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I have seen contradictory answers online, and I would like a full clarification of the difference or defining difference between the following, where $P$ is a statement about natural numbers:

  • Assume $k \in \mathbb {N}$ and $P(k)$
  • Assume $\exists k{\in}\mathbb {N}: P(k).$
ryang
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Nav Bhatthal
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    In (1) $k$ is a fixed number. The variable $k$ is free. In (2) there is no free variable. It is a statement. The letter $k$ may be replaced (twice) by any other letter. – Anne Bauval Aug 07 '23 at 07:39
  • Is saying something is free and fixed not contradictory? – Nav Bhatthal Aug 07 '23 at 07:50
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    Let $Q(k)$ be: $k\in \Bbb N$ and $P(k).$ In $Q(k),$ the variable $k$ is free. This is a syntactical notion. In the sentence "Assume $Q(k)$", you are talking about some fixed predefined $k.$ – Anne Bauval Aug 07 '23 at 07:55
  • So that is the correct assumption we should use for induction? What is wrong with using $(2)$ for induction? Thanks Anne! – Nav Bhatthal Aug 07 '23 at 07:57
  • For proving an induction step $$(\forall k\in\Bbb N)(P(k)\implies P(k+1)),$$the correct assumption is (1). – Anne Bauval Aug 07 '23 at 08:00
  • Thanks, and why in particular is $2$ wrong, they seem very familiar to me but I need to read more on free variables and stuff, do you have reading material you think is particular good? – Nav Bhatthal Aug 07 '23 at 08:00
  • Sorry I don't. (2) is wrong because (1) is right. And (1) is right because any proof of $(\forall k\in X) R(k)$ begins with "Assume $k\in X,$ let us prove $R(k)$", and any proof of $A\implies B$ begins with "Assume $A,$ let us prove $B$". – Anne Bauval Aug 07 '23 at 08:06
  • I just can't justify why they are different, $(1)$ and $(2)$ that is. – Nav Bhatthal Aug 07 '23 at 08:20
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    In (1) you are talking about some $k,$ for which you intend to prove $P(k+1)$ (as a result of the 2 proof rules of my previous comment). In (2) you are only assuming that $P(~)$ holds for some (unknown and unnamed) natural number, and this assumption is of no use in an induction proof. – Anne Bauval Aug 07 '23 at 08:28
  • Is the number not named $k$ In $(2)$? And doesn't "for some" mean "there exists", meaning "for some" is $2$ and not $1$? https://math.stackexchange.com/questions/1378906/whats-logical-symbol-for-for-some – Nav Bhatthal Aug 07 '23 at 08:40
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    There is no named number in (2) (as I said, "the letter $k$ may be replaced..."). For some = there exists is indeed in (2) and not (1). In (1) you fix some $k$ and you make an assumption (and then a proof) about that $k.$ In (2) you only make an (unusable) assumption about the property $P.$ – Anne Bauval Aug 07 '23 at 08:51
  • Thanks Anne this really helped. Now I do not mean to sound like a broken record but what is the defining difference between $1$ and $2$ that means the $k$ cannot be replaced in $1$ and hence it is different to $2$? – Nav Bhatthal Aug 07 '23 at 08:55

3 Answers3

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I assume that you are bringing this up in the context of an inductive proof, whose inductive step usually starts out with something like

"Assume that $P(k)$ for some number $k$".

Now, the use of 'some' here makes it sound like your (2): $\exists k \in \mathbb {N} : P(k)$

However, this is certainly not the assumption of the inductive step. Consider: suppose we use that assumption. We could then also suppose that $k_0$ is one of those numbers, and maybe we can even show that that means that $k_0 +1$ has property $P$ as well. OK, so what have we then proven? All we would have proven is that there is some number such that it and its successor have property $P$. In logic: we would have shown that $\exists k \in \mathbb {N} : P(k) \land P(k+1)$. Or, discharging the assumption, we obtain $$\exists k \in \mathbb {N} : P(k) \to \exists k \in \mathbb {N} : P(k) \land P(k+1)$$ But that is not what the inductive step tries to show.

The inductive step tries to show that $\forall k \in \mathbb {N} : P(k) \to P(k+1)$. We do that by picking some arbitrary number, assume that it has property $P$, and use that to show that $P(k+1)$.

So here is the difference: instead of assuming $\exists k \in \mathbb {N} : P(k)$, we simply assume $P(k)$ once $k$ has been introduced as an arbitrary number. So, the assumption is really just $P(k)$. Once we obtain $P(k+1)$, we can discharge the assumption and get $P(k) \to P(k+1)$, and by noting that $k$ was an arbitrarily chosen number, we can generalize this for all numbers, and thus obtain $$\forall k \in \mathbb {N} : P(k) \to P(k+1)$$

So, the assumption is really more like your (1), where $k$ is a free variable. Indeed, when the inductive step starts out with:

"Assume that $P(k)$ for some number $k$"

you should really read that as:

"Assume that $P(k)$ for some arbitrary number $k$"

or, better yet, as:

"Let $k$ be some arbitrary number and assume that $P(k)$"

It is $k$ being an arbitrary number (as reflected by it being free) that allows you to conclude with the universal statement that you want.

Bram28
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    Thanks so much. $(1)$ is definitely the correct inductive assumption! – Nav Bhatthal Aug 07 '23 at 14:16
  • @Mathguy Yes! Hope this analysis was helpful. – Bram28 Aug 07 '23 at 15:22
  • $x,y$ are real numbers here. I would like to ask how do you make sense of $\forall x: y = 5x$ (where $y$ is free) and the formula $y = 5x$ (where $x,y$ are free). I know that $\forall x \exists ! y: y = 5x$ is true always. – Nav Bhatthal Aug 08 '23 at 09:21
  • @Mathguy Much depends on the context in which these expressions are used. In the context of an inductive proof, the use of $k$ in $P(k)$ is understood to be an arbitrary natural number. In some other proof context, the $y$ and $x$ used in $y = 5x$ could likewise refer to arbitrary contexts. But more likely is that you will find the expression $y = 5x$ within some mathematical context where the $x$ and $y$ are specific numbers whose value we may not necessarily know. Or, it could be our description or even definition of a function. Symbols can be used in many, many, different ways! – Bram28 Aug 08 '23 at 11:48
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"Assume $\exists k \in \mathbb{N} : P(k)$"

Assume there exists at least one natural number $k$ such that $P(k)$ holds true. The specific natural number for which $P$ is true is not named. You are merely told that one exists for which $P$ is true.

"Assume $k \in \mathbb{N}$ and $P(k)$."

$k$ is assumed to be a specific natural number for which $P(k)$ holds true.

RyRy the Fly Guy
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  • Why is the number in $2$ not named when we use the symbol $k$ in both instances but it only "Names" the number in the first way? – Nav Bhatthal Aug 07 '23 at 08:54
  • are both statements a part of the same proof? does one follow after the other? – RyRy the Fly Guy Aug 07 '23 at 09:02
  • I am talking about when we start the proof, the two different statements being the first one I write down. If I write $\exists k$ why does that mean that the number I am referring to is not named, I used a letter $k$ surely? (Here I have not used $k$ before this instance) – Nav Bhatthal Aug 07 '23 at 09:03
  • When the variable $k$ is bound to the existential quantifier $\exists$, it does not represent a specific object in the domain; rather, it represents any arbitrary or random object in the domain. And the existential quantifier effectively tells you that $P$ holds true for at least one of the arbitrary objects, without naming them. In the other statement, the existential quantifier has been removed and you are being asked to treat $k$ no longer as an unnamed object but as one of the specific objects for which $P$ is true. – RyRy the Fly Guy Aug 07 '23 at 09:12
  • Technically, a different letter should be used whenever a quantifier is eliminated and a variable, in this case $k$, is instantiated. It helps to avoid the kind of confusion you're currently experiencing. But sometimes people do not use a different letter. Instead, they simply ask you to stop treating the letter as a variable and start treating it as a constant. – RyRy the Fly Guy Aug 07 '23 at 09:15
  • Alright, so when we start the proof, we have a choice on $1$ or $2$, if we chose $2$ we have effectively deduced nothing useful, if we chose $1$ then we can construct a proof by induction, got it! And if we mistakenly write down $2$, to be extra neat we should write down $1$ but with a different letter instead. – Nav Bhatthal Aug 07 '23 at 09:17
  • it might be helpful if you actually explain what you're trying to prove and write down the first few lines of your proof. then we can give you some feedback – RyRy the Fly Guy Aug 07 '23 at 09:27
  • Actually, by reading your older comments, its making much more sense, do you know any good things to read to get more understanding on logic/free/bound variables and stuff like this? So I can answer questions like, why does $\exists$ not name a number on my own? – Nav Bhatthal Aug 07 '23 at 10:09
  • Yes! see https://www.amazon.com/Concise-Introduction-Logic-Patrick-Hurley/dp/1305958098/ref=pd_lpo_sccl_2/140-1517274-6961447?pd_rd_w=kwYFj&content-id=amzn1.sym.116f529c-aa4d-4763-b2b6-4d614ec7dc00&pf_rd_p=116f529c-aa4d-4763-b2b6-4d614ec7dc00&pf_rd_r=1DEBD776H1KXDYJK0P9P&pd_rd_wg=wvywt&pd_rd_r=95e80f0d-f250-4a38-919b-d7af61c02eb8&pd_rd_i=1305958098&psc=1 – RyRy the Fly Guy Aug 07 '23 at 10:24
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  • $k \in \mathbb {N}$ and $P(k)$

In this logic formula, depending on the context, $k$ could be either

  1. a constant with a predefined value
  2. an arbitrary constant
  3. a free variable (strictly speaking: a variable whose occurrences here are all free).
  • $\exists k{\in}\mathbb {N}: P(k).$

In this logic formula, $k$ is a

  1. bound variable (strictly speaking: a variable whose occurrences here are all bound).

Notice that in case #3, the logic formula is just a propositional function rather than a statement.

Follow-up comments

Among the $\text“k∈\mathbb N$ and $P(k)\text”$ cases, which one are we using for mathematical induction (“Assume $k \in \mathbb {N}$ and $P(k)\text”)\,?$

The induction step begins by letting $\boldsymbol{k}$ be an arbitrary natural number, then assumes that $\boldsymbol{P(k)}$ is true.

Eventually, the induction step generalises the foregoing conditional to conclude that $\forall n{\in}\mathbb N \;\big(P(n)\implies P(n+1)\big),\tag*{}$ where $\boldsymbol n$ is clearly a bound variable. It is common to instead write $\forall k{\in}\mathbb N \;\big(P(k)\implies P(k+1)\big),\tag*{}$ but note that in this statement, $\boldsymbol k$ is a bound variable and actually a different object from the arbitrary constant $\boldsymbol k$ in the preceding lines.

“Assume $\exists k{\in}\mathbb {N}: P(k)\text”$

As observed in Bram28's excellent answer—which has marvellously addressed the root of your Question—this sounds like a misinterpretation of the frequently-seen induction-hypothesis construction, “Assume that $P(k)$ for some natural number $k,\text”$ which actually means, “Assume that $P(k)$ for an arbitrary natural number $k,\text”$ rather than, “Assume that $\exists k{\in}\mathbb {N}\: P(k).\text”$ Most plainly: “Let $k$ be an arbitrary natural number, and assume that $P(k).\text”$

ryang
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