MATH1061/7861, Mon 17 Aug 2020 Question 1. Prove or disprove: true
Proof
Voct2
TT fiIz
Proof
where oct2
integer 4g
I.sc 2 is
an
odd
y
it follows that 1 an odd integer usingthe
4yyDiocetproot Las sume x
If x is an integer such that x + 2 = 4y for some integer y, then x / 2 is an odd integer. i
I is odd integer
i.sc z
Question 2. Prove or disprove:
For each integer n ≥ 2, the product of the first n prime numbers minus 1 is prime.
z
integer y 4y d
lets be an
for some
Then
A Since l is an integer2ly Dtl 2cg tl’s
xt2 ix
4g 2 2y l
2.1 integer
ppaps Pn 1 Ege 7 Ies n2 23lis
statement is taek
Challenge A. Prove or disprove:
For each integer n ≥ 2, the product of the first n prime numbers plus 1 is prime.
Il 19
f
2y 2 1
definition integer I
FASE
of
odd
which is notprime
A 3 nee
is prime
23 5 1 29 2.3.57 i 209
Question 3. Prove or disprove:
∀n ∈ N, either n is composite or n + 1 is composite.
FALSE
l is a counterexample since n L is not comp
n
z is a counterexample one nn.EE ToTIIip
Ef
statement is Question 4. Prove or disprove:
∀n ∈ N, either n is even or n + 1 is even. Tmetementeguiavalat to
out 2 is not comp
iYglesias
and ntl
odd and
Fai KER
EnsignG operator
26
for 11
x
2k k
false
nel is ever
n is noteven and nel is not
then pvq up q
Tn EIN if n is not even bycontradiction Lts true Rtes false
Proof
let new Assume
ie
n
n 2 forsome
nai
1 8
l 2e
11
i
z
integer
2k 2k
2 2e 2b
a
2e 11
some lek 241
c ke
2 integer
This
a contradiction is eamnoteven
is i n
assumptions were
net is even a
o
wrong
Question 5. Consider the lemma: ∀n ∈ N, if n is odd then 3n + 2 is odd. ◦ Prove this by contradiction.
◦ Prove this by direct proof.
Question 6. Consider the following statement:
For all m ∈ N, if m, m + 2 and m + 4 are all prime, then m = 3. Write an appropriate first sentence for a proof by contradiction.
Challenge B. Finish the proof for question 6!