Download Class Notes 10/14/05 - URI Math Department

Chapter 2: Section 2.1
We provide a direct proof for each of the statements in Theorem 1 about integers. Use the definition of the odds as integers
n that can be written in the form n = 2k + 1 for some integer k and
evens as those integers n that can be written in the form 2k for some
integer k to prove:
Theorem 1.
(1) The sum of 2 odds is even.
(2) The sum of 2 evens is even.
(3) The sum of an odd and an even is odd.
(4) The product of 2 odds is odd.
(5) The product of an even and an odd is even.
(6) The product of 2 evens is even.
Proof.
(2)
(3)
(4)
(5)
(6)
(1) Let m, n be two arbitrary odd numbers. Then there
exists two integers, k1 and k2 such that m = 2k1 + 1 and n =
2k2 + 1. Then m + n = 2k1 + 1 + 2k2 + 1. Regrouping, we get
m + n = 2(k1 + k2 ) + 2 = 2(k1 + k2 + 1) which is even because
it is of the form of 2 times the integer: k1 + k2 + 1.
complete this case on your own.
complete this case on your own.
complete this case on your own.
complete this case on your own.
complete this case on your own.
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We provide two examples of a proof by contradiction.
Theorem 2. If a, b, and c are odd integers, then ax2 + bx + c = 0 has
no rational solution.
Proof. Suppose x is a rational solution, where x = pq is in reduced form.
If both p and q were even, x would not be in reduced form, so at least
one of p and q is odd.
µ ¶
µ ¶2
p
p
+b
+c = 0
a
q
q
µ 2¶
µ ¶
p
p
a 2 +b
+c = 0
q
q
µ ¶
µ 2¶
p
p
2
q (a 2 + b
+ c) = q 2 · 0
q
q
ap2 + bpq + cq 2 = 0
Case 1. p even, q odd.
1
2
p2 is even, so a · p2 is even.
b · p is even, so bpq is even.
q 2 is odd, and so is cq 2 .
Thus, summing the first two terms of ap2 + bpq + cq 2 , we have an
even plus an even which is even and adding in the odd term to the even
term, we get an odd number.
But the right hand side of the equality is zero which is even.
We have reached a contradiction, because on the left we have an odd
and on the right we have an even. An odd number cannot equal an
even number.
Case 2. p odd, q even.
complete this case on your own.
Case 3. p odd, q odd.
complete this case on your own.
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Theorem 3. Among numbers y1 , y2 , . . . , yn , at least one is greater than
equal to the average of all n of them.
Proof. By contradiction. Suppose not.
n
The average can be expressed as y1 +y2 +···+y
. Call this number a.
n
By assumption, ∀i ∈ [n], yi < a.
That is, y1 < a, y2 < a, . . . , yn < a.
Adding up the Left hand sides of each inequality, we get y1 + y2 +
· · · + yn . Adding up the right hand sides of the inequalities, we get a,
n times, P
that is n · a.
Thus, ni=1 yi < n · a. We label this inequality (1).
But,
Pn
n
X
yi
a = i=1
and n · a =
yi .
n
i=1
Now, by (1), we have
n
X
i=1
yi <
n
X
yi .
i=1
We have reached an obvious contradiction. Thus, our assumption is
false which implies that the theorem is true.
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We prove an implication by proving the contrapositive instead.
Theorem 4. Let f (x) = mx + b, m 6= 0. If x 6= y then f (x) 6= f (y).
3
Proof. We will assume that f (x) = f (y) and show that x = y.
We assume mx + b = my + b. Subtracting b from both sides, we have
mx = my. Dividing both sides by m, (which we may do since m 6= 0)
we obtain, x = y.
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