Download CIS 260 Recitations 3 Feb 6 Problem 1 (Complete proof of example

CIS 260
Recitations 3
Feb 6
Problem 1 (Complete proof of example 11 on page 91) Show that there exist irrational
numbers x and y such that x y is rational.
First, we remind ourselves what the definition of a rational/irrational number is: A real
number r is rational is there exist integers p and q  0 such that r  p . A real number
q
that is not rational is called irrational.
Claim 1. If a 2 is even, then a is even.
[p: a 2 is even; q: a is even; We need to show that p  q  p  q . For a proof by
contradiction we assume that the statement we need to show is true is false and show that
this leads to a contradiction. So our assumptions is that (p  q ) is true, or equivalently
that p  q is true.]
Proof of claim 1: (by contradiction) Assume a is odd (and a 2 is even). Then by
definition a  2k  1 for some integer k . When we square both sides, we obtain that
a 2  (2k  1) 2  4k 2  4k  1  2(2k 2  2k )  1  2s  1
By definition this means that a 2 is even, which leads to contradiction with the
assumption that it is odd. This shows that if a 2 is even then a is also even.
Claim 2. 2 is irrational.
Proof of claim 2: (by contradiction) Assume that 2 is rational. By definition this means
p
that there exist p and q  0 such that 2  where p and q have no common factors
q
(they are in lowest terms). We square both sides of the equation and obtain
p2
, or p 2  2q 2 (*). By definition this means that p 2 is even. [Note that here we
2
q
assume as a known fact that the square of an integer is an integer.]. As we already
demonstrated in Claim 1, if p 2 is even, then p is even. Then again by definition p  2k
(**) for some integer k .
2
When we substitute (**) in (*), we obtain 4k 2  2q 2 , or q 2  2k 2 which by the
definition of even number means q 2 is even. So q is also even (from Claim 1).
We have shown that both p and q are even, or equivalently that they are both divisible by
2. This contradicts our assumption that p and q have no common factors.
Our assumption that
2 is rational led to contradiction, so
2 must be irrational.
Proof of problem 1:
s 2
2 is irrational. (from claim 2). Let’s examine the possibilities for
2
(it is either rational or irrational).
1. If s is rational, then we have found irrational numbers x and y such that x y is
rational. Specifically, x  2 and y  2 .
2. If s is irrational, we can let x  2
2
and y  2 . Then x y  2 
2
, or by definition
1
x y is rational.
In both possible cases, we showed that x and y with the desired property exist.
Problem 2 (Problem 16, page 85) Prove that if m and n are integers and mn is even,
them m is even or n is even.
Proof (by contradiction) Assume that ( m and n are integers and mn is even and) both m
and n are odd. Then by definition of even number there exist integers k and s such that
m  2k  1 and n  2s  1 . Then
mn  (2k  1)( 2s  1)  4ks  2s  2k  1  2(2ks  s  k )  1  2 p  1 , where p is an
integer. [Note here we are using the fact that addition and multiplication of integers result
in an integer as a fact known to be true]. By definition, we showed that mn is odd, which
contradicts the initial assumption that mn is even. So the initial assumption was false, or
equivalently, it is true that if m and n are integers and mn is even, them m is even or n is
even.
Problem 3 (Problem 41, page 86) Prove that these four statements about an integer n
are equivalent.
(i) n is even
(ii) n  1 is odd
(iii) 3n  1 is odd
(iv) 3n is even
Proof: We will show the equivalence of the four statements by showing that [
(i)  (ii) and (ii)  (iii) and (iii)  (iv) and (iv)  (i) [See proofs of equivalence on page
82].
1. If n is even, then n  1 is odd. (Direct proof) If n is even, by definition n  2k for
some integer k . Then n  1  2k  1 and it is odd by definition.
2. If n  1 is odd, then 3n  1 (**) is odd. (Direct proof) If n  1 is odd, by definition
n  1  2k  1 for some integer k , or n  2k (*). By substituting (*) in (**) we get
3n  1  3(2k )  1  2(3k )  1  2s  1 . By definition, 3n  1 is odd.
3. If 3n  1 is odd, 3n is even (Direct proof)
4. If 3n is even, n is even. (Proof by contradiction). Assume n is odd. Then n  2k  1
for some integer k and 3n  3(2k  1)  6k  3  6k  2  1  2(3k  1)  1  2s  1 which
is equivalent to saying that 3n is odd. We have reached a contradiction. If 3n is even, n
is even.
Problem 4 (Problem 24, page 85) Show that at least 3 of any 25 days chosen must fall
on the same day of the week.
[This is in fact a conditional statement: If we choose 25 days, than at least 3 of them will
fall on the same day of the week.]
Proof (by contradiction) Assume that we find a way to choose 25 days such that no more
than 2 days fall on the same day of the week. Given that there are 7 days in the week, we
have chosen no more than 2X7=14 days. This contradicts the assumption that we have
chosen 25 days (25>14). So the initial assumption is false, if we choose 25 days, than at
least 3 of them will fall on the same day of the week.
[This type of argument is called the pigeonhole principle. It states that if you have n
pigeonholes and n  1 pigeons, then if all pigeons need to be accommodated in the
pigeonholes, there will be at least one pigeonhole with more than one pigeon. In our
problem 4, the days of the week are the pigeonholes, and the days are the pigeons we
need to accommodate]