So I'm trying to come up with a solution to this task: Each natural number x greater than 1 has log x digits. Is this true? And if it isn't, what would be the correct answer?  After plugging several numbers as "x" I'm assuming there is in fact a mistake, I however don't know what's the right answer to this task.
8712
I read recently that you can find the number of digits in a number through the formula $\lfloor \log_{10} n \rfloor +1$ What's the logic behind this rather what's the proof?
10408
Prove that for any natural number $n&gt;1$, we have $(2n)!&lt;(n(n+1))^n$    I tried using induction, but I failed in that approach, I rather found it was untrue, but on several case testing, I found it to be true. I am out of ideas here, please help. Thank you.
40667
I'm reading through The normal number of prime factors of a number $n$. I'm confused by a remark on the : let $f(n)$ represent the number of distinct prime factors of $n$. Then  we can shew (by purely elementary reasoning) that, if $\epsilon$ is any positive number, we have $$ f(n) &lt; (1 + \epsilon) \frac{\log n}{\log \log n}$$ for all sufficiently large values of $n$ and $$ f(n) &gt; (1 - \epsilon) \frac{\log n}{\log \log n}$$ for an infinity of values; so that the maximum order of $f(n)$ is $$ \frac{\log n}{\log \log n} $$  I don't follow their &quot;purely elementary reasoning.&quot; What am I missing?
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Some very wise people here have just told me that  $\aleph_0 = \mathbb{N}$, i.e. that the cardinality of the set of natural numbers is just the set of natural numbers itself. Is this now the general consensus in mathematics, real analysis in particular? Or have I got completely the wrong end of the stick as usual?
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I've been asked to solve this and I've tried a few things but I have trouble eliminating $x$. I first tried taking the natural log: $$x\ln \left( 2\right) =2\ln \left( x\right)$$ $$\dfrac {\ln \left( 2\right) }{2}=\dfrac {\ln \left( x\right) }{x}$$ I don't know what to do from here so I decided to try another method: $$2^{x}=2^{\log _{2}\left( x^{2}\right) }$$ $$x=\log _{2}\left( x^{2}\right)$$ And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple&hellip;
89309
$Let$ $a,b,c&gt;0$  $$\begin{cases}\log_a(b^x)=2\\\log_b(c^x)=2\\\log_c(a^x)=5\end{cases}$$  $$x=?$$  So my attempt is just to use the logarithmic definition:  $$\log_a(b^x)=2\iff b^x=a^2$$  By similar logic,  $$a^x=c^5$$ $$c^x=b^2$$  So, if add everything together, we should get:  $$a^x+b^x+c^x=a^2+b^2+c^5$$  Looks to me like x is equal 2 different numbers at the same time which is strange, what am I doing wrong here?  I'm going to be a maths student in the upcoming year, this is taken from the Tel-Aviv university preparation material - shouldn't be too complex.
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lists this as a limit representation of the natural log:  $$\ln{z} = \lim_{\omega \to \infty} \omega(z^{1/\omega} - 1)$$  Is there a quick proof of this? Thanks
35146
Can anyone tell me what is the maximum number of consecutive composite numbers possible? I mean can I get say 1000 consecutive natural numbers. Is there any general theorem that when I have a n-digit number there will always be p consecutive composite numbers?
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How is well-ordering in real line possible?  I know that the axiom of choice provides possible well-ordering, but intuitively, this does not seem to make sense.  How can you compare 1.111111.... and 1.111111... when we can never know the full digits of each number?
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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?
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How find the asymptotic behavior for $x(n)$  if $x^{x^x} = n$?  I supposed that $x = O(\log\log{n})$ and took logarithm two times.  So I get $x = O(\frac{\log\log{n}}{\log\log\log{n}})$  Is it right? How to improve this estimate?
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I just read somewhere that there are as many even numbers as natural numbers! Is there an intuitive way of coming to terms with this fact? Because honestly, it's quite baffling. Also, does the concept of odd/even apply to negative integers as well? By definition, it should. But I have never seen a negative multiple of two being used as an example of an even number.
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This question came from  and from  where it's noted that "integers have a finite count of digits", so that the "number" in the title is not at all a number (not integer nor rational or real) . This statement seems to me justified because I suspect that if we admit an infinite count of digits we build a set of "numbers" that is not countable (but I've not a proof). I'm wrong?    Reading comments and answers I'm a bit confused. So I add a more specific question: Is the string in the title a number in some model of Peano Axioms?
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Does every number not ending with zero have a multiple without zero digits at all?   We consider numbers in their usual base 10 representation. we don't consider numbers ending with $0$, because all multiples of such numbers end with $0$.   It can be proven that all powers of $2$ have such a multiple: Indeed, for every number of the form $2^n$, we can find a multiple $l$ of it whose last $n$ digits are nonzero (This can be done by repeatedly multiplying a number by $10^k+1$, as explained ), and then forget all other digits; these last $n$ digits constitute a number that is still divisible by $2^n$, since it is obtained from $l$ by subtracting a multiple of $10^n$ which is in particular a multiple of $2^n$.  The same argument shows that every power of $5$ has a multiple without zeros, but this does not seem to generalize to other numbers.  My intuition is actually that the claim is not true, since the density of numbers without any zero digits approaches $0$ for big numbers. This is because the probability of a random number with at most $n$ digits to contain only nonzero digits is $(\frac{9}{10})^n$. Thus big numbers "nearly always" contain a zero. This hints on the existence of a counterexample, but does not prove its existence.
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Say you have a number $x^{\sqrt 2}$.  Is there any way to represent this number so that there's no root (or irrational) as the exponent (so that it's easier to understand for me)? I just can't wrap my head around this.  I was thinking something like $$x^{2^{1/2}}$$ and extending on that idea, but I don't know how.  Also, what about for $x^{\sqrt 3}$? Or $x^{\sqrt n}$?
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I'm a little bit confused! what is the domain of this function:  $$ \ln(x) ^{ \ln(x) } $$  this function, in fact, is:  $$ \exp(\ln(\ln(x))\cdot\ln(x)) $$  so the domain would be: $$ x&gt;1 $$  But: $x$ can "also" take on the value (for example) :$$\sqrt[3] {e^{-1}}$$  then there would be some other numbers, not included in $x&gt;1$, in the domain of the function.
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Problem :   Let $n$ be a positive integer, for a real number $x$, let $[x]$ denote the largest integer not exceeding x and $\{x\} =x -[x]$ Then $$\int^{n+1}_1 \frac{(\{x\})^{[x]}}{[x]}dx$$ is equal to   (a) $\log_e(n)$   (b)  $\frac{1}{n+1}$   (c)  $\frac{n}{n+1}$   (d)  $1+\frac{1}{2}+\cdots +\frac{1}{n}$   Please suggest how to proceed in such cases not getting any clue how to proceed in cases of fractional part and greatest integral cases. please guide will be of great help thanks.