MATH1061/7861, Wed 19 Aug 2020 Question 1. Consider the lemma: ∀n ∈ N, if n is odd then 3n + 2 is odd.
nel i n 2kt
ne IN kioetze
◦Prove this by contradiction. ◦Prove this by direct proof.
od
w
a GD Aumssume PGD deduce
to f tE
IF
q0
D DIRECT ins
i
TE
Since KEI 342 is also an integer
3ntz 2ett where lez
3m12
iE 362ktDt2
6kt3 2
6kt
4
the 213kt2 t I
d 3 2 is odd iW
I
Question 1. Consider the lemma: ∀n ∈ N, if n is odd then 3n + 2 is odd.
Un EN
plug In
FALSE
Contradiction i Assuartin
The on
Assume theorem is false
i Fn EN sie n is odd but3n
Proof
3 iseveni
If
BUT I is odd i
Fktf 31247
3nez.z 26th
i 2k Bnt 2
i 26 61 5
i 5 2kbe
2
Ues Ct7L RHSc 2
ebg egoledeceacontradiction
s.t 7 22 sis n
indef
Challenge. Prove the following statement by contradiction:
For all m ∈ N, if m, m + 2 and m + 4 are all prime, then m = 3.
Question 2 (pre-class). Prove or disprove:
a ◦The sum of any two irrational numbers is irrational.
b ◦For any two irrational numbers r and s, the number r/s is irrational.
as 52TEE It Ntta
O Fa
Fr in In
Question 3. Which proof technique(s) would you use to prove that: ◦∀r ∈ Q, (2r2 – 3r + 1) ∈ Q. Direct
◦√2 is irrational.
◦∀r ∈ Q, if s is irrational then r + s is irrational. CONTRA ◦∀m,n ∈Z, if mn is odd then m and n are both odd.
Tr Tm the PG
by
fat
QED
we 8 deduced
Question 4 (pre-class). Prove that the sum of any rational number and any irrational number is irrational. Question 5. Prove the following lemma by contraposition:
For every integer n > 1, if no integer k in the range 1 < k ≤ √n divides n, then n is prime. a
Proof Use contraposition
as i wa
Q mei kns
4
kduasn.I
u not prime and n I we know niscopositt ne ab where a 1 bsl a BEI
Assume n is
notprime
Sine i
f
DetoectrelinAssume P Contrapose Assume
no skin Q ftp.hee
7 Concludes
ab
rn
i n n Impossible b
i I Ra Ern E K
She o onfa.b is in ray KEEF
KEI St ILKER
or
proof by contraposition Question 6. Prove that ∀r ∈ Q, (2r2 - 3r + 1) ∈ Q.
I
Ern