A Treatise on Solid Geometry by Percival Frost Joseph Wolstenholme

By Percival Frost Joseph Wolstenholme

This Elibron Classics e-book is a facsimile reprint of a 1863 version by way of Macmillan and Co., Cambridge - London.

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5. If a, b, c and d are positive integers such that a < b and c < d, prove that ac < bd. Solution “Now I see,” said Tweedledee. “We have a + k = b and c + = d for some positive integers k and . It follows that bd = (a + k)(c + ) = ac + (a + ck + k ) since multiplication is distributive over addition. Now all of a, c, k and are positive integers. By the Closure Property for multiplication of positive integers, a , ck and k are all positive integers. By the Closure Property for addition of positive integers, a + ck + k is also a positive integer.

1 Exercises 1. Test whether 778562140 is divisible by (a) 5; (b) 2; (c) 4; (d) 8. 2. Test whether 824112284 is divisible by (a) 5; (b) 2; (c) 4; (d) 8. 3. For the test of divisibility of 64, how many digits at the end of the given large number must be examined? 4. For the test of divisibility of 25, how many digits at the end of the given large number must be examined? 5. In a stationery store, the cost of every item was an integral number of cents. Carson bought 5 pencils at 12 cents each, 1 pen at 50 cents, as well as 10 notebooks.

Second, b divides c. This means that c = b for some positive integer . Hence c = b = (3k) = 3(k ). By the Closure Property for multiplication of positive integers, k is a positive integer. By definition, 3|c. We are finally in position to introduce the last of the three basic properties. Transitive Property. Let a, b and c be positive integers. If a|b and b|c, then a|c. Proof. We are given two conditions. First, a divides b. This means that b = ak for some positive integer k. Second, b divides c.

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