Let x be a positive real number
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San Jose State University *
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Course
108
Subject
Mathematics
Date
Feb 20, 2024
Type
docx
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Uploaded by BaronSquirrelPerson826
1.
Let x be a positive real number. Prove that if x –(1/x)>1, then x > 2 by
a)
A direct proof
Here x > 0
Let x - (1/x)>1
This shows x > 2
Consider x-(1/x)>1
x
2
-1>x
x
2
-x+1/4-1/4-1>0+1
(x-1/2)
2
-9/4>0
(x-1/2-3/2) (x-1/2+3/2)>0
Since x >0 => x-2> 0
x-2> 0
x> 2
Case i: x-2<0 if x+1<0
x<2 if x<-1
x<-1
But x is a positive real number
Contradiction!
Case ii. (x-2) > 0 if (x-1)>0
x > 2 if x>1
x >2
x-1/x>1 => x>2
Proved!
b)
Contrapositive
Let x≤2
This Shows x-1/x≤1
x≤2 => 1/x≥ 1 => -1/x≤-1 x≠0
x-1/x≤x-1 ≤ 2-1 ≤1
x-1/x≤1
x≤2 => x-1/x≤1
Proved by contrapositive!
c)
Contradiction
Suppose x is not greater than 2, which means x ≤ 2
Given x-1/x>1
1/x>1
-1/x≤-1
x-1/x≤x-1≤2-1=1
x-1/x≤1
Contradiction to x-1/x>1
The assumption is wrong that x-1/x>1 => x>2. So, proved!
2.
Criticize the following proof. Claim ∑ i=1 infinity 1/i< infinity
Proof. Let S(n) = Claim ∑ i=1 n
1/i. Then consider the statement assert S(n) < infinity. This statement is true since ∑ i=1 1
1/i=1/1= 1 < infinity. And if it’s true for n, then it’s true for n+1, since the next sum S(n+1) = S(n) +1/n+1, and this is the sum of two infinite numbers. So, S (1) is true, and S(n) implies S(n+1). So, we’ve therefore proved that Claim ∑ i=1 infinity 1/i< infinity by induction. Let S(n) = ∑ i=1 n
1/i.
Base Case: When n=1, the statement is true for n=1 since ∑ i=1 1/1 =1 < infinity
Inductive hypothesis: Suppose n+1.
Then, substitute n by n+1 for the next sum to be S(n+1) = S(n) +1/n+1, and this sum holds for n=n+1. So, since S (1) is true, S(n) implies S(n+1). So, ∑ i=1 infinity 1/i< infinity is proved by induction.
3.
Evaluate the proposed proof of the following result. 1+3+5+ ... + (2n-1) =n
2
.
Proof. Proceed by induction. Since 2(1) -1 =1
2
, the formula holds for n=1. Assume t that 1+3+5+ ... + (2k-1) =k
2
for a positive integer k. Prove that 1+3+5+ ... + (2k+1) =(k+1)
2
. Observe that 1+3+5+ ... + (2k+1) =(k+1)
2
1+3+5+ ... + (2k-1) + (2k+1) =(k+1)
2
.
k
2
+(2k+1) = (k+1)
2
(k+1)
2
=(k+1)
2
The only mistake in the proof is that it assumes the statement for n=k+1. We’ll need to prove the statement for n=k+1 assuming the statement is true for n=k.
So, assume 1+3+5+ ... + (2k-1) =k
2 and prove 1+3+5+ ... + (2k-1) + (2k+1) =(k+1)
2
.
Correct Statement is:
The problem’s objective is to evaluate the given for 1+3+5+ ... + (2n-1) =n
2
for all positive integers n using induction.
Principle Mathematical Induction.
The statement P(n) is true for all positive integers n if
(1) P (1) is true and
(2) If P(n) is true then P(k+1) is true too
Consider the statement
P(n): 1+3+5+ ... + (2n-1) =n
2
.
Substitute n-1 then
P (1): 2(1)-1=1=1
Therefore, the statement is true for n=1
Assume that the statement is true for n=k and then show that P(n+1) is true.
P(k): 1+3+5+ ... + (2k-1) =k
2
. This past solution is correct
To show that P(n+1) is true take LHS of the equation then prove RHS but in provided solution.
RHS is taken together with LHS while proving P(n+1) is true.
Next, show that
P(n+1): 1+3+5+ ... + (2n+1) =(n+1)
2
Consider the left side of equation,
LHS=1+3+5+ ... + (2n+1)
=1+3+5+ ... + (2n-1) +(2n+1)
=n
2
+(2n+1) since P(n) is true
= n
2
+2n+1
= (n+1)
2
RHS
Thus, P(n+1) is true
Hence, by the principle of mathematical induction 1+3+5+ ... + (2n-1) =n
2
for all positive
integers n.
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4) For sets A and B, find a necessary and sufficient condition for (A x B) ∩ (B x A) = Empty Set.
Verify that this condition is necessary and sufficient.
The condition is that Sets A and B can’t contain any same elements.
Ex. A= {9,5} and B = {0,6} Then A x B = {(9,0), (9,6), (5,0), (5,6)} and B x A = {(0,9), (6,9), (0,5), (6,5)}. Then (A x B) ∩ (B x A) = empty set. So, it worked because A and B do not have any same elements.
5) Show that for sets A, B, and C, A x (B-C) = (A x B) – (A x C)
Given A, B and C are the sets.
To prove A x (B-C) = (A x B) – (A x C)
Use this proof to get A x (B-C) ⊆
(A x B) – (A x C)
Allow (x, y) ϵ A x
(B-C)
x
ϵ A and y ϵ (B-C)
A x (B-C) = {x, y | x ϵ A x y ϵ
(B-C)}
= {x, y | x ϵ A ᴧ y ϵ (B-C)} = {x, y | x ϵ A ᴧ (
y ϵ B ᴧ
y ϵ C)} Definition difference
= {x, y | (x ϵ A ᴧ y ϵ B) ᴧ (x ϵ A ᴧ y ϵ C)} Distributive Law
= (x, y) ϵ A x B and (x, y) ϵ (A x C)
= (x, y) ϵ (A x B) – (A x C)
Hence, A x (B-C) ⊆
(A x B) – (A x C)
Next, we should prove (A x B) – (A x C) ⊆
A x (B-C)
Let (x, y) ϵ [(A x B) – (A x C)]
(x, y) ϵ (A x B) and (x, y) ϵ (A x C)
(x ϵ A and y ϵ B) and (x ϵ A and y ϵ C)
x ϵ A and (y ϵ B and y ϵ C)
x
ϵ A and y ϵ (B-C)
(x, y) ϵ A x
(B-C)
(A x B) – (A x C) ⊆
A x (B-C)
So, A x (B-C) = (A x B) – (A x C)
6) Summarize Gower’s blog posts.
According to arithmetic fundamental theorem, the fundamental theorem is not so obvious
because many numbers that are composite can be in prime factorization. Some prime numbers are 2, 3, 5, 7, 9, and 11, meaning they may factor out any composite natural number. All positive integers even if they’re not prime can be factorized in a way as the products of primes. After reading the proving the fundamental theorem of arithmetic blog, there is a unique sequence of prime numbers because the terms do have exactly two
factors. This blog posts gave a notice stating that the product of odd numbers s always odd, which is so encouraging.
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