Ii before we decide whether the number is irrational

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(ii) Before we decide whether the number is irrational, let us reduce the fraction under the square root: 54 150 = 27 · 2 75 · 2 = 27 75 = 9 · 3 25 · 3 = 9 25 . 11
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Now we can simplify: 54 150 = 9 25 = 9 25 = 3 5 . Thus 54 150 is rational . (iii) Suppose Q := 3 π is rational. Then π = Q/ 3 will be rational which is a contradiction. So 3 π is irrational . (iv) - 2 / 17 is the ratio of two integers, so - 2 / 17 is rational . The correct answer is (e) . Solution of problem 1.9: (1) is True. We can construct a one-to-one cor- respondence all natural numbers ending with 7 { all negative integers } explicitly by pairing: a natural number n ending with 7 the negative integer - ( n - 7) / 10 . (2) is False. If we have a one-to-one correspondence f between A and B and we have a one-to-one correspondence g between A and C , then we can construct a one-to-one correspondence between B and C . Indeed, let x be a number in B , then f pairs b with a unique element y in A . On the other hand g will pair y with a unique element z in C . So we can define a correspondence h between B and C by pairing x and z . The correspondence h will be automatically a one-to-one correspondence because of the uniquencess of the pairings f and g . 12
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(3) is True. A rational numbers whose denominator is a power of 3 can be written uniquely as a reduced fraction m 3 n where n is an integer with n 0, and m is an integer that is not divisble by 3. Such a number is between 0 and 1 if and only if 0 m 3 n . So for instance the numbers whose numerators are all equal to 1 and whose denominators are powers of 3 are in our set. But we can easily construct a one-to-one correspondence all natural numbers of the form 1 3 n , n = 1 , 2 , . . . { all natural numbers } by pairing 1 / 3 n with n . Therefore the set of all numbers of the form 1 / 3 n is infinite. But our set contains the set of all of the form 1 / 3 n is infinite, so it also must be infinite. Solution of problem 1.10: The sets A and B have the same cardinality: we can match a natural number n with the number that is made out of n digits, each of which is equalto 5. The set C has a bigger cardinality. We can see this by Cantor’s diagonalization argument. Indeed, suppose that we can find a one-to-one correspondence between the set C and all natural numbers. Let c n denote the number in C that corresponds to the natural number n . We will construct a number a in C that is not equal to any one of the c n ’s. Define a to be the number between 0 and 1 for which: n -th digit of a after the decimal point =
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