Introduction  This probably sounds a bit weird to ask, but the question is, is there such a base that is base 1.  One thing we know is that this base will contain one symbol only, but will this symbol represent something i.e. a value, or nothing?  If we try to see a connection between other bases, for instance base 10, 4, 3, 2, we see following:   base 10 {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} base 4  {0, 1, 2, 3} base 3  {0, 1, 2} base 2  {0, 1}   So base 1 should have the symbol $0$, in order to fit this pattern.  Number representation  Let's say that we introduce a new symbol, which will be used in base 1, namely $\alpha$. We will probably think that if we list the first numbers in base 1, we will get something like:  $$\alpha,\alpha\alpha, \alpha\alpha\alpha, \alpha\alpha\alpha\alpha$$  Which is quite logical. But, if we observe other systems, such as base 10, we will see that 10 is actually built up of two symbols. Similarly, for radix 2, $10_2$ is equal to 2, which appears to be the base.  So, would it not now be logical to claim that the definition of 1 (base 10) in base 1 is $\alpha\alpha$, or with other words, two symbols?  Number conversion  If we would, for example, convert $\alpha\alpha\alpha\alpha$ to radix 10, we would get following: $$(\alpha\alpha\alpha\alpha)_1=\alpha*1^3+\alpha*1^2+\alpha*1^1 + \alpha*1^0 = (4\alpha)_{10}$$  Sure enough, $$(\alpha\alpha)_1 = 1_{10} = (2\alpha)_{10}$$ So I suppose this statement is therefore true as well:  $$(\alpha\alpha\alpha\alpha)_1=(2\alpha)_{10} + (2\alpha)_{10} = 1 + 1 = 2$$  But the interesting part here is that our symbol $\alpha$ is nothing, i.e. does not have any value at all, which implies that all numbers do not exist as values.  Conclusion  I do not now how correct this is, so it would be great if someone could comment upon what is written here. The question is finally, is this plausible?  Thank you,  Artem
120624
Base $10$ uses these digits: $\{0,1,2,3,4,5,6,7,8,9\};\;$ base $2$ uses: $\{0,1\};\;$ but what would base $1$ be?    Let's say we define Base $1$ to use: $\{0\}$. Because $10_2$ is equal to $010_2$, would all numbers be equal?  The way I have thought Base 1 might be represented is tally marks, $0_{10}$ would be represented by nothing. So, $5$ in Base 1 would be represented by $00000$? Or we could define Base 1 to use: $\{$|$\}$ and $5$ would be |||||?
69311
I know that it is possible to represent each natural number using the concept of successor. But how do you assign meanings into the representation? I mean, when presented as a set, it's just a set unless we agree on some definition that some form of sets refer to natural numbers. Or is it something impossible at ZFC?  Thanks.
3282
Answer: Ternary strings have symbols 0, 1, and 2. If there is exactly one 1, then there are 3 positions the one can be in and 2*2*2 ways to fill the other 3 blanks with a 0 or a 2. So the answer is 3*2*2*2 = 24.  I don't understand why there can only be 3 positions the one can be in?
506
In the decimal system: $...-3, -2, -1, 0, 1, 2, 3...$ In the binary system: $...-11, -10, -1, 0, 1, 10, 11...$ In the unary system: $-111, -11, -1, \text{ ??? }, 1, 11, 111...$   I was only able to find a related question but none with an answer so far:
85138
I wish to use numbering like 1\alpha, 1\beta,... in some subequations. I know that this is obtainable by redefining the \theequation like the following  (which gives lower-case Roman numerals, 1.i, 1.ii,...)  \begin{subequations} \renewcommand{\theequation}{\theparentequation{}.\roman{equation}} \begin{align}   a1 &amp;= b1 \\   a2 &amp;= b2 \end{align} \end{subequations}   The problem is I don't know the equivalent of \roman that will yield the Greek alphabet.  A related question (which might subsume the above) is: can someone complete the set of commands  \arabic, \alph, \Alph, \roman, \Roman, ...
42642
Let $\Phi$ be a root system in a real inner product space $E$. Define $\alpha^\vee = \frac{2\alpha}{(\alpha, \alpha)}$. Then the set $\Phi^\vee = \{\alpha^\vee: \alpha \in \Phi \}$ is also a root system.  Let $B$ be a base for the root system $\Phi$, ie. $B$ is a basis for $E$ and each $\alpha \in \Phi$ can be written as $\alpha = \sum_{\beta \in B} {k_\beta} \beta$ such that $k_\beta$ are integers of same sign.     Question: How to prove that $B^\vee = \{ \alpha^\vee: \alpha \in B\}$ is a base for the root system $\Phi^\vee$?   It is easy to see that $B^\vee$ is a basis for $E$. For the other property, here is what I have so far. Let $\alpha = \sum_{\beta \in B} {k_\beta} \beta$ be a root. Then  $$\alpha^\vee = \sum_{\beta \in B} \frac{k_\beta (\beta, \beta)}{(\alpha, \alpha)} \beta^\vee$$  so to prove the claim, it is necessary to prove that $\frac{k_\beta (\beta, \beta)}{(\alpha, \alpha)}$ is an integer. I don't see how that follows. In the case where $(\alpha, \beta) = 0$ the proportion $\frac{(\beta, \beta)}{(\alpha, \alpha)}$ could be anything, so I am a bit confused.
82315
What do we mean when we say &quot;Let $x$ be an element of the set $\mathbb{R}$&quot;?  Does $x$ represents only a single  element of set $\mathbb{R}$? or does $x$ represent all the element of set $\mathbb{R}$ simultaneously at the same time? People say that if $x\in\mathbb{R}$ then $x$ is any real number; that means $x$ represents all real numbers. But if $x$ is any real number then let's say $x=1$; so $x$ is one, then how it can represent all real numbers? Please help me I am very confused.
67803
Alright, so the idea of an exponent, $x$, is that you are multiplying its base by itself $x$ number of times. With base $5$ and $x=3$, we have that $5^3$ = $5 \cdot 5 \cdot 5$  I understand that the logarithm with base $a$ of $x = c$,  tells us that  $$a^c = x$$  and for $c =$ positive; values for $x$ are greater than $1$, and for $c =$ negative; values for $x$ are less than $1$,  and for $c = 0$, values for $x$ are...$1$.  So in short, I understand how, by means of observation of the graph of $f(x) = \log x$, we can see that $f(1) = 0$, BUT, I see no other way to understand why $x^0 = 1$, apart from the graph and everything around that very point.  I honestly cannot get my head around the idea, "$5$ times itself $0$ times... is one".  Is it that there is no fundamental answer for this but that we simply know by the graph? Or can I truly understand $x^0 = 1$ on its own?
25667
DISCLAIMER: I do not deal with in-depth mathematics on a daily basis as some of you may, so please pardon my ignorance or lack of coherence on this topic.  QUESTION: What is the significance of starting the Fibonacci sequence with $0,1$ ?  For instance, if I picked any two random integers, say 2 and 7, to start a sequence would I actually be creating some multiple or derivation of the Fibonacci sequence?  Is there a general mathematical explanation for the relationship between any sequence represented by $a[0] = x, a[1] = y, a[n] = a[n-1] + a[n-2]$ and the Fibonacci sequence?  Or, back to my example sequence, is there a general mathematical relationship between:  $2,7,9,16,25,41,66,107,173,280...$  and  $0,1,1,2,3,5,8,13,21,34...$  Perhaps the Golden Ratio explains it somehow?  Any help would be appreciated.
1461
First, my apologies if this has already been asked/answered.  I wasn't able to find this question via search.  My question comes from Rudin's "Principles of Mathematical Analysis," or "Baby Rudin," Ch 1, Example 1.1 on p. 2.  In the second version of the proof, showing that sets A and B do not have greatest or lowest elements respectively, he presents a seemingly arbitrary assignment of a number $q$ that satisfies equations (3) and (4), plus other conditions needed to show that $q$ is the right number for the proof.  As an exercise, I tried to derive his choice of $q$ so that I may learn more about the problem.  If we write equations (3) as $q = p - (p^2 - 2)x$, we can write (4) as   $$ q^2 - 2 = (p^2 - 2)[1 - 2px + (p^2 - 2)x^2]. $$  Here, we need a rational $x &gt; 0$, chosen such that the expression in $[...]$ is positive.  Using the quadratic formula and the sign of $(p^2 - 2)$, it can be shown that we need   $$ x \in \left(0, \frac{1}{p + \sqrt{2}}\right) \mbox{ for } p \in A, $$  or, for $p \in B$, $x &lt; 1/\left(p + \sqrt{2}\right)$ or $x &gt; 1/\left(p - \sqrt{2}\right)$.  Notice that there are MANY solutions to these equations! The easiest to see, perhaps, is letting $x = 1/(p + n)$ for $n \geq 2$.  Notice that Rudin chooses $n = 2$ for his answer, but it checks out easily for other $n$.  The Question:  Why does Rudin choose $x = 1/(p + 2)$ specifically?  Is it just to make the expressions work out clearly algebraically?  Why doesn't he comment on his particular choice or the nature of the set of solutions that will work for the proof?  Is there a simpler derivation for the number $q$ that I am missing?
24436
Today in Coding/Cryptography class, we were talking about basic definitions, and the professor mentioned that for a set $A=\left \{ \left. a, b, \dots, z \right \} \right.$ (the alphabet) we can define a set $A^{*}=\left \{ \left. a, ksdjf, blocks, coffee, maskdj, \dots, asdlkajsdksjfs \right \} \right.$ (words) as a set that consists of all finite sequences of the elements/letters from our $A$/alphabet. My question is, is this set $A^{*}$ countably or uncountably infinite? Does it matter how many letters there are in your alphabet? If it was, say, $A=\left \{ \left. a \right \} \right.$, then the words in $A^{*}$ would be of form $a, aa, aaa, \dots$ which, I think, would allow a bijection $\mathbb{N} \to A^{*}$ where an integer would signify the number of a's in a word. Can something analogous be done with an alphabet that consists of 26 letters (Latin alphabet), or can countability/uncountability be proved otherwise? And as mentioned before, I am wondering if the number of elements in the alphabet matters, or if all it does is change the formula for a bijection.   P.S. Now that I think of it, maybe we could biject from $\underset{n}{\underbrace{\mathbb{N}\times\mathbb{N}\times\mathbb{N}\times\dots\times\mathbb{N}}}$ to some set of words $A^{*}$ whose alphabet $A$ has $n$ elements? Thanks!
41848
I here define the hierarchy of basic mathematical operators and their respective "inverse" operation (see ).  $$ \begin{array}{c|c|c|}   &amp; \text{Operator} &amp; \text{"Inverse"} \\ \hline \text{Incrementation} &amp; a+1 &amp; a-1 \\ \hline \text{Addition} &amp; a+b &amp; a-b \\ \hline \text{Multiplication} &amp; ab &amp; \frac{a}{b} \\ \hline \text{Exponentiation} &amp; a^b &amp; \sqrt[b]{a} \\ \hline \text{Tetration} &amp; ^ba &amp; \sqrt[b]{a}_s \\ \hline \text{} \vdots &amp;   \vdots &amp; \vdots \\ \hline \end{array} $$  Now, it is clear that each "level" is simply the previous one except the process is done several times.  Adding two integers is like incrementing one integer many times.  Multiplication is addition of the same number many times.    My question is: does anything precede incrementation?
52574
For no particular reason I stumbled upon the sequence/set of all numbers where all substrings of the decimal representation is prime (A085823). It's quite easy to see that this must be a finite set. My approach was that the digits must alter between 3 and 7 eventually (since a prime can't end with 2 or 5 for larger than that and a prime can't consist of just two non-1 digits, those would be dividable by 2, 5 or 11) and we can see where it ends.  But if we change the base that approach fails. For example in base 8 a prime can end in 3, 5 or 7 AFAICS so such numbers can end in a altering sequence of 3, 5 and 7s but having tree digits to alter between there's always a way to vary the sequence.  I tried to write a python snippet to produce such numbers in different bases and for those I've tested the program seem to hang (ie don't seem to find any more numbers). Is it true that there are only a finite number of such numbers for every base?
55572
Motivation: If operation is change, and there is no change, then there is no operation. Then the meaning of  1+0 =1 is not in the equality itself but in the comparison with other equalities as  1+2=3. The following definition does not use the meaningless equality  1+0 =1, or generally a*e=a. I guess that trying to prove statements using the next definition, matters/problems will arise that do not arise with the standard definition. Let be the non empty set G = {a,b,c,d,.....z} with the operation *, satisfy the following 3 conditions:  a(bc) = (ab)c, for every a, b, c ∈ G (associative law). For every a,b there is c (unique) such that ac = b. (c  is the &quot;path&quot; from a to b). For every a,b there is d (unique) such that bd = a. (d is the &quot;path&quot; back from b to a).  P.S.: the meaning of 2) and 3) is the possibility to &quot;go&quot;/&quot;move&quot; from any member to any member, through the same path (if a-&gt;b-&gt;c then c-&gt;b-&gt;a as in the Klein group) or through other path (if a-&gt;b-&gt;c then c-&gt;X-&gt;a). Prove: G,*  is a group. Proof: Associativity: From 1) Identity element: Let's prove for every a there is e such that ae=ea=a. Proof: According to 2) For every a,a there is e such that ae = a. (3) According to 2) for every e,a there is x such that ex=a, that means ae =ex =a (4) Let's proof x=a. Multiply by a on the left, such that the result is some g ∈ G: ae = ex = a (4) &lt;---&gt; (a)ae = (a)ex = aa =g &lt;---&gt; aae = (ae)x = aa =g (associative) &lt;---&gt; &lt;---&gt; aae = ax = aa =g (3). (5) According to 2) for every a,g there is x (unique) such that aa=g and ax=g , then a and x are the same, and it's proved x=a. Then replacing in (5): ae = ea = a. Inverse element: Let's prove that for every a, there is b such that a ⋅ b = e and b ⋅ a = e According to 2) for every a,e there is b such that ab = e. (6) According to 2) for every b,e there is X such that bX=e. (7) Let's proof that X=a ab = bX = e (6),(7) Multiply be a on the right: aba =ea a(ba) = ea; Associativity ea = a; Identity element a(ba) = a; then ba = e bX=e according to (7) ba = bX = e; according to 2) for every b,e there is x unique such that bX = e then x=a and it's proved that for every a, there is b such that a ⋅ b = e and b ⋅ a = e.
78773
The decimal (base 10) number "2" can also be represented as the binary (base 2) number "10".  Let's use binary "10" (equivalent to decimal "2") as an example. I could see a few different ways to go here. Assume that the base doesn't need to be specified, and is understood from the context of the conversation (e.g. two programmers talking about memory addresses would understand that they were using hexadecimal).  It could be "ten", since that is what it looks like. One might even argue that ten, as a concept, refers to a one followed by a zero irrespective of the radix. In other words, ten means "a quantity exactly equal to the base it's represented in".  On the other hand, you could argue that "ten" refers specifically to the quantity; in other words, "1010" in binary, "10" in decimal, and "12" in octal would all be pronounced "ten," and "10" in binary should be pronounced "two".  So how would you pronounce the following numbers?  "10" binary ("2" decimal)  "10" octal ("8" decimal)  "10" hexadecimal ("16" decimal)  "1F" hexadecimal ("31" decimal)
1308
What is the meaning of the notation $\exp(\text{expression})$ ?  I think that it's something of the form $a^\text{expression}$ but does it mean that the base $a=e$ or can it be any base?
116626
I am trying to understand this example of strong induction. I know normal induction. In normal induction, if base case is true then we assume some number $n$ to be true. Afterwards, we prove $n+1$ is true.  In strong induction, as you can see below, you prove more than 1 base case. I am not sure why. Then you prove $n+1$ is true...  I don't know why or how they choose however many base cases to prove or why proving extra base cases matters. Could someone please explain the inductive step of $3 + n - 2$? Where did it come from?