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0 votes
0 answers
49 views

No positive integers $(a,b)$ satisfying $2^b -1\mid 2^a +1$ (AoPS Vol 2)

I posted this proof on AoPS but I'm having trouble understanding what's wrong with it. I was hoping this community could offer a different perspective? Suppose the claim is false. That is to say, ...
Sage Stark's user avatar
1 vote
2 answers
642 views

Well-Ordering Principle "proof"

Theorem. Well-Ordering Principle. Every non-empty subset of natural numbers has a least element. I have seen some proofs for the theorem, but is very "complex"proof really needed here? My ...
mathslover's user avatar
  • 1,484
1 vote
1 answer
272 views

Confusion in well ordering principle

Well-ordering principle states that every non-empty set of positive integers contains a least element. I have a set S which is a subset of natural numbers. Now by well-ordering principle I can ...
Singh's user avatar
  • 2,138
-1 votes
2 answers
269 views

Using the WOP to check divisibility

The well-ordering principle states that every non-empty set of positive integers contains a least element. I need to prove that 9|$n^3+(n+1)^3+(n+2)^3$ , $n∈N$ using this principle. Regards
Jewgah's user avatar
  • 29
1 vote
1 answer
145 views

Proof related with Well ordering principle

In each of the following scenarios use the Well Ordering Principle to answer the questions: a) Is there a process that is capable of generating an infinitely long sequence of numbers $a_1,a_2,a_3\dots$...
Ethan's user avatar
  • 5,283
0 votes
3 answers
202 views

Help solving this proof using Well Ordering Principle

Let $a_1,a_2, ... , a_n ∈ \Bbb{N} $ . Prove that there exists $ l ∈ \Bbb{N} $ such that $a_i | l$ for all $i ∈ \{1,2,...,n\}$ and if $x ∈ \Bbb{N} $ is such that each $a_i$ divides $x$, then $l | x$. ...
somber_singularity's user avatar
1 vote
1 answer
56 views

proof of Well Ordering Principle over positive integers

Theorem If $A$ is any nonempty subset of $\mathbb{Z}^+$, there is some element $m \in A$ such that $m \leq a$, for all $a \in A$ where such $m$ is the minimal element of $A$. (AA: Dummit and Foote) ...
rankNull's user avatar
3 votes
2 answers
328 views

Well ordering principle

I am trying to understand why is the well ordering principle stated as an axiom of integers. In the process, I found a "proof" of the principle (which is obviously wrong) and want to ...
volatile's user avatar
  • 259
0 votes
1 answer
81 views

Application of the Well-Ordering Principle

I am suppose to show that there is no infinite sequence of strictly decreasing non-negative integers and know that the Well-ordering principle will have to come into play. Also, will induction need to ...
Kyogre's user avatar
  • 193
1 vote
1 answer
257 views

well-ordering principle for natural numbers from the definition of real numbers.

Define the set of real numbers $\mathbb{R}$ by means of the following axiom: There exists a totally ordered field $(\mathbb{R},+,\cdot,\leq)$ which is Dedekind complete. We also assume that $a\leq b$ ...
user avatar
0 votes
2 answers
101 views

Show that the set {1/6, 1/7 , 1/8,.....} does not have a least element

Show that the set $\{\frac 16,\frac 17 ,\frac 18,\dots\}$ does not have a least element and conclude that no set containing this set is well ordered. I am not sure how can I show this ... The set ...
dr.rise's user avatar
  • 619
2 votes
1 answer
115 views

Is this proof using well ordering principle correct? Are all parts necessary?

Example. Suppose the Royal Canadian Mint was to introduce a 3 cent coin like the British three pence to replace the 1 cent penny. Prove that 7 cents is now the largest quantity unable to be made with ...
Jacob House's user avatar
1 vote
0 answers
63 views

Question about choice of set used in proof of Division Algorithm with the Well Ordering Principle

When we prove the existence part of the division algorithm, namely "Given natural numbers $a, n$, there exist integers $q$ and $r$ such that $a = nq + r$ with $0\leq r < n$" , I constructed the ...
quietkid's user avatar
  • 151
0 votes
2 answers
187 views

Prove that all integers can be represented as powers of 2 multiplied by an odd number [closed]

I would like to prove that $\forall n \in \mathbb{N},n\geq1$ $, n = 2^k\times m$ with $k \in \mathbb{Z}$ and $ m $ odd and $m \in \mathbb{N}$.
Ibrahim Abouhashish's user avatar
6 votes
1 answer
2k views

Proof of Lehman's Lemma with the well ordering principle

I was reading an article on the well ordering principle and there was a problem that asked to use the well ordering principle to solve Lehman's Lemma: That there are no positive integer solutions to ...
MITjanitor's user avatar
  • 2,718