Difficulty with "Exists" & "Let" or "Arbitrary"

[CGandC]

1. when writing a proof, I stumbled upon cases where I wondered if the following two are equivalent?:
a. " There exists $n \in N$ such that ___ "
b. " Let $n \in N$ be arbitrary such that ___"

Are these statements the same? Is $n$ in both of them the same?

2. Suppose I have the following set

And I want to prove that there exists a real number $m$ such that for all $a \in C$ it is satisfied that $a \leq m$.
So writing of the proof goes as follows:
We'll take $2 \in R$. Let $b \in C$ , meaning there exists $n \in N$ such that $b = \frac{n+1}{2^n}$ . So we'll prove that $\frac{n+1}{2^n} \leq 2$ .

I saw in the answers that the proof of $\frac{n+1}{2^n} \leq 2$ is done with induction on $n$.

The question is: how is it possible to do proof by induction on $n$ here if in the writing of the proof I said " there exists $n \in N$ " and not " arbitrary $n \in N$ " or " Let $n \in N$ "?
I know that I do proof by induction if I have a claim such as $\forall n \in N. P(n)$. But here in the writing of the proof I have a claim of $\exists n \in N. P(n)$ which I have to prove, which isn't the same as the previous claim.

 

[fresh_42]

1. when writing a proof, I stumbled upon cases where I wondered if the following two are equivalent?:
a. " There exists $n \in N$ such that ___ "
b. " Let $n \in N$ be arbitrary such that ___"

Are these statements the same? Is $n$ in both of them the same?

No. T´hey are not.

The existence quantor means: there is one instance, as in "there exists a natural number which is divisible by 3 and 1 only.

Let there be (an arbitrary) is the all quantor. It means e.g.: Given an arbitrary natural number $n$, there is another natural number $n+1$ following $n$. Since we haven't assumed any restrictions on $n$ other than being a natural number, this implies that it is true for all natural numbers. $n$ here is a place holder for any, or all. In the previous example, only $n=3$ satisfied the condition.

2. Suppose I have the following set View attachment 272036
And I want to prove that there exists a real number $m$ such that for all $a \in C$ it is satisfied that $a \leq m$.
So writing of the proof goes as follows:
We'll take $2 \in R$. Let $b \in C$ , meaning there exists $n \in N$ such that $b = \frac{n+1}{2^n}$ . So we'll prove that $\frac{n+1}{2^n} \leq 2$ .

You should have highlighted meaning! We have given any $b\in C$. This means we do not pose any restrictions on $b$. Nevertheless, and all we know is, that it is contained in $C$. This means that $b$ is of the form $\dfrac{n+1}{2^n}$ for some $n\in \mathbb{N}$, since all members of $C$ look this way. We have not determined which one in $C$, except that we've arbitrarily chosen one of them to work with, which we called $b$.

I saw in the answers that the proof of $\frac{n+1}{2^n} \leq 2$ is done with induction on $n$.

The question is: how is it possible to do proof by induction on $n$ here if in the writing of the proof I said " there exists $n \in N$ "
and not " arbitrary $n \in N$ " or " Let $n \in N$ "?

There exists an $n$ refers to the shape of the elements we are talking about, not a specific one. This makes $n$ arbitrary in the induction proof, and specified if we talk about $b$. It would had been better to write $b(n)= \dfrac{n+1}{2^n}\in C$ to demonstrate that $b$ is parametrized or labelled by $n$. However, $b(n)$ and therewith $n$ were arbitrary. Only the shape was specified. The same was true in my examples under point $1)$. $3$ is also divisible by $-3$ and $-1$ but we were talking about natural numbers only. The same happens here: we are talking about $C$ only, which makes $b=b(n)$ arbitrary except for its shape. The label $b$ is only given in order that we can address those elements, as we address natural numbers by $n$.

I know that I do proof by induction if I have a claim such as $\forall n \in N. P(n)$. But here in the writing of the proof I have a claim of $\exists n \in N. P(n)$ which I have to prove, which isn't the same as the previous claim.

I'm a little afraid that this explanation might have confused you a bit. If so, please ask. And btw., you are right: it would have been better if we used logical symbols:
$$
\forall \,b\in C\, : \, b<2\in \mathbb{R} \Longleftrightarrow \forall\, b(n)\in C\, : \,b<2\Longleftrightarrow \forall \, n\in \mathbb{N}\, : \,P(n) \Longleftrightarrow \forall \,n\in \mathbb{N} \, : \, \dfrac{n+1}{2^n}<2
$$

 

[Mark44]

1. when writing a proof, I stumbled upon cases where I wondered if the following two are equivalent?:
a. " There exists n∈N such that ___ "
b. " Let n∈N be arbitrary such that ___"
Are these statements the same?

As already stated, no they are not the same. In the first, it is asserted that there is a specific number n that satisfies the given condition, and possibly others as well.
In the second, any integer n will satisfy the given condition.

 

[sysprog]

1. when writing a proof, I stumbled upon cases where I wondered if the following two are equivalent?:
a. " There exists $n \in N$ such that ___ "
b. " Let $n \in N$ be arbitrary such that ___"

Are these statements the same? Is $n$ in both of them the same?

They're not the same $-$

the statement

$\exists n [(n \in \mathbb N) \wedge Pn]$

means that there is at least one natural number $n$ that has property P,

while the statement

$\forall n [(n \in \mathbb N) \Rightarrow Pn]$

means that every natural number $n$ has property P.

2. Suppose I have the following set

And I want to prove that there exists a real number $m$ such that for all $a \in C$ it is satisfied that $a \leq m$.

Symbolizing what you posit as to be proven:

$\exists m [(m \in \mathbb R) \wedge (\forall a [(a\in C) \Rightarrow (a \leq m)]$,

then, noting that in the expression defining membership in $C$, the numbers generating the values for $a \in C$ are the natural numbers $n \in \mathbb N$, and as we ascend through the natural numbers, beginning with $n=1$ we get:

$\frac {(1+1)^2} {2^1} = \frac {2^2} {2^1} = \frac 4 2 = 2$,

and:

$(n=2) \rightarrow (a =\frac {3^2} {2^2} = \frac 9 4 = 2.25)$,

but after $n=2$, the divisor grows faster than the dividend, so the quotient decreases for every natural number greater than 2:

$(n = 3) \rightarrow (a= \frac {16} 8 = 2)$,

$(n=4) \rightarrow (a= \frac {25} {16} = 1.5625)$,

$(n = 5) \rightarrow (a= \frac {36} {32} = 1.125)$,

(at $n=6$ the fraction is no longer top-heavy:)

$(n = 6) \rightarrow (a= \frac {49} {64} = 0.756625)$,

$(n = 7) \rightarrow (a= \frac {64} {128} = \frac 1 2 = 0.5)$,

$(n = 8) \rightarrow (a= \frac {81} {256} = 0.31640625)$,

$(n=9) \rightarrow (a= \frac {100} {512} = 0.1953125)$,

$(n=10) \rightarrow (a=\frac {121} {1024} = 0.1181640625$

So it's clear that for greater values than $2$, with the divisor continuing to double, while the dividend continues to be a $1$-greater number squared, the $2.25$ value of $a$ when $n=2$ can only diminish for $n>2$, and that's the mathematical induction step (informally stated) that justifies the conclusion that

$\exists m [(m \in \mathbb R) \wedge (\forall a [(a\in C) \Rightarrow (a \leq m)]$;

however, if we take the problem as it is here stated, then because the problem statement calls for a real number $m$, and the highest value we can get for $a$ that is generated by a natural number $n$ is 2.25 , which occurs when $n=2$, the correct answer cannot be $m=2$; instead, the correct answer would be $m=2.25$,.

So writing of the proof goes as follows:
We'll take $2 \in R$. Let $b \in C$ , meaning there exists $n \in N$ such that $b = \frac{n+1}{2^n}$ . So we'll prove that $\frac{n+1}{2^n} \leq 2$ .

Your prior statement of the problem clearly calls for the universal quantifier for $n$ here; not for the existential quantifier:

And I want to prove that there exists a real number $m$ such that for all $a \in C$ it is satisfied that $a \leq m$.

(emphasis added)

All members $a \in C$ are produced, respectively, by all members ##n \in \mathbb N.

I saw in the answers that the proof of $\frac{n+1}{2^n} \leq 2$ is done with induction on $n$.

That statement is true if and only if $n$ is not equal to $2$, i.e. when $n=1$, $a=2$; when $n=2$, $a=2.25$; as $n$ increases after that, $a$ decreases $-$ when $n=3$, $a=2$, with further diminution of $a$ for all further incrementations of $n$.

So:

$((n \in \mathbb N) \wedge (\frac{n+1}{2^n} \leq 2)) \iff \neg(n=2)$

I think that there may be some confusion regarding whether the quarry is $m=n=2$, i.e $m$ is equal to the natural number $n$ by which the greatest number $a \in C$ is generated, i.e. $m=2$ or rather is, as the problem as statement appears to call for, the real number $m$ to which the greatest $a \in C$ is less than or equal, i.e. $m=2.25$, which is generated when $n=2$.

The question is: how is it possible to do proof by induction on $n$ here if in the writing of the proof I said " there exists $n \in N$ " and not " arbitrary $n \in N$ " or " Let $n \in N$ "?

The problem statement uses both types of quantifier: there exists a real number $m$ such that for all $a \in C$, $a \le m$.

I know that I do proof by induction if I have a claim such as $\forall n \in N. P(n)$. But here in the writing of the proof I have a claim of $\exists n \in N. P(n)$ which I have to prove, which isn't the same as the previous claim.

Please post the problem and answer exactly as you encountered it.

 

[disregardthat]

The major difference between 1. and 2. is that 1. is a sentence (with some truth value), while 2) is not.

When we say: "Let n be [something] satisfying P(n)" for some proposition P, we are not positing anything.

In a proof by induction, the goal is to prove A: $P(1)$ and B: $P(n) \Rightarrow P(n+1)$. For B, one usually starts out with "Let n be [something] satisfying P(n)". Note that in your example (the one with "let $b \in C$, there exists $n \in N$ [...]"), you're not formulating the beginning of the proof of the inductive step. The inductive step is as follows:

Let $n \in N$ be some positive integer satisfying $\frac{(n+1)^2}{2^n} \leq 2$. We are not positing that something is true (in fact, it's not meaningful to assign any truth value to it when n is indeterminate). We are rather supposing it to be true for an indeterminate n for the sake of proving $P(n) \Rightarrow P(n+1)$. To complete the proof of the inductive step, prove that $\frac{((n+1)+1)^2}{2^{n+1}} \leq 2$.