Download HW 2 Solutions

MC215 – 2/18/14 –SOLUTIONS TO HOMEWORK #2 – 64 POINTS TOTAL
1. Consider the following claim:
The square of any odd integer can be written as + for some integer .
a. Reformulate this claim as an "If-then" statement.
b. For a direct proof of this, what do you assume? As we do in class, please state your
assumption in complete detail using variables.
c. For a direct proof of this, what must you prove? Again, please state this in complete
detail using variables.
d. Give a direct proof of this claim. You should give a "final version" proof, written as a
paragraph or short essay that proves the claim rigorously, and which also is written in a
clear, grammatically correct style, understandable to the reader.
SOLUTION TO #1: (16 pts: a-c 3 each, d 7)
a. If n is an odd integer then = 8 + 1 for some integer m.
b. Assume: n is an odd integer. This means that there is an integer k such that = 2 + 1.
c. Prove: = 8 + 1 for some number .
d. Proof. Let be any odd integer, i.e., = 2 + 1 for some integer . It follows from = 2 + 1
that = 2 + 1 = 4 + 4 + 1. We can rewrite 4 + 4 + 1 = 4 + 1 + 1. Since and + 1 are consecutive integers, one is even and the other is odd, hence + 1 = 2 for
some integer . It follows that = 4 + 1 + 1 = 8 + 1, proving the claim. (Another way
to do this would be to use two cases, depending on whether is even or odd.)
2. Consider the following claim about real numbers x, y and z:
If + ≤ , then ≤ / or ≤ /.
a. For a proof by contrapositive of this claim, what do you assume and what must you prove? As
in parts 1b and 1c, please state in complete detail.
b. Give a proof by contrapositive of this claim. As in part 1d, you should give a "final" version of
your proof.
c. State the converse of this claim as simply as possible. Is it true? Either prove it's true or give
a counterexample to prove it's false.
SOLUTION TO #2: (15 pts: a-4, b-5, c-6)
a. The contrapositive of this statement is: For any real numbers , , and , if > / and > /
then + > . Hence we assume that > / and > /, and we prove that + > .
b. Proof. We prove the contrapositive, so suppose , , and are real numbers and > /2 and
> /2. Then adding these two inequalities yields the desired result, + > .
c. The converse of this claim is: For any real numbers , , and , If ≤ / or ≤ /, then
+ ≤ . Although there are some values of , , for which the converse is true, in general it
is false. For example, = 1, = 10, = 3 satisfy the condition (since ≤ /2), but + =
11 > 3 = .
3. Use proof by contradiction to prove that the number ! " is irrational. Hint: If we let = log & 7, we want to prove that is irrational. But the equation = ! " can be rewritten
as 3( = 7 (that's the definition of log & 7). Thus you can prove (by contradiction) that if ! = ",
then is irrational. For this problem, you should give only the final proof.
SOLUTION TO #3: (8 pts)
Proof. As mentioned in the problem, we let = log & 7, which is equivalent to 3( = 7, and we show
by contradiction that must be irrational. So assume instead that is rational. Then there are
integers ), *, with * ≠ 0, such that = )/*. Thus 3,/- = 7. Raising both sides of this equation to
the power * gives 3, = 7- . This says that the number 3, , which is factored into primes, equals the
number 7- , which is factored differently into primes. The Fundamental Theorem of Arithmetic says
that every integer > 1 can uniquely factored into primes. But this number is not 1, since * > 0. This
is a contradiction, and so we conclude that is irrational.
4. Prove the following “if and only if” statement. As we discussed in class, an if and only if
statement is equivalent to two If-Then statements. Prove this statement as two separate IfThen statements. For each one, (a) state clearly what the if-then statement is, and (b) say which
type of proof you are doing (direct, contrapositive, or contradiction), and then (c) give the
proof.
An integer . is even if and only if !. + / is odd.
SOLUTION TO #4: (10 pts: 5 for each proof)
The two if-then statements are: (i) If an integer is even, then 3 + 5 is odd, and (ii) If 3 + 5 is
odd, where is an integer, then is even.
(i)
(ii)
If an integer . is even, then !. + / is odd. Proof. We give a direct proof, so suppose is
an even integer. Then = 2 for some integer , and it follows that 3 + 5 = 6 + 5 =
23 + 2 + 1, demonstrating that 3 + 5 is odd.
If !. + / is odd, where . is an integer, then . is even. Proof. We prove the
contrapositive, which says that if is an odd integer, then 3 + 5 is even. So assume is
odd, so that = 2 + 1, for some integer . Then 3 + 5 = 6 + 8 = 23 + 4,
showing that 3 + 5 is even.
5. This problem entails making a conjecture and then proving that it is true.
a. Write down a list of the first 10 or so perfect squares, more if you need them, and look at
the first digit of each one (e.g., the first digit of 92 = 81 is 1). You will begin to notice that
not all of the 10 possible digits (0, 1, 2, …, 9) seem to appear. Based on your observation,
complete the statement of the following conjecture:
If an integer 2 is a perfect square,
then the first digit of 2 is one of the following digits:
b. Prove that the completed conjecture is true. Hint: If 3 is the first digit of an integer ,
where 0 ≤ 3 ≤ 9, then = 10 + 3, for some integer . What does this say about the
square of ?
c. Prove that if we look at perfect cubes, instead of perfect squares, then the first digit can
be anything from 0 to 9.
SOLUTION TO #5: (15 pts: a-3, b-8, c-4)
a. If we write down the first 10 or so perfect squares, the first digit seems to always be one of
0, 1, 4, 5, 6, 9. So our conjecture is:
If an integer 2 is a perfect square,
then the first digit of 2 is one of the following digits: 0, 1, 4, 5, 6, 9
b. Proof: Let ) be a perfect square, which means that ) = , for some integer . Let 3 be the
remainder when is divided by 10. Then = 10 + 3, where and 3 are integers, and
0 ≤ 3 ≤ 9. Thus ) = = 10 + 3 = 100 + 203 + 3 . The first digit of ) will equal the
first digit we get by adding up these three summands. The first two terms of the sum are
multiples of 10, so they both add 0 to the first digit. Thus ) = and 3 have the same first
digit. But there are only 10 possible values for 3 : 0 = 0, 1 = 1, 2 = 4, 3 = 9, 4 = 16, 5 =
25, 6 = 36, 7 = 49, 8 = 64, and 9 = 81. By inspection this confirms that the first digit of ),
which is the same as that of 3 , can only be 0, 1, 4, 5, 6,or 9.
c. To see that the first digit of a perfect cube can be anything from 0 to 9, we just give ten perfect
cubes have all the digits from 0 to 9 as their first digits: 0& = 0, 1& = 1, 2& = 8, 3& = 27, 4& =
64, 5& = 125, 6& = 216, 7& = 343, 8& = 512, and9& = 729. By inspection we see that all the
integers from 0 to 9 can be the first digit of a perfect cube.