Download Logic (Mathematics 1BA1) Reminder: Sets of numbers Proof by

Reminder: Sets of numbers
N: set of positive (or natural) integers: 0, 1, 2, · · ·
Logic (Mathematics 1BA1)
Z: set of integers: 0, ±1, ±2, · · ·
Rozenn Dahyot
Q: set of rational numbers (or informally fraction). A rational number is a ratio or
Room 128 Lloyd Institute
quotient of two integers, usually written as the fraction a/b, where b is not zero
School of Computer Science and Statistics
Trinity College Dublin, IRELAND
and, a and b have no common factors.
https://www.cs.tcd.ie/Rozenn.Dahyot/
[email protected]
R: set of real numbers.
Week 5 - February 2007
C: set of complex numbers.
Proof by contradiction
Proof by contradiction
Theorem proof by contradiction:
(¬p ⇒ F) ≡ p
Proof of Theorem proof by contradiction (¬p ⇒ F) ≡ p :
We wish to prove a proposition p to be true. If we assume p to be false (i.e ¬p is
We start from a theorem presented earlier in the course p ⇒ F ≡ ¬p (in Theorems
Implication and boolean constants):
True), and then derive some contradiction from this fact (i.e that ¬p ⇒ F ), we see
that our original supposition is false and so p is true.
≡
Remember that axioms and theorems are boolean expressions that are always
≡
true (whatever the values of the boolean variables involved in the expression).
p ⇒ F ≡ ¬p
¬p ⇒ F ≡ ¬¬p
¬p ⇒ F ≡ p
hSubstitution p := ¬pi
hTheorem Double Negation ¬¬p ≡ pi
So you can also rewrite or understand the theorem proof by contradiction :
(¬p ⇒ F) ≡ p ≡ T
Proof that
√
or
(¬p ⇒ F) ≡ (p ≡ T)
2 is irrational using contradiction I
The proposition is p:
Lets assume ¬p:
√
A and B) such that:
√
Proof that
√
2 is irrational using contradiction II
2 is irrational and we have to show that p is true.
2 is rational. It means that ∃(A, B) ∈ Z2 (e.g. it exists 2 integers
we just shown that the expression ¬p ⇒ F is True i.e. that assuming
√
A
2=
B
where A and B have no common factors. This is coming from the definition of
2 rational it
leads to a contradiction (ie something false). Using the theorem proof by
contradiction, we have proven by equivalence that p ≡ T.
rational numbers.
Hence
2
√
√
2 is not a rational number.
2
Consequently, it follows A = 2B .
Thus A2 is even, we can deduce that A is even 1 , so A = 2a and 2 a2 = B2
Thus B is also even and B = 2b. 2 is therefore a common factor. This is a
contradiction with our hypothesis (that A and B have no common factor)!
1
We will show that using the proof by contrapositive in tutorial.
Proof by contrapositive
Proof that x + y > 2 ⇒ (x > 1 ∨ y > 1) I
Lets define p : x + y > 2 and q : (x > 1 ∨ y > 1), we can prove by contrapositive
that p ⇒ q is true by showing that ¬q ⇒ ¬p is true.
By contrapositive, this is equivalent2 to proving (x < 1 ∧ y < 1) ⇒ x + y < 2
The proof method is taken from Theorem Contrapositive : p ⇒ q ≡ ¬q ⇒ ¬p.
hence
Theorem proof by contrapositive p ⇒ q ≡ ¬q ⇒ ¬p:
x+y
h Assumptions x < 1 and y < 1 i
<
It means that you can prove p ⇒ q to be True by proving its contrapositive that
1+1
¬q ⇒ ¬p is true.
=
2
h Arithmetic i
we have just shown that (x < 1 ∧ y < 1) ⇒ x + y < 2 is true which by
contrapositive is the same as showing x + y > 2 ⇒ (x > 1 ∨ y > 1) is true.
2
Remember the theorem De Morgan: ¬(p ∨ q) = ¬p ∧ ¬q.