + ab2+ • • • + abn . The proof is by an easy induction. For n = 1, we have ab1 = ab1. Given that a(b, + b2+ • • • + bn) = ab, + ab2+ - - • + abn , it follows that a(b, + b2+ • • • + bn + b,,+1) = a[(b, + b2+ • • • + bn) + bn+1] = a(b, + b2+ • • • + bn) + abn+, = (ab1 + ab2+ • • • + abn) + abn+i = ab1 + ab2+ • • • + abn + abn+, .

Here we are really appealing to a theorem in the theory of numbers, to be proved in Appendix B at the end of the book. 7 1. A positive integer n is even if n = 2k, where k is an integer; n is odd if n = 2/ + 1, where/ is an integer. Show that every positive integer is either even or odd. [Hint: Let S be the set of all positive integers which are either even or odd. What you need to show is that S = N. ] 2. Show that if n is odd, then n 2 is odd. 3. Show that if n 2 is even, then n is even. 4. Show that if = p/q, then p is even.

We would now like to prove the well-known fact that the rational numbers form a field. Under the scheme that we have been using in this chapter, the proof involves an unexpected difficulty. Following a procedure which is the reverse of the usual one, we have defined the positive integers in terms of the real numbers; and at the present stage we do not officially know that sums and products of integers are always integers. This can be proved, but we postpone the proof until the end of this chapter; in the meantime we regard the closure of the integers as a postulate.