• Support PF! Buy your school textbooks, materials and every day products Here!

Help -- with proofs please

  • Thread starter Saterial
  • Start date
  • #1
54
0

Homework Statement


Prove the following propositions:
1) ∀x ∈ (0, 1), ∃y ∈ (0, 1), x < y
and
2) ∀x, y ∈ R, if x < y, then ∀b ∈ (0, ∞), ∃a ∈ (0, ∞),
x + ab < y.

Can anyone help me out with either one?

I have a few others that I can get but I can't get these two. Mainly because these don't have a specific method to use so I don't know which to use. Where as previously the ones said prove by contradiction etc.

Methods that can be used:
Direct Proof
Contraposition
Contradiction


Homework Equations


1) ∀x ∈ (0, 1), ∃y ∈ (0, 1), x < y
and
2) ∀x, y ∈ R, if x < y, then ∀b ∈ (0, ∞), ∃a ∈ (0, ∞),
x + ab < y.

The Attempt at a Solution


How would I know with method to use for these?
 

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
7,008
1,233
You should list some relevant assumptions and theorems about the real numbers that are given in your course material (or give a link to a document that lists them.) Some of them would have to be used in a proof. Unless we know what's available, it's hard to guess what type of proof to use.
 
  • #3
54
0
This is the list here.
 

Attachments

  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,706
5,028
This is the list here.
I would think up a very simple constructive proof, without worrying about the axioms initially, then see if you can express it in terms of the axioms.
 
  • #5
525
345
Number 1 is a slice of the proof that deals with proving a supremum - the least upper bound. (there is no way to prove a supremum exists, though, that is axiomatic)
#1 For every x in the open interval, there exists a y such that x < y. Assume the contrary:
[itex]\exists x\in (0,1)\colon\forall y\in (0,1), y \leq x[/itex] This is to say that there exists a maximum element (x) in an open interval. Is there a way you can think of this is contradictory?
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,706
5,028
Number 1 is a slice of the proof that deals with proving a supremum - the least upper bound. (there is no way to prove a supremum exists, though, that is axiomatic)
#1 For every x in the open interval, there exists a y such that x < y. Assume the contrary:
[itex]\exists x\in (0,1)\colon\forall y\in (0,1), y \leq x[/itex] This is to say that there exists a maximum element (x) in an open interval. Is there a way you can think of this is contradictory?
That doesn't seem to me to be any closer to the axioms.
 
  • #7
525
345
Unsure, what you mean by "any closer to the axioms".
The problem statement essentially claims there is no maximum element in an open interval - the way to prove it is to assume the contrary and show that the assumption leads to contradiction.
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,706
5,028
Unsure, what you mean by "any closer to the axioms".
The problem statement essentially claims there is no maximum element in an open interval - the way to prove it is to assume the contrary and show that the assumption leads to contradiction.
I mean that it doesn't look like progress towards a proof. Did you try my suggestion?
 
  • #9
525
345
I am not going to solve the problem. I am only hinting to the OP to use proof-by-contradiction for #1.

As for your suggestion:
think up a very simple constructive proof, without worrying about the axioms initially, then see if you can express it in terms of the axioms.
Assuming you mean the Axiom of real numbers i.e Every upper bound set has a supremum - I cannot think of a way to show that every
(a+b)/2 is indefinitely contained within the open interval Unless, I have the axiom at my disposal, which is why they are called axioms, makes it convinient.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,804
931

Homework Statement


Prove the following propositions:
Both of these have a conclusion that says "there exist". The best "method" for showing something exist is to find it!
1) ∀x ∈ (0, 1), ∃y ∈ (0, 1), x < y [/quote]
[itex]\all x\in (0, 1)[/itex] means x< 1. Can you find a number, say, half way between 0 and 1?

and
2) ∀x, y ∈ R, if x < y, then ∀b ∈ (0, ∞), ∃a ∈ (0, ∞),
x + ab < y.
Since x< y, there exist an infinite set of numbers between them. Choose one and subtract x from it to get ab. There are, of course, an infinite set of numbers, a and b, that will give the same product, "ab". Pick any one a an example.

Can anyone help me out with either one?

I have a few others that I can get but I can't get these two. Mainly because these don't have a specific method to use so I don't know which to use. Where as previously the ones said prove by contradiction etc.

Methods that can be used:
Direct Proof
Contraposition
Contradiction


Homework Equations


1) ∀x ∈ (0, 1), ∃y ∈ (0, 1), x < y
and
2) ∀x, y ∈ R, if x < y, then ∀b ∈ (0, ∞), ∃a ∈ (0, ∞),
x + ab < y.

The Attempt at a Solution


How would I know with method to use for these?
You determine that by knowing the basic definitions and thinking. That requires practice.
 
  • #11
Stephen Tashi
Science Advisor
7,008
1,233
1) ∀x ∈ (0, 1), ∃y ∈ (0, 1), x < y
Before we prove that, should we worry about whether [itex] \exists y \in (0,1) [/itex] ?

i.e. before we worry about whether something in [itex] (0,1)[/itex] satisfies a certain condition, what's the proof that there is anything at all in that set ? A related question is: What's the proof that [itex] 0 < 1 [/itex] ?

Have some theorems been proven related to these questions?
 
  • #12
525
345
Before we prove that, should we worry about whether [itex] \exists y \in (0,1) [/itex] ?
i.e. before we worry about whether something in [itex] (0,1)[/itex] satisfies a certain condition, what's the proof that there is anything at all in that set ?
An empty set has cardinality 0, an interval within real numbers has cardinality c, hence it's not empty - contains elements. Furthermore, an interval is defined to be a set, per the definition of a set - there are no identical elements in a set - therefore 0,1 are different (the ordering of real numbers is axiomatic) and between any 2 real numbers, there exists a set of real numbers with cardinality c - proof by Cantor's process of diagonalization. I highly doubt that introductory analysis demands this.
 
  • #13
Stephen Tashi
Science Advisor
7,008
1,233
An empty set has cardinality 0, an interval within real numbers has cardinality c, hence it's not empty - contains elements.

We need to know whether the Saterial's course work has given him any theorems or further assumptions beyond what's on the page that was given. We can utilize facts and terminology from general mathematical knowledge to prove the exercises, but that doesn't help do the assignment if the purpose is to use only a limited set of assumptions and theorems. If this assignment is of average difficulty, it would only be given after some helpful theorems have been established.

We can consider the following question: Suppose the relation "<" in the page of assumptions given by Saterial was defined to be exactly the reverse of the usual definition of "<" that is employed on the real numbers. Would any of the assumptions on that page be inconsistent with others? If the assumptions are consistent by either interpretation of "<" then it isn't safe to base a proof on our ideas about "<" as the usual interpretation. If the reverse interpretation makes some of the assumptions inconsistent,then it would be a hint about which of the assumptions would be useful in the proof.

The proposition
1) ∀x ∈ (0, 1), ∃y ∈ (0, 1), x < y
is true if the set (0,1) is empty because it is a vacuous implication.

The question of whether (0,1) is not empty or empty divides the proof into two cases.
 
  • #14
Stephen Tashi
Science Advisor
7,008
1,233
Saterial,

Have you had theorems or exercises that prove basic results such as:

(-1)(x) = -x
(x)(0) = 0

?

-------------
Assumption 04 is inconsistent with the reverse interpretation of "<". So it is probably a key to proving 0 < 1
 

Related Threads for: Help -- with proofs please

  • Last Post
Replies
15
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
20
Views
2K
Replies
4
Views
2K
Replies
2
Views
2K
Replies
2
Views
6K
Top