10
$\begingroup$

I'm being asked to prove that if $k∈ \Bbb Z$, $k+1>k$. Judging from our instructions, it appears (I am unsure) as though I cannot use the law of induction to solve this. A hint gives that the proof depends on $1∈\Bbb N$. I was thinking of approaching this by using if $k∈\Bbb N$ then $x+k∈\Bbb N$. Problem is, we were not given $k∈\Bbb N.$ Is this solvable without induction?

We are given associativity, commutivity, and the identity elements for addition and multiplication, the additive inverse, the properties of $=$, and basic set theory and logic.

Order for the integers is given by Let $m,n,p∈\Bbb Z$. If $m

Also assumed, and possibly relevant is if $x∈\Bbb Z$, then $x∈\Bbb N$ or $-x∈ \Bbb N$ or $x=0$.

  • 3
    You'd need to expand on what definitions you're working from. – 2017-01-22
  • 0
    Welcome to Math SX! What's your definition of the order on $\mathbf Z$? – 2017-01-22
  • 0
    Might also be helpful to you and to us if you familiarized yourself with MathJax/LaTeX markup. – 2017-01-22
  • 4
    What is your definition of "$>$"? And to @TheCount 's point, [here's some guidance on MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) – 2017-01-22
  • 3
    Can't you substract $k$ in the inequation and assume that $1 > 0$? – 2017-01-22
  • 0
    I suppose I could, but I don't see how that requires use of $1∈N$ P.S. I tried to make everything in-line equations, is that correct? In regards to Joffan's question on the definition of >, the proposition Order for the integers is given by "Let m,n,p∈Zm,n,p∈Z. If m – 2017-01-22
  • 0
    Depending on what one assumes as obvious, to proof this can range from trivial to very hard. – 2017-01-22
  • 0
    I've tried to improve the question for anyone else who reads this. If there is more I can do to improve it, please let me know. I apologize for being so fail at asking question on here, in my defense, this is my first time. Thank you so much for the help. – 2017-01-22
  • 1
    I think that some of the comments above are asking for a definition of $x>y$. In my answer, I am using $x>y$ defined as $x-y\in\mathbf{N}$, but it's not necessarily the only way to define it. I do not think your "if $m$ (without some extra work) because it already seems to assume that $>$ is defined. – 2017-01-22
  • 0
    Yes, @mvw, if you pretend you don't know *anything* which you do in actual fact know with certainty, then proving any particular fact will be impossible. The "proof" of this is by inspection. This question is stupid, and I mean that in the most literal possible sense—technically, epistemologically. (A *stupidity* is an *unknownness.* This question assumes unknownness of everything, and unavailability of direct observation to find out.) – 2017-01-22
  • 0
    @Wildcard no, it's not a stupid question. You can derive the properties of the integers rigorously by a smallish set of axioms; some of the properties are quite "intuitive", but if they are not part of the axioms, you have to derive them. It's easy, but not immediate, to prove that $1\neq 0$, for example. Check out https://www.math.ucdavis.edu/~osserman/classes/115A-F14/notes/A-axioms.pdf – 2017-01-22
  • 0
    @Anonymous, rigorously? No, not really. The very statement in that PDF which states *what the axioms are about* uses "a" "the" **"two"** "we" and "nonempty," to say nothing of "set" and "subset." These statements are entirely meaningless without **first** having a concept of quantity and counting, and "nonempty" in particular *requires* distinguishing $1$ from $0$ in order to *understand* the word. I reiterate: Assuming that nothing is known except what is "explicitly stated" makes the relay of ideas via implicit symbols impossible—and *all symbols are implicit.* Precisely *stupid.* – 2017-01-22
  • 1
    @Wildcard this is degenerating, so I'll drop off after this comment (please do not take it as rudeness). However: note that the 0 and the 1 that people introduce when axiomatizing the integers have no meaning except being, respectively, the additive and multiplicative identity. Of course, you need formal logic on top of the axioms etc. But the essence is, you can't base yourself on "intuition" when working at the level the original question was asked. – 2017-01-22
  • 0
    @Anonymous, I don't take it as rudeness at all. I hope you understand that I'm using the word "stupid" in a precisely defined sense (involving the unknownness of consideration) rather than a mere label. My approach is similar to intuitionism, but I can't find any named branch of intuitionism which I entirely agree with. Fundamentally, though, I'm defining "intuition" as the *understanding* of something independent of *mechanical* means—and further asserting that *it is impossible to even understand a written question without intuition.* Understanding is a function of the mind, not symbols. – 2017-01-22

3 Answers 3

12

As stated in the comments, the answer may depend to some extent on what set of axioms for the integers you are using; it also depends on the exact definition of $"a>b"$. But under most axiomatizations, assuming that $"a>b"$ is defined as $"(a-b)\in \mathbb{N}"$ (where $\mathbb{N}$ is the set of positive integers), something like the following proof by contradiction would work:

$[(k+1)\ngtr k]~~~ \equiv~~~~ [(k+1)-k \notin \mathbb{N}]~~~$(definition of $a>b$)

$\implies~~~ [(1+k)-k \notin \mathbb{N}]~~~$ (commutativity of addition)

$\implies~~~[1+(k-k) \notin \mathbb{N}]~~~$ (associativity of addition)

$\implies~~~[1\notin \mathbb{N}]$

Note that $1\in\mathbb{N}$ is often not part of the axioms of the integers, and in general must itself be proved. The original question suggests it has "already" been proved, so here we can take it as a given. Otherwise, a rough outline of the proof that $1\in\mathbb{N}$ would be (under most axiomatizations of the integers, such as the one here, that I particularly like):

  1. Prove that for any integer $a$ we have $a\cdot 0=0$. Proof sketch: $0=a\cdot 0 - a\cdot 0$ $= a\cdot (0+0) - a\cdot 0$ $=a\cdot 0$.
  2. Prove that for any integer $a$ we have $-a=a\cdot(-1)$. Proof sketch: using the above $0 = a \cdot (1+(-1)) = a+ a\cdot(-1)$.
  3. Combine the above with the axiom of closure of $\mathbb{N}$ under multiplication to obtain $-1\notin\mathbb{N}$.
  4. Exploit the trichotomy axiom (every integer $a$ must satisfy exactly one of the following three: either $a\in\mathbb{N}$, or its additive inverse $-a\in\mathbb{N}$, or $a=0$ i.e. $a$ is an additive identity) to deduce that $1\neq 0$ since otherwise for one $a\in\mathbb{N}$ we'd have $a=a\cdot 1= a\cdot 0=0$.
  5. Exploit the trichotomy axiom again to conclude that since $1\neq 0$ and $-1\notin\mathbb{N}$, then $1\in\mathbb{N}$.
  • 0
    I hope this is indirect. – 2017-01-22
  • 0
    Eh, wrong symbol... fixed it (thanks @mvw). – 2017-01-22
  • 5
    You could start with 1 \in N with no negation or contradiction. – 2017-01-22
  • 0
    I don't get how the first $\Rightarrow$ follows from associativity, as we have $k-(k+1)=(k-k)-1\neq(k-k)+1$. – 2017-01-22
  • 0
    Oh, I guess it should be $[(k+1)-k\not\in\mathbb N]$ in the second formula. Then it should work out. – 2017-01-22
  • 0
    @HermannDöppes yes, I fixed it. I've made so many mistakes in writing down this simple formula, I can't give up now :) – 2017-01-22
  • 1
    I tried to fix the unbalanced parens in the second and third line, but there seems to be a questionable rule enforcing edits to change at least six characters. – 2017-01-22
  • 0
    @SvenMarnach Your edits on my posts are subject to that rule. My edits on my posts are not - luckily :) Fixed it, and thanks for pointing it out. – 2017-01-22
  • 0
    @djechlin I could certainly go without contradiction as you say, but I'd rather start with the goal, and end with $1\notin \mathbf{N}$ so that I can then continue with "what if we have not proved *that* yet?". I guess it's a matter of taste :) – 2017-01-22
  • 0
    @Anonymous Once you reach 2k reputation points, the rule is lifted even for edits to other people's posts. The idea of the rule is that edits by low-rep users needs to be reviewed, and they want to avoid that overhead for trivial edits. However, in some cases editing even a single character can significantly improve a post, in particular on math StackExchange, so I think it's the wrong way of trying to avoid trivial edits. – 2017-01-22
8

If you assume that $1 > 0$ is true, so it's trivial because, given $k \in \mathbb{Z}$, $$ \mbox{ the inequation } \quad k + 1 > k \qquad \mbox{ is equivalent to } \qquad 1 > 0 $$ In general, if $x , y , z \in \mathbb{R}$, $$ \mbox{ the inequation } \quad x > y \qquad \mbox{ is equivalent to } \qquad x + z > y + z $$ and you can replace "$>$" by "$\geq$", "$<$", "$\leq$" or "$=$".

  • 0
    can't this be extended to well ordered fields? – 2017-01-22
  • 0
    Yes it can be done but must always maintain some properties – 2017-01-22
  • 0
    Perhaps $1 ∈ N$, is supposed to be used to indicate that 1>0? – 2017-01-22
  • 0
    Not necessarily because $\mathbb{R}$ is also a well ordered field; for example $\pi > 0$ and $\pi \in \mathbb{R} \setminus \mathbb{N}$. – 2017-01-22
  • 1
    @B.Simmons when you were given the axiomatic definition of the integers, there were some axioms defining the set of the "positive" integers, too (which is what you denote by $\mathbf{N}$). The usual way to define the relation $>$ between two integers is to say "we write x>y to mean that the sum between $x$ and the additive inverse of $y$ belongs to the set of positive integers". But you need to say *somewhere* what you mean by $x>y$. – 2017-01-22
  • 3
    I don't know if there is a proof of the inequation $1 > 0$, but if you assume it, you have that $k + 1 > k$ for all $k \in \mathbb{N}$ by the explanation in my answer. – 2017-01-22
  • 0
    How on Earth could you possibly have 1 *not* greater than 0? $1 \gt 0$ is true by definition. If these symbols aren't defined, they are meaningless; if they *are* defined, there is no question left to be answered. I understand certain mathematicians' propensity for pondering the imponderable and trying to "unknow" everything and then prove it without knowing it—but this particular emperor is without garb. – 2017-01-22
1

You should search for it in Theory of Numbers. There are axioms for proving it for Natural numbers. Peano's Axioms this can be used to prove it. I hope you solve it.

A set of $N$ objects is called a set of natural numbers if it satifies Peano's Axioms:

  • Axiom 1: $1$ belongs to $N$
  • Axiom 2: to each element $n$ belonging to $N$, there corresponds a unique element $n'$ called sucessor of $n$.
  • Axiom 3: For each n belonging to $N$, we have $n' \ne 1$.
  • Axiom 4: if $m,n$ belong to $N$ then $m' = n'$ implies that $m=n$, or $m\ne n$ implies that $m' \ne n'$.
  • Axiom 5: et $M$ be a set of elements of $N$, i.e. $M$ is a subset of $N$ then $M=N$ provided the following two conditions are satisfied
    1. $1$ belongs to $M$
    2. If $k$ belongs to $M$ then $k'$ belongs to $M$, $k'$ being sucessor of $k$.