In many mathematical textbooks the definitions are given in the following form (considering even numbers as an example): &quot;A natural number n is called even iff it is a multiple of 2&quot;. Now as we know a definition should serve only as an abbrevation. That is I can use the following form for defining even numbers. &quot;Even number = a natural number that is a multiple of 2&quot;. Is there any difference between the above definitions? I was thinking that the first makes the definition more strict. For example, suppose that I state: &quot;$1.5$ is an even number&quot;. According to the first definition this statement is meaningless because the term &quot;even&quot; is defined for natural numbers. According to the second definition the statement should be false because $1.5$ isn't a natural number. Which of the above definitions is more appropriate and why? Edit I have read the question with the proposed answer before I asked this question. But there is a difference because I am asking what notation should I use and if there is a difference between equality sign and &quot;iff&quot; whereas the other question is about the difference between &quot;if&quot; and &quot;iff&quot;.
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In some books the word "if" is used in definitions and it is not clear if they actually mean "iff" (i.e "if and only if"). I'd like to know if in mathematical literature in general "if" in definitions means "iff".  For example I am reading "Essential topology" and the following definition is written:     In a topological space $T$, a collection $B$ of open subsets of $T$ is said   to form a basis for the topology on $T$ if every open subset of $T$ can be written as a union of sets in $B$.   Should I assume the converse in such a case? Should I assume that given a basis $B$ for a topological space, every open set can be written as a union of sets in $B$?  This is just an example, I am not asking specifically about this sentence.
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This question is about asymptotic notation in general. For simplicity I will use examples about big-O notation for function growth as $n\to\infty$ (seen in algorithmic complexity), but the issues that arise are the same for things like $\Omega$ and $\Theta$ growth as $n\to\infty$, or for little-o as $x\to 0$ (often seen in analysis), or any other combination.  The interaction between big-O notation and equals signs can be confusing. We write things like $$\tag{1} 3n^2+4 = O(n^2)$$ $$\tag{2} 5n^2+7n = O(n^2)$$ But we're not allowed to conclude from these two statements that $3n^2+4=5n^2+7n$. Thus it seems that transitivity of equality fails when big-O is involved. Also, we never write things such as $$\tag{3} O(n^2)=3n^2+4,$$ so apparently commutativity is also at risk.  Many textbooks point this out and declare, with varying degrees of indignation, that notations such as $(1)$ and $(2)$ constitute an "abuse of notation" that students are just going to have to get used to. Very well, but then what are the rules that govern this abuse? Mathematicians seem to be able to communicate using it, so it can't be completely random.  One simple way out is to define that the notation $O(n^2)$ properly denotes the set of functions that grow at most quadratically, and that equations like $(1)$ are just conventional abbreviations for $$\tag{4} (n\mapsto 3n^2+4)\in O(n^2)$$ Some authors even insist that writing $(1)$ is plain wrong, and that $(4)$ is the only correct way to express the estimate.  However, other authors blithely write things like $$\tag{5} 5 + O(n) + O(n^2)\log(O(n^3)) = O(n^2\log n)$$ which does not seem to be easily interpretable in terms of sets of functions.  The question: How can we assign meaning to such statements in a principled way such that $(1)$, $(2)$ and $(4)$ are true but $(3)$ is not?
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A friend of mine was trying to explain to me how all infinities are equal. For example, they were saying that there are the same amount of numbers between $0$–$1$ as there are between $0$–$2$.  The way they explained it, you could prove that there are the same amount of numbers by getting a match to each number.  For example, for any number in the range from $0$–$2$, you can find a matching number in the range $0$–$1$ by dividing the number by 2. In every case, the ending number will end up between $0$–$1$. In addition, if you multiplied any number in the range of $0$–$1$ by 2, you will always end up with a number between $0$–$2$ Therefore, there should be the same infinite amount of numbers between  $0$–$1$ as there are between $0$–$2$.  Is this principle accurate? The problem that I had with it was, if in-fact there are the same amount of numbers between 0 and 1 as there are between 0 and 2, why is 2 greater than 1?
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Prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer.      For clarity: the denominator is the only part being squared.   My thought process:   The numerator is the product of the first $n$ even numbers and the product of the first $n$ odd numbers.   That is, $(2n!) = (2n)(2n-2)(2n-4)\cdots(2n-1)(2n-3)(2n-5).$ In effect, the product of even numbers can be cancelled out with $n!$ resulting in the following quotient:  $$ \frac{(2^n)(2n-1)(2n-3)}{(n!)}\;.$$  To me this looks even thanks to the powers of $2$. But I am not convinced for some reason.   Was this approach to naive?   Sorry for the poor notation, I don't know any coding languages, my apologies.
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Plato puts the following words in Socrates' mouth in the Phaedo dialogue:     I mean, for instance, the number three, and there are many other examples. Take the case of three; do you not think it may always be called by its own name and also be called odd, which is not the same as three? Yet the number three and the number five and half of numbers in general are so constituted, that each of them is odd though not identified with the idea of odd. And in the same way two and four and all the other series of numbers are even, each of them, though not identical with evenness. ()   The  is that there exists an Idea (or Form) called Odd and odd numbers are merely specific instances of Odd.  The are not, themselves, identical to the concept of Oddness.  But I bolded a throwaway phrase that prompts my question: Are half of all integers odd?    Plato likely did not consider  nor did he likely consider either zero or negative integers among his set of "numbers in general".  I don't see his proof (if he had or was aware of one), but I would imagine it be something along the lines that for every even number N there is an odd number N+1.  Therefore half of all numbers greater than 1 are odd.  I'm aware that , which makes me think there is one extra even number than odd numbers.  My thinking is that if half of all positive numbers are odd and half of all negative numbers are odd, than leaving out zero, half of all integers are odd.  But when you add in the only non-positive, non-negative number, which is even, you have an extra even number.  (The programmer in me wants to add the concepts of -0 and +0 for symmetry.  The rest of me thinks that's nuts!)  However, I think my imagined platonic proof still works as long as zero is definitely even and won't work if it's either odd or neither parity.  Are either of these proofs valid?
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This question is me trying to understand (again) why there can be no one-to-one correspondence between the sets of natural and real numbers.  The source of confusion is this: if we abstract completely from the properties and operations on numbers and just think of them as strings of digits, it appears to be possible to list all the real numbers between $0$ and $1$ and all positive integers having the same digits in the same (or reverse) order, for example:  $$0.1 \to 1$$  $$0.01 \to 10$$  $$0.0101 \to 1010$$  $$0.110101 \to 101011$$  I use binary expansion for simplicity.  Since irrational numbers have infinite expansions, we would have to consider some kind of 'infinite natural numbers', like:  $$101110001\dots$$  or  $$\dots101110001$$  I think the main (and only) reason why such numbers don't exist is this requirement:   We need the sets of real and natural numbers to be ordered. We need to be able to do arithmetics on real and natural numbers.   Obviously, if we consider any of two types of 'infinite natural numbers' and the usual absolute value, we find that it's impossible to order them, or do arithmetics with them.  The only way is to change the norm (to $p-$adic norm for example).  While with the real numbers it's still possible to order them, and do arithmetics with them when they have infinite amount of digits in their expansion (or at least in principle).     Is my reasoning correct? Are the two requirements (ordering and arithmetics) enough to explain why natural numbers with infinite amount of digits do not exist, while real numbers do?   This confusion is not restricted to me alone, see  for example
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Are all prime numbers greater than 2 odd?   I wasn't allowed to assume this was true.   AFSOC that there exists an even prime $n$. Then by definition of even, $n = 2q$, where $q$ is a positive integer. But if $n$ is even, it is divisible by $2$, hence that contradicts $n$'s prime-ness. QED.   What? Is there some wacky prime that can't be assumed odd?
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For example, can we say: $\infty=\lim\limits_{n\rightarrow\infty} n &lt; \aleph_0$?  These are two different types of structures. The limit being like the length, extension, or just generic magnitude and the other being cardinality of a set. Can we compare magnitude to cardinality?  Intuitively, we can reach $\aleph_0$ by counting the natural numbers on the number line and in the process will be approaching $\infty$. Which leads me to believe $\infty\leq\aleph_0$. But I can't see why it should be a strict inequality. I feel like they should be of equal magnitude.  I saw on a recent comment that $2^\infty=\infty$, but are those infinities really the same? It seems not to me. Of course we (usually) have that  $2^{\aleph_0}=\aleph_1$ where ${\aleph_0}$ and $\aleph_1$ are clearly two very distinct infinities, countable vs uncountable at least. Maybe one might argue that as far as the concept of magnitude is concerned, all infinities are "equal".
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Consider any integer, suppose a number 3. We can break any such integer down into a sequence of 1's. So, we can write: $$ 3\equiv1+1+1 $$ 3 represents what we get when we add 1 to 1 to another 1. Conceptually, what I am getting confused with, is the fact that notionally, at least according to this definition, there is one unit at the beginning, to which we add another, and yet a third. There seems to be a sequence here, which would imply there being an order. Am I missing something obvious?
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I had a task where I had to figure if the argument was true or not.  $A=\{ n ∈ ℤ \mid n^2 &lt; 5 \}, \quad  B=\{ 7, 8, \{2\}, \{2, 7, 8\}, \{\{7\}\} \}$  The first one was $\{-1, 2\} ∈ A$ and the answer to this was not true  since the set is not an integer.  The second one was $\{-1, 2\} ⊂ A$ and the answer to this was true since  -1 and 2 are elements of $A$.  I just can't seem to understand the difference between ∈ and ⊂ in this task and why the first argument was not true. Also it confuses me that the answer to the first one was that not true because the set is not an integer but aren't -1 and 2 integers?  There is other parts of this task that also add more confusion:     {2, 7, 8} ∈ B Answer: True since {2, 7, 8} is an element of set B      {2, 7, 8} ⊂ B Answer: Not true, for example 2 is not an element in set   B
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Given a set of $n+1$ numbers out of the first $2n$ natural numbers, $\{1,2,\ldots,2n\}$, prove that there are two numbers in the set, one of which divides the other.  I can't tell if I'm reducing the problem in the right way.  Would it be something like:  Induction Hypothesis: A set of $n+1$ numbers from the first $2n$ natural numbers contains two numbers, one of which evenly divides the other.  Base Case: $n=1$, $\{1,2\}$, 2 is divisible by 1.  Consider a set of $n+2$ numbers, $A$, from the first $2n+2$ natural numbers. There are three cases:   Each number in $A$ is less than or equal to $2n$.  The proof follows from the induction hypothesis. One number in $A$ is greater than $2n$.  The remaining set of $n+1$ numbers must each be less than or equal to $2n$.  The proof follows from the induction hypothesis. $A$ contains both $2n+1$ and $2n+2$. I'm not sure how to solve this case.   I'm working through Udi Manber's Introduction To Algorithms: A Creative Approach for personal development, so in keeping with the spirit of the book, I'm trying to use induction.  Without induction, the question is identical to
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An arithmetic function is defined as follows:$f(1)=1$,  $f(2k)=k$ and $f(2k+1)=f(k)+f(k+1)$. When (for which $k$) is $f(k)$ even?  While it is obvious that $f(4n-1)=f(4n)=2n$, therefore $f(k)$ is even for $k=4n$ and $k=4n-1$, which is possible to prove using induction (isn't it?), I don't know how to prove the converse.  I tried to solve the recurrence or make use of constant difference between certain elements of the sequence $\ldots$ What am I missing?  Thank you.
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My sister just asked this question to me: "Is an irrational number odd or even?" I told her that decimals are not odd or even and that does imply that not recurring and non repeating decimals will also not be odd or even. But I want a rigorous math proof for this. Can someone help me with this proof? Also, can I do this proof by contradiction?
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I am not very well versed on set theory or syntax, but I thought I knew the basics.  However, in a book about databases I am reading now, the author uses $2^x$ to signify "a set of $x$."  For example, $2^{\text{dogs}}$ is a set of $\text {dogs}$, etc.    The author never really explained this or why he does it, I just picked up the meaning from context.    I am not sure why the exponent operator is used, nor am I sure what the number $2$ has to do with it.  The sets being represented are NOT powers of $2$ (in size)... they come in all sizes.  Is this a valid notation?  I have not seen it anywhere before...
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I've come to this: $$f: \mathbb{N} \to \{\ldots, -6,-4,-2,0,2,4,6, \ldots\},\qquad f(n) = \begin{cases} 2n &amp; \text{ if } n \text{ is odd} \\  -n &amp; \text{ if } n  \text{ is even} \end{cases}$$  I don't know what to do with this though. I never know how to format a proof correctly.
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I posted this question nearly 10 days ago, but am still really not satisfied with the answers I got, I have no prior education in abstract algebra, group theory, or other abstractions, and most of the answers I got involved defining sets, groups and other things I am unfamiliar with, my question was,  How does one know the notion of real numbers is compatible with the axioms defined for complex numbers, ie how does one know that by defining an operator '$i$' with the property that $i^2=-1$, we will not in some way contradict some statement that is an outcome of the real numbers.  For example if I defined an operator x with the property that $x^{2n}=-1$, and $x^{2n+1}=1$, for all integers n, this operator is not consistent when used compatibly with properties of the real numbers, since I would have $x^2=-1$, $x^3=1$,thus $x^5=-1$, but I defined $x^5$ to be equal to 1.  How do I know I wont encounter such a contradiction based upon the axioms of the complex numbers.  I reiterate here again, Please don't use terminology, or abstract math, that hasn't been taught in say a thorough collage level calculus 1 course.  Here's a link to the same question I asked 8 or so days ago:
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I have a question about how to interpret arbitrary elements. When we say:  &quot;let $x$ be an arbitrary integer&quot;  &quot;let $x$ be any integer&quot;  &quot;let $x$ be an integer&quot;   at the beginning of proofs, are we saying:  &quot;let the symbol $x$ represent one of the integers, although it is not clear which specific integer the symbol $x$ represents&quot;.  If not, please let me know how I should think about the above statements. Thank you for your time.  Edit: I had a follow up question to the one I asked earlier. What do mathematicians mean by “arbitrary element” or “any element”. I can understand when mathematicians say “let $x$ be $4$” as this means “the symbol $x$ represents $4$”. Could someone please elaborate on why the specific terms “arbitrary” and “any” are used?