Download Math 335 Homework Set 4

Math 335 Homework Set 4
Exercise 3.2.9. The one-to-one correspondence is x 7→ 6x going from N to 6N and y 7→ y/6
going in reverse from 6N to N.
Exercise 3.2.30. Yes, the collection of individual candy bars has the same cardinality as
the set of natural numbers. We can send package k’s contents to the numbers 2k − 1 and
2k. For example, when k = 1, we send the contents to 1 and 2. To go in reverse, we send
the odd number 2k − 1 to the first candy bar in package k and we send the even number 2k
to the second candy bar in the same package.
Exercise 3.2.31. Yes, we have infinitely many nuts. In the first n piles we can find 1 +
2 + . . . + n = sn nuts. The set of nuts has the same cardinality as the set of the natural
numbers. To see the correspondence, just put all the nuts straight in a line from the first
pile onwards. Then write the natural numbers next to each nut. We could be more precise
though. There are sn−1 nuts going from pile 1 all the way to pile n, not included. Then we
send the nuts 1, 2, . . . n from pile n to sn−1 + 1, sn−1 + 2, . . . sn−1 + n = sn .
Exercise 3.3.12. For the first number we take the natural numbers and place them all in a
line in a consecutive manner. The string of digits becomes our decimal number. For number
k of the sequence we take the natural numbers, multiply each of them by k and place them
in the same manner, with the digits forming our k-th decimal number. For the diagonal
argument we can take the k-th digit of the k-th decimal number and switch it to something
else. The sequence that we form in this way differs from the k-th decimal number in the
k-th digit so it’s not on the list.
Exercise 3.3.13. If the set of all such colorings had the same cardinality as the natural
numbers we would be able to put them all in a list. Using the diagonal argument we can
construct a sequence of red and blue circles that is not on the list. Take the k-th circle
from the k-th colored sequence from the list and flip its color from R to B or from B to R
depending on what it is. This new colored circle is the k-th entry of our new sequence. Since
the sequence disagrees with the k-th sequence at the k-th circle, this new sequence is not on
the lisdt. Therefore we have shown that the set of colorings of the circles cannot be put into
one-to-one correspondence with the set of natural numbers and must therefore have a larger
cardinality.
Exercise 3.3.14. No, it doesn’t. This is really the same question as question 3.3.13 except
that we don’t have the letters R and B but H and T.
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