Download Solution 4 - WUSTL Math

Solutions to Homework 4, Math 310
(1) Give a direct proof to show that if a, b are integers which are
squares of integers, then ab is the square of an integer.
Proof. We show that if a, b are integers which are squares of
integers, then ab is the square of an integer.
We may write a = m2 , b = n2 for some integers m, n by
hypothesis. Then ab = m2 n2 = (mn)2 and since mn is an
integer, by closure, we see that ab is the square of the integer
mn. This proves the result.
(2) Write the converse to the statement appearing in the first problem. Is the converse true or false? If true, give a proof and if
false, give a counterexample.
The converse to the statement above is, if a, b are integers
and ab is the square of an integer then so are a and b.
The converse is false. For example, you may take a = b = 3,
then ab = 9 is a square but neither a nor b is a square.
(3) Give a constructive proof to show that the equation, x5 − x4 +
x3 − x2 + x − 1 = 0 has a solution in integers.
Proof. We show that the equation x5 − x4 + x3 − x2 + x − 1 = 0
has a solution in integers.
One easy way to do this is to notice that 1 is a solution of
the equation. Substituting 1 for x in the equation, we have,
15 − 14 + 13 − 12 + 1 − 1 = 1 − 1 + 1 − 1 + 1 − 1 = 0.
Another way could be to write the expression as follows:
f (x) = x5 − x4 + x3 − x2 + x − 1
= (x5 − x4 ) + (x3 − x2 ) + (x − 1)
= x4 (x − 1) + x2 (x − 1) + (x − 1) distributivity
= (x4 + x2 + 1)(x − 1) distributivity
Thus we see that f (1) = (14 + 12 + 1)(1 − 1) = 3 × 0 = 0. So,
1 is a solution of the given equation.
(4) Using results you have studied, show that if x, y are real numbers, then | xy |≤| x2 + y 2 |.
Proof. Any two real numbers can be considered as the coordinates of a point in the plane and thus from trigonometry, we
may assume that they are of the form x = r cos θ, y = r sin θ
1
2
for some real number r ≥ 0 and an angle θ. Thus,
| xy | =| r cos θ · r sin θ |
=| r2 cos θ sin θ |
= r2 | cos θ sin θ |
(1)
≤ r2
since | cos θ |≤ 1 and | sin θ |≤ 1 and so their product is atmost
one.
On the other hand, we have,
| x2 + y 2 | =| r2 cos2 θ + r2 sin2 θ |
= r2 | cos2 θ + sin2 θ |
=r
(2)
2
since cos2 θ + sin2 θ = 1. Now, using equations (1) and (2), we
get that | xy |≤| x2 + y 2 |.
(5) Use intermediate value theorem (and thus a non-constructive
proof) to show the following: A hiker walks up a hill starting
at 6am and reaching the top at 6pm. He walks back from the
top to the bottom starting at 6am and reaching the bottom at
6pm along the same path. Show that there exists some point
in his path where he was at the same time, going up or down.
Proof. Let us denote by the variable t the time and by the
variable x the distance along the path starting from the bottom
of the hill. Then we have two continuous functions, f, g : I =
[6am, 6pm] → [0, A], where A is the distance to the top of the
hill, defined as follows. f (t) for t ∈ I is the distance from the
bottom the hiker has travelled on his way up. Similarly, g(t) is
the distance from the bottom the traveller reached on his way
down at time t. Then h(t) = f (t) − g(t) is continuous. We also
have,
h(6am) = f (6am) − g(6am) = 0 − A = −A
h(6pm) = f (6pm) − g(6pm) = A − 0 = A
By intermediate value theorem, therefore there exists a t ∈ I
so that h(t) = 0. This means that at this time t, f (t) = g(t)
or in other words that the hiker was at the same distance from
the bottom on his way up and down at time t. This establishes
the result we set out to show.
3
I make a remark about problem solving. In the above, we
have been told that we are supposed to use the Intermediate
value theorem. We know that this theorem is a theorem about
continuous functions. So, from the data given, we should somehow create a continuous function to make use of the theorem.
Any function must have an input and an output. In our problem, there are not too many choices. More or less the only
function is the distance travelled in time t. Once you reach this
conclusion, the rest is straight forward.
(6) Show that if a, b are postive real numbers and ab 6= (a + b)2 /4,
then a 6= b.
Proof. We will prove the contrapositive of the above statement.
Since the negation of a 6= b is a = b and ab 6= (a + b)2 /4 is ab =
(a + b)2 /4, the contrapositive is: If a = b then ab = (a + b)2 /4.
So, assume that a = b. Then ab = a2 and (a+b)2 /4 = (2a)2 /4 =
a2 . Thus, we get ab = (a + b)2 /4, establishing the result.
(7) (a) If y is an irrational number and x 6= 0 is a rational number,
show that xy is irrational.
Notice that this statement can be written as,
((y, irrational) ∧ (0 6= x ∈ Q)) ⇒ (xy is irrational).
This can also be written as
(∀y, irrational)(∀x ∈ Q, x 6= 0)(xy irrational).
Why is the above not same as
((∀y, irrational) ∧ (∀x ∈ Q, x 6= 0)) ⇒ (xy irrational)?
Proof. We will prove this statement by contradiction. The
negation of the statement is, there exists an irrational number y and a rational number x 6= 0 such that xy is rational.
So, we assume this and reach a contradiction. If x is rational, we can write x = p/q where p, q are integers and
q 6= 0 by definition of a rational number. Since x 6= 0,
we see that p 6= 0. But, then x−1 = q/p is also a rational
number, by definition of a rational number. Then, we can
write y = x−1 (xy) and if xy is rational, since rational numbers are closed under multiplication, we get that y is also
rational. This is a contradiction which proves our result.
(b) If x is a real number and y is an irrational number, show
that either x + y or −x + y is irrational.
4
Proof. We will show this by contradiction. So, assume the
above statement is false. Then writing the negation, we
see that we have a real number x, an irrational number
y with both x + y and −x + y rational. Since the set
of rational numbers is closed under addition, we see by
adding, (x + y) + (−x + y) = 2y is rational. Since the
product of two rational numbers is also rational, we see
that y = 2y × 12 is also rational. This contradicts our
assumption that y is irrational. This proves our result. (c) Give an example of x, y as before, so that both x + y and
−x + y are irrational.
Taking any irrational number y and letting x = 2y, we see
that x + y = 3y is irrational, using the first part of the
problem. Similarly, −x + y = −y is also irrational.
(d) Give an example of x, y as before so that one of x+y, −x+y
is irrational and the other is rational.
For, this we may take y to be any irrational number and
let x = y. Then x + y = 2y is irrational, but −x + y = 0 is
rational.
(8) If a, b, c are integers with a2 + b2 = c2 , then show that either a
or b is even.
Proof. We will prove this by contradiction. If the statement is
false, we have odd integers a, b with a2 + b2 = c2 for an integer
c.
Writing a = 2m + 1, b = 2n + 1 for some integers m, n, by
definition of odd numbers, we get,
a2 + b2 = (2m + 1)2 + (2n + 1)2
= 4m2 + 4m + 1 + 4n2 + 4n + 1
= 2(2m2 + 2n2 + 2m + 2n + 1)
= 2(2(m2 + n2 + m + n) + 1)
From this we see that c2 = a2 + b2 is even. We know that if
c is even, then so is c. So, writing c = 2d for an integer d, we
see from the above,
2
c2 = 4d2 = 2(2(m2 + n2 + m + n) + 1).
Cancelling a 2, we get,
2d2 = 2(m2 + n2 + m + n) + 1.
5
But the left hand side above is even and the right hand side
is odd. This is the contradiction we desire, proving the result.