If any number times zero is zero and any number time infinity is infinity, then what do you get when you multiply zero times infinity? Do they cancel one another out and equal any number since any number = 0 to infinity) and any number infinity to 0? "also can 0 and infinity be one an others inverse?
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I did a bit of math at school and it seems like an easy one - what am I missing?  $$n\times m = \underbrace{n+n+\cdots +n}_{m\text{ times}}$$  $$\quad n\times 0 = \underbrace{0 + 0 + \cdots+ 0}_{n\text{ times}} = 0$$  (i.e add $0$ to $0$ as many times as you like, result is $0$)  So I thought an infinite number of $0$'s cannot be anything but $0$? But someone claims different but couldn't offer a reasonable explanation why. Google results seemed a bit iffy on the subject - hopefully this question will change that.
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I know what you're thinking: "of course it has, for example, it can be used to tell you how many times you can go around a circle". But that isn't really true, now is it? You'd be dead or the world would go under long before an infinite amount of loops had been reached.  Are there any practical applications for the concept of infinity? Is it a useful concept in maths at all?  I know that Donald E. Knuth has argued that for all practical purposes, a very, very large number has the same effect as infinity, in his book "Things a Computer Scientist Rarely Talks About" (can't remember the exact quote, nor find it online, unfortunately).  Examples are appreciated.
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Problem Statement:- The number of zeroes at the end of the integer   $$100!-101!+\ldots-109!+110!$$   I am having a bit of a trouble in thinking how do I proceed. A little push in the right direction would be appreciated.   And if you are posting a full solution do use the spoiler tag, as sometimes I cant stop myself from seeing the whole solution and lose the chance of thinking through it by myself with a push in the right direction from you guys.  Also, I dont know whether I am using the right tag feel free to correct it if its wrong.
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One common math puzzle I've seen around asks for how many zeros are in the product of "100!"  Usually, the solution everyone gives goes something like try to match pairs of 5s and 2s that factor out of the numbers, which ends up being 24 zeroes (you can factor a 5 out of 20 of the numbers, and factor 2 5s out of 4 of the numbers; you can factor more than  24 2s out).   This however as far as I know gives the number of trailing zeroes at the end of the number, but does not account for the zeroes that are within the number. My question is, is this answer correct anyways? Can there be zeroes that aren't trailing that are inside? Why or why not and if there can be can we somehow figure out how many are within the product?   Thanks
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Well, just another idea came up into my mind and i have no idea how to solve it :D Is there infinitely many prime numbers, which are not repunits and their inverse is also prime? (For example, inverse of 31 is 13 which is also prime. i didn't have any other name to describe the function!) P.S:I now know they are called s.
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I'm trying to prove the statement in the title, a.e existence of ordinal substraction. I think it can be done with transfinite induction, but do any of you have a better/easier way?  thanks
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Since $\infty&gt;0$, so $1/\infty &gt;0$, thus I think $1/\infty$ should be infinitesimal, but the calculus book says  $$\lim_{x \to \infty} \frac{1}{x}= 0$$  So is $1/\infty$ zero or infinitesimal ?  P.S. I mean are $1/\infty$ and $\lim_{x \to \infty} 1/x$ the same thing here?
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I was surprised when I heard zero is plural, and even 1.00001 is plural.  Then, what about following numbers ?  (1) 1.00  which means the value measured  between 0.995-1.004, having possibility that it was exactly 1.  (2) 0.99999.....  which is mathematically equal to 1. Also, are others that are mathemattically equal to 1 than this, treated completely same as 1?  (3) -1  Is this singular? Or negative numbers are generally plural?  (4) i = square root of -1  I know it's weird to count things with imaginary numbers, nonetheless, if you were to count, which form, plural or singular, would you be to use?  That's it.
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Why is $a^0=1$  $\forall a \in Z, a\neq0$. I understand $2^4=2\cdot2\cdot2\cdot2$ How can I express $a^0$. I am serious about the practical proof of this
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For any $\epsilon$, if $\epsilon&gt;0$ and $|x|&lt;\epsilon$, then $x=0$.  I understand that supposing  $\epsilon=\frac{x}{2}$ will lead to a contradiction, but let’s take a correct case: Let $\epsilon=3$, then $x$ would have a whole set of values. Can you explain what is going on?
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Ok, here is the problem statement:     Prove that if $S$ is any finite set of real numbers, then the union of $S$ and the integers is countably infinite.   This seems pretty obvious to me, knowing that 2 countable sets are countable.  But is there some step by step way to prove this?  Like do I need to prove bijectivity or something?  Thanks!
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If a nonzero element $a$ of a ring $\mathbb{D}$ does not have a multiplicative inverse, then $a$ must be a zero divisor.   If $\mathbb{D}$ has finitely many elements, then the statement is true. However, I don't see why, as for the infinite case, 7 isn't a zero divisor only because every element $a &gt; 0$ when multiplied by a $b &gt; 0$ gives you $ab &gt; 0$ which has nothing to do with infinity or finite amounts of elements.
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How many integers from 1-999 do not have any repeated digits?  The answer is explained in this , but why is the last set 9*9*8? Why not 9*9*9?
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Why is any number (other than zero) to the power of zero equal to one? Please include in your answer an explanation of why $0^0$ should be undefined.
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I came up with this little simple exercise, stating:  $x^{x^{x^{\dots}}}$ infinite times is equal to $2$, find $x$.   As we're dealing with infinity, we can just separate the first $x$ and get $x^2=2 \implies x=\sqrt2$  But then I thought what if we change $2$ to $4$? Now we get $x^4=4 \implies x=\sqrt[4]4=\sqrt2$  Something went wrong, can someone explain?
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I know that there are questions asking like "intersection of a infinite collection of sets" and I can understand that the answer for that one is a null set, but I got a question here, in which all sets are finite and nonempty. Please take a look at the pic below.  It's a True/False question.  What I'm confused about is that the "infinity" sign in the interception. Since all sets are finite, does it mean that there are many sets in the chain are the same? (because the symbol used is for subsets not for proper subsets). Please give me some hints how I should think about this question.  Thank you
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I would say $\infty - \infty=0$ because even though $\infty$ is an undetermined number, $\infty = \infty$. So $\infty-\infty=0$.