Inferences
To prove a trait or investigate the truth is already known, can be used patterns of argumentation based on the principles of logic. Conclusions drawn from some statements that are assumed true called assumptions that premise. Said to be a valid inference or implication of conjunctions validd if the premises with the conclusion is a tautology. Conversely, if the premises do not provide enough information to support the conclusions drawn, it says invalid inferences.
Principles that are used to draw conclusions, among other modes ponens, modus tollens, dansilogisme.
1. Tollens Modus
Inferences using ponnens mode based with the principle of "if p à q and p is true q must be true". principle can be formulated as follows
To prove a trait or investigate the truth is already known, can be used patterns of argumentation based on the principles of logic. Conclusions drawn from some statements that are assumed true called assumptions that premise. Said to be a valid inference or implication of conjunctions validd if the premises with the conclusion is a tautology. Conversely, if the premises do not provide enough information to support the conclusions drawn, it says invalid inferences.
Principles that are used to draw conclusions, among other modes ponens, modus tollens, dansilogisme.
1. Tollens Modus
Inferences using ponnens mode based with the principle of "if p à q and p is true q must be true". principle can be formulated as follows
Premise 1: p à q
Premise 2: p
Conclusion: q
Read the above principles:
"If p à q true and p is true then q is true"
Legitimate modes ponens can be proved by the truth table compound statement “((p à q) ^ p) à q”
Premise 2: p
Conclusion: q
Read the above principles:
"If p à q true and p is true then q is true"
Legitimate modes ponens can be proved by the truth table compound statement “((p à q) ^ p) à q”
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p
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q
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p à q
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(p à q) ^ p
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((p à q) ^ p) à q
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T
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T
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T
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T
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T
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T
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F
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F
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F
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T
|
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F
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T
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T
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F
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T
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|
F
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F
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T
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F
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T
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In the table shows that in the fifth column is the truth value "true" entirely. Therefore ((p à q) ^ p) à q is a tautology
Example
1. Premise 1: if the equilateral triangle abc the length ab = ac = bc
Premise 2: equilateral triangle abc
-------------------------------------------------- ----------------------
Conclusion: so long-term ab = ac = bc length
2. Determine what inferences are valid under
a. If Ardi tried it.
Ardi successfully
----------------------------------------------------------------------------------
1. Premise 1: if the equilateral triangle abc the length ab = ac = bc
Premise 2: equilateral triangle abc
-------------------------------------------------- ----------------------
Conclusion: so long-term ab = ac = bc length
2. Determine what inferences are valid under
a. If Ardi tried it.
Ardi successfully
----------------------------------------------------------------------------------
So, Ardi is trying
b. If Desi feel sad then cry.
Desi feel sad
-------------------------------------------------- ----------------
So, Desi is crying
Completion:
a. Conclusion is invalid. Because according to the principle modes ponens, should premise 2: ardi try and conclusions: ardi successfully.
b. Valid conclusions. Because in accordance with the principle mode of p ponens: Desi feel sad and q: Desi crying.
Premise 1 : p à q
Premise 2 : ~p
Conclusion : ~p
Desi feel sad
-------------------------------------------------- ----------------
So, Desi is crying
Completion:
a. Conclusion is invalid. Because according to the principle modes ponens, should premise 2: ardi try and conclusions: ardi successfully.
b. Valid conclusions. Because in accordance with the principle mode of p ponens: Desi feel sad and q: Desi crying.
Premise 1 : p à q
Premise 2 : ~p
Conclusion : ~p
2. Modus Tollens
On the modus tollens inferences based on the principle "if pà q is true and q is not true.” principle can be formulated as follows.
On the modus tollens inferences based on the principle "if pà q is true and q is not true.” principle can be formulated as follows.
Premise 1 : p à q
Premise 2 : ~p
Conclusion : ~p
Premise 2 : ~p
Conclusion : ~p
q is true and correct, then ~ P ~ p is true.àThis principle
is read if p correctness of the modus
tollens can be by students using a truth table kontrraposisi.
Example:
1. Premis 1 : If the rectangle of ABC siku-siku on the point of B, then AC = AB + BC
Example:
1. Premis 1 : If the rectangle of ABC siku-siku on the point of B, then AC = AB + BC
Premis 2 : AC ≠ AB + BC
----------------------------------------------------------------------------------
Conclution : So, the rectangle of ABC is not siku-siku
on the point of B
2. By the taken inferences from below, determine which
one is valid and which are not valid
a. All even numbers divisible by two.
Thirteen is not even number
a. All even numbers divisible by two.
Thirteen is not even number
--------------------------------------------------
--------------------
So, thirteen divisible by two.
So, thirteen divisible by two.
b.If the wind blows the trees swaying
Trees are not swaying.
-------------------------------------------------- --------------------
So, the wind doesn’t blow.
Completion:
a. Conclusion is invalid. Because according to the principle of modus tollens, premise 2 should be: thirteen is not divisible by two and the conclusion is: thirteen instead of an even number.
b. Valid conclusions. because in accordance with the principle of modus tollens yaittu p: wind blowing hard, q: the trees swaying.
Premise 1: p à q
Premise 2: ~q
Conclusion: ~p
a. Conclusion is invalid. Because according to the principle of modus tollens, premise 2 should be: thirteen is not divisible by two and the conclusion is: thirteen instead of an even number.
b. Valid conclusions. because in accordance with the principle of modus tollens yaittu p: wind blowing hard, q: the trees swaying.
Premise 1: p à q
Premise 2: ~q
Conclusion: ~p
3. Syllogism
Drawing conclusions by syllogism based on the principle "if p à q and q à r then p à r." principle can be formulated as follows.
Drawing conclusions by syllogism based on the principle "if p à q and q à r then p à r." principle can be formulated as follows.
Premise
1: p à q
Premise 2: q à r
Conclusion: p à r
Premise 2: q à r
Conclusion: p à r
Truth of the syllogism can be seen from the table of truth((p à q)^( q à r))à( p à r) is a tautology.
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P
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q
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r
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p à q
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q à r
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p à r
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(p à q)^( q à r)
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((p à q)^( q à r))à( p à r)
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T
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T
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T
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T
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T
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T
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T
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T
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T
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T
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F
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T
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F
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F
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F
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T
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T
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F
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T
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F
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T
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T
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F
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T
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T
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F
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F
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F
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T
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F
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F
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T
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F
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T
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T
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T
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T
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T
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T
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T
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F
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T
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F
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T
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F
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T
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F
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T
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F
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F
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T
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T
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T
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T
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T
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T
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F
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F
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F
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T
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T
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T
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T
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T
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From
table 5.17, it appears that the eighth column has a truth value "T"
(true) all.
Therefore, ((p à q)^( q à r))à( p à r) is a tautology. The following is contaoh deduction using the principle of the syllogism.
Premise 1: if the math teacher did not go to school, then students mingle
Therefore, ((p à q)^( q à r))à( p à r) is a tautology. The following is contaoh deduction using the principle of the syllogism.
Premise 1: if the math teacher did not go to school, then students mingle
Premise 2: If the students mingle then they excited
_____________________________________________
Conclusion: Thus, if the math teacher did not go to school, then students excited.
Competency Test 9
Among the under-drawing inferences, determine safe legitimate (valid) and Amna are not valid (invalid).
1. Premise 1: Every integer is a rational number.
Premise 2: -8 integers.
Conclusion: So, -8 rational numbers.
_____________________________________________
Conclusion: Thus, if the math teacher did not go to school, then students excited.
Competency Test 9
Among the under-drawing inferences, determine safe legitimate (valid) and Amna are not valid (invalid).
1. Premise 1: Every integer is a rational number.
Premise 2: -8 integers.
Conclusion: So, -8 rational numbers.
2. Premise 1: Every natural numbers are natural numbers
Premise 2: -7 instead of natural numbers
Conclusion: So, -7 instead of natural numbers.
3. Premise 1: If a child is late for school amak he will get a warning.
Premise 2: Andy gets a warning.
Conclusion: So, Andi late for school.
4. Premise 1: If everyone is so diligent exercise
healthy body
Premise 2: Roni healthy body.
Conclusion: So, Roni diligent exercise.
5. p à ~ q 7. p à ~ q 9. p à ~ q
p q r à q
______ _______ ______
΅ ~ q ΅ ~ p ΅ p à ~ r
6.~ p à q 8.
~ p à q 10. ~p à ` q
~ p ~ q q à r
______
______ _______
΅ q ΅ p ΅ p à r
G. Nature of Mathematical Proof of the Direct and Indirect Evidence
Evidentiary nature in mathematics show the truth of nature in mathematical logic. In this section we will study some properties of proof in mathematics related to mathematical logic, namely proof by direct evidence, evidence by evidence reversed, and the proof by mathematical induction.
1. Direct
Evidence by Evidence
Proof by direct evidence used to prove properties in Mathematics with
implications of p -> q. This proof uses the truth value of statements
(implications), ie if the unknown p is true (true antecedent) and the
implication is true, then with the right measures, surely the resulting q-value
true (the consequent is true).
Example:
Prove that if x + 2 = 5 then x = 3
Proof:
Known x + 2 = 5. Then it will be proved that x = 3. Since x + 2 = 5 then
x + 2-2 = 5-2 or x = 3. Thus, it is evident that if x +
2 = 5 then x = 3.
2. Proof
of Evidence by Reversed
Of proof with evidence of inverted using the principle of modus tollens. There are two proof with proof reversed, ie contraposition and contradiction.
a. Contraposition
Proof by contraposition used to prove the nature of mathematics that has implications for p -> q. Recall, that the truth value of an implication with the truth value of kontraposisinya. Therefore, the proof by contraposition of the nature of mathematics with the implication p -> q is done by showing the truth of the nature of Mathematics ~ q -> p.
Suppose that the nature of mathematics will prove p -> q. Authentication is done by proving ~ q -> ~ p. In this case the unknown ~ q is true and the implication is true, then the steps are correct, the resulting ~ p must be true.
Example:
Prove that if x and y odd then x + y even number.
Proof:
Contraposition of the implication "If x and y odd then x + y even number" is "If x + y is not an even number then the x or y is not an even number.
X + y is not known even number, it means that x + y odd number. Therefore, the x or y is an odd number means x or y is not an even number. So it proved bahawa if x and y odd then x + y even number.
b. Contradiction (reductio ad Absordum)
Proof by contradiction to prove properties can diguanakn Mathematics which is a necessary implication. To prove the nature of mathematics that is an implication p -> q, assumed not q. Furthermore, if the resulting contradiction (something wrong, say no because of the unknown p is p), means the assumption is wrong. Therefore, the assumption must diingkar. Thus obtained q.
To prove the mathematical properties of nature verupa p, assumed no p. Furthermore, if the resulting contradiction (something wrong eg an even number), then the assumption is wrong. Therefore, the assumption must diingkar.
Example:
1. Prove that 2 + 4 = 6
Proof:
Suppose that 2 + 4 ≠ 6 then 2 + 4-4 6 -4 or 2 ≠ ≠ 2. This is a contradiction with the proviso that 2 = 2. Modality 2 + 4 is not equal to 6 must diingkar so 2 + 4 = 6. So it proved 2 + 4 = 6.
2. Prove that √ 2 irrational numbers.
Proof:
Suppose √ 2 a rational number then there are numbers x and y so that ASII √ 2 = x with x and y have no factor fellowship.
(√ 2) = (x / y)ó √ 2 =( x / y )
ó 2 = x / yó
ó x = 2y
Of proof with evidence of inverted using the principle of modus tollens. There are two proof with proof reversed, ie contraposition and contradiction.
a. Contraposition
Proof by contraposition used to prove the nature of mathematics that has implications for p -> q. Recall, that the truth value of an implication with the truth value of kontraposisinya. Therefore, the proof by contraposition of the nature of mathematics with the implication p -> q is done by showing the truth of the nature of Mathematics ~ q -> p.
Suppose that the nature of mathematics will prove p -> q. Authentication is done by proving ~ q -> ~ p. In this case the unknown ~ q is true and the implication is true, then the steps are correct, the resulting ~ p must be true.
Example:
Prove that if x and y odd then x + y even number.
Proof:
Contraposition of the implication "If x and y odd then x + y even number" is "If x + y is not an even number then the x or y is not an even number.
X + y is not known even number, it means that x + y odd number. Therefore, the x or y is an odd number means x or y is not an even number. So it proved bahawa if x and y odd then x + y even number.
b. Contradiction (reductio ad Absordum)
Proof by contradiction to prove properties can diguanakn Mathematics which is a necessary implication. To prove the nature of mathematics that is an implication p -> q, assumed not q. Furthermore, if the resulting contradiction (something wrong, say no because of the unknown p is p), means the assumption is wrong. Therefore, the assumption must diingkar. Thus obtained q.
To prove the mathematical properties of nature verupa p, assumed no p. Furthermore, if the resulting contradiction (something wrong eg an even number), then the assumption is wrong. Therefore, the assumption must diingkar.
Example:
1. Prove that 2 + 4 = 6
Proof:
Suppose that 2 + 4 ≠ 6 then 2 + 4-4 6 -4 or 2 ≠ ≠ 2. This is a contradiction with the proviso that 2 = 2. Modality 2 + 4 is not equal to 6 must diingkar so 2 + 4 = 6. So it proved 2 + 4 = 6.
2. Prove that √ 2 irrational numbers.
Proof:
Suppose √ 2 a rational number then there are numbers x and y so that ASII √ 2 = x with x and y have no factor fellowship.
(√ 2) = (x / y)ó √ 2 =( x / y )
ó 2 = x / yó
ó x = 2y
Therefore x is an even
number. Consequently x is an even number because
if x odd then x is also an odd number.
Since x is an even number so that there are natural numbers z x = 2z. Furthermore,
x = 2y ó (2z) = 2yóX = 2y
ó 4z = 2y
ó2z = y
x = 2y ó (2z) = 2yóX = 2y
ó 4z = 2y
ó2z = y
It appears that, y2 is an even number. Consequently,
y is an even number. Since x2 and y2 is an even
number, then definitely have fellowship factors. This is a contradiction with
the above provisions that say that x and y have no factor fellowship.
Consequently modality should diingkar. So it proved that the irrational
numbers √ 2.
Note
Note the use of words assume, for example, and example. In a mathematical proof, suppose the word is often used to prove the contradiction. This word is used to deny a statement. While the example and the example refers to the shape pemisalan.
3. Proof by Mathematical Induction
Proof by mathematical induction is used to prove the nature of mathematics that includes the original numbers. Suppose to be proved that for every n natural numbers, apply P (n). The steps are as follows:
a. proved valid P (n) for n = 1
b. P (n) are true for n = k. furthermore, demonstrated that
p (n) is true for n = k +1
c. a and b, it was concluded that for every real number n applies
P (n)
Example:
1. Prove that for any natural numbers n, 4n - 1 is divisible by 3.
Proof:
a. For n = 1, 4n - 1 = 41-1 = 4-1 = 3
b. Assumed to be true for n = k, means 4k - 1 is divisible by 3.
Furthermore, for n = k + 1 valid
+1-1 = 4k (4k x 4) - 1
= (4k x (3 + 1)) - 1
= ((4k x 3) + (4k +1)) - 1
= (3 x 4k) + (4k-1)
Since 3 x 4k and 4k-1 is divisible by 3 then the 4k +1 - 1 = (3 x 4k) + (4k-1) is divisible by 3.
c. of a and b, it was concluded that for any n natural numbers, apply 4n-1 divisible by 3.
Note
Note the use of words assume, for example, and example. In a mathematical proof, suppose the word is often used to prove the contradiction. This word is used to deny a statement. While the example and the example refers to the shape pemisalan.
3. Proof by Mathematical Induction
Proof by mathematical induction is used to prove the nature of mathematics that includes the original numbers. Suppose to be proved that for every n natural numbers, apply P (n). The steps are as follows:
a. proved valid P (n) for n = 1
b. P (n) are true for n = k. furthermore, demonstrated that
p (n) is true for n = k +1
c. a and b, it was concluded that for every real number n applies
P (n)
Example:
1. Prove that for any natural numbers n, 4n - 1 is divisible by 3.
Proof:
a. For n = 1, 4n - 1 = 41-1 = 4-1 = 3
b. Assumed to be true for n = k, means 4k - 1 is divisible by 3.
Furthermore, for n = k + 1 valid
+1-1 = 4k (4k x 4) - 1
= (4k x (3 + 1)) - 1
= ((4k x 3) + (4k +1)) - 1
= (3 x 4k) + (4k-1)
Since 3 x 4k and 4k-1 is divisible by 3 then the 4k +1 - 1 = (3 x 4k) + (4k-1) is divisible by 3.
c. of a and b, it was concluded that for any n natural numbers, apply 4n-1 divisible by 3.
2. Prove that for any n natural numbers, apply 1 + 2 + ... +n = ½ n (n + 1)
Proof:
a. For n = 1, then 1 = ½ (1) (1 + 1) ↔ 1 = ½ (2) ↔ 1 = 1
b. Considered true for n = k, then 1 + 2 + ... + k = ½ k (k + 1). Next, for n = k + 1
Apply
1 + 2 + ... + k + (k + 1) = ½ k
(k + 1) + (k + 1)
ó 1 + 2 + ... + k + (k + 1) = (k + 1) (½ k + 1)
ó 1 + 2 + ... + k + (k +
1) = (k + 1) [½ (k + 2)]
ó 1 + 2 + ... + k + (k + 1) = ½ (k + 1) (k + 2)
ó 1 + 2 + ... + k + (k + 1) = ½ (k + 1) [(k+1) + 1]
c. Of a and b, it was concluded that for any n natural
numbers, applies
1 + 2 + ... + n = ½ n (n + 1)
C ompetition Test 10
1. With direct evidence, prove that
a. If x and y are rational numbers then the rational number x + y;
b. If m and n are even numbers then even numbers m + n;
c. If x – 6 = 11 then x = 17.
ó 1 + 2 + ... + k + (k + 1) = ½ (k + 1) (k + 2)
ó 1 + 2 + ... + k + (k + 1) = ½ (k + 1) [(k+1) + 1]
c. Of a and b, it was concluded that for any n natural
numbers, applies
1 + 2 + ... + n = ½ n (n + 1)
C ompetition Test 10
1. With direct evidence, prove that
a. If x and y are rational numbers then the rational number x + y;
b. If m and n are even numbers then even numbers m + n;
c. If x – 6 = 11 then x = 17.
2. With indirect evidence, prove that
a. If x and y are even numbers then xy divisible by 4;
b. If 4x = 6 then x = 4;
c. 4 + (-5) = -1;
d. 3 x 5 = 15;
e. √3is an irrational number.
3. By mathematical induction, prove that
a. 12 + 22 + ... + n2 = ⅙n (n + 1) (2n + 1);
b. Summing the first n odd numbers equal to n2;
c. 2 + 5 + 8 + 11 + ... + 3n – 1 = ½n (3n + 1);
d. 9n – 1 divisible by 8.
Task
Try
To deepen your horizons on this mathematical logic, by seeking information in the media close to the neighborhood. For example, the library or on the internet.
Reflection
Have you everlogic / reason to take a decision (conclusion) according
to the true path? Contemplate, what benefits you get after studying this material.
Resume
1. Statement is a sentence that has truth value true or false, but not all at once right and wrong.
1. Statement is a sentence that has truth value true or false, but not all at once right and wrong.
2. Compound statement is a statement which is formed from two or more statements with a specific logic circuit, among others, the following
A a. Conjunction, itsconjunctions"and" is denoted by “ Λ”.
b. Disjunction, itsconjunctions"or" is denoted by “ V”.
c. Implication, its conjunctions “if...then ...” is donated by “→”.
d. Biimplication, its conjunction “... if and only if...” is donated by “↔”.
3. The implicationis called logical implication tautology, denoted “⇛”, as for Biimplication is called logical biimplication tautology, denoted “↔”
4. Principles are used to draw conclusions
a. Ponnens Modus b. Tollens Modus c. Syllogism
Premis 1: p → q premis 1: p→q premis 1: p→q
Premis 2 : p premis 2: ~q premis 2 : q→r
v q ~p p→r
Exercise daily test V
I. Choose one right answer.
1. If the statement p is true and q is false, the statement below true value is ...
a. P → q
b. ~p → q
c. ~p Λ q
d. ~p ↔ ~q
e. ~p V q
2. Implication p→~q value (having the same meaning)...
a.~q → p d. q → ~p
b. ~p →q e. ~p Λ q
c. ~(q→p)
3. Statement pair p ↔ q that satisfy the statement p and q, for x real number is ...
a. p: x is odd numbers q: 2x is even numbers
b. p: x is positive numbers q: 2x is positive numbers
c. p: x is odd numbers q: 2x + 1 is odd numbers
d. p: x is negative numbers q: 2x + 1 is negative numbers
e. there is noight answer
4. Sentence below which statement is ...
a. Solo is thecapital of Indonesia
b. Take two pieces only!
c. Where do you go?
d. Hi, don’t angry!
e. Ok!
5. Negation of the statement "all students like mathematics" is ...
a. All students liked the addition of Mathematics
b. Some students do not like Mathematics
c. Some students like Mathematics
d. No one else who likes Mathematics
e. No student who does not like Mathematics
6. statements“ ~(p V q)→ p” would be worthfalseif ...
a. p is true and q is false
b. p is true and q is true
c. p is false and q is false
d. p is false and q is true
e. p and q have true value is same
Arum
Wulandari (3115115684)
Evi
Syahida (3115115716)
Siti
Nurfidannaufal (3115115692)
Pendidikan
Matematika Non Reguler
Universitas
Negeri Jakarta
2011
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