jordan Posted July 2, 2004 Posted July 2, 2004 I'm not sure if there is a thread on this. If so, I don't know where it is. The question is, what is 00 and why?
Guest Methyl Posted July 2, 2004 Posted July 2, 2004 Well in theorey shouldnt it be 1? As everything to the zero is 1. Yet surely on paper you would expect it to be 1. But doesnt it become an infinate number? Im not really sure about it but something just seems to twig lol.
jordan Posted July 2, 2004 Author Posted July 2, 2004 I understand why division by 0 is undefined, but we aren't deviding by 0, or are we in a sense? How did they decide it's undefined?
Dave Posted July 2, 2004 Posted July 2, 2004 http://mathforum.org/dr.math/faq/faq.0.to.0.power.html I think it's more or less generally regarded for 0^0 to be 1 - I think it's needed for some power series, namely those with a radius of convergence 0.
jordan Posted July 2, 2004 Author Posted July 2, 2004 Since my question seems to have been answered well enough (thanks all) I have a quick question probably not worth a new thread. I just came across Skewe's number again ([math]10^{{10}^{10}^{34}}[/math]). To solve this number, I would have to work from the bottom up (starting with [math]10^{10}[/math]) and not top to bottom ([math]10^{34}) right. I'm aware this is probably very easy, but I thought it would be working from the top down. That means [math]10^{34}[/math] or ten followed by 34 0's. But then working down and taking 10 to that power only adds one zero, doesn't it? And then [math]10^{that-big-number}[/math] would add 1 more 0 which isn't too impressive. On the other hand, if you work for the bottom up, the answer is much bigger. But these seems counterintuitive, like working outside the parenthesis first. Am I missing something?
Dave Posted July 3, 2004 Posted July 3, 2004 When you have stacked indexes, you start from the top. i.e. [math]2^{2^2} = 2^4 = 16[/math].
Dave Posted July 3, 2004 Posted July 3, 2004 In retrospect, that was a rather bad example. [math]2^{3^4} = 2^{81} = 2417851639229258349412352[/math] (according to my TI-89 ) So you can imagine how big that number is going to be.
jordan Posted July 3, 2004 Author Posted July 3, 2004 So then 10^10^10^34 is as follows: 10^10^(10^34) or 10^10^(10 followed by 34 0's) as 10^34 is 10 followed by 34 0's. So then 10^10^(10 followed by 34 0's) would become 10^(10 followed by 35 0's) as 10^(10 followed by 34 0's) is 10 followed by 35 0's. Then 10 to that power makes it 10 followed by 36 0's or 10^36. There has to be an error in there somewhere, but I've looked at this too long and now can't come up with anything new. Sorry for the elementry question.
Dave Posted July 3, 2004 Posted July 3, 2004 Hmm. You're misinterpreting 10^(10 followed by 34 0's). Think of it this way: [math]10^{10^{34}} = \underbrace{10^{34} \cdot 10^{34} \cdots 10^{34}}_{\displaystyle 10^{34} \mbox{ times}}[/math] It's rather a large number.
Martin Posted July 3, 2004 Posted July 3, 2004 something looks wrong here So then 10^10^10^34 is as follows: 10^10^(10^34) or 10^10^(10 followed by 34 0's) as 10^34 is 10 followed by 34 0's. So then 10^10^(10 followed by 34 0's) would become ... Hmm. You're misinterpreting 10^(10 followed by 34 0's). Think of it this way: [math]10^{10^{34}} = \underbrace{10^{34} \cdot 10^{34} \cdots 10^{34}}_{\displaystyle 10^{34} \mbox{ times}}[/math] It's rather a large number. it would seem that [math]10^{10^{34}} = \underbrace{10 \cdot 10 \cdots 10}_{\displaystyle 10^{34} \mbox{ times}}[/math]
Dave Posted July 3, 2004 Posted July 3, 2004 Yes, I'd realised that about 20 seconds after I'd posted it. However, I was too lazy to change it Bloody stacked powers always confuse me.
Martin Posted July 3, 2004 Posted July 3, 2004 Bloody stacked powers always confuse me. that I doubt the trouble comes when you have to do a lot of copy/paste operations and forget to erase things---LaTex is a marvel but can be a pain in the butt as well
Martin Posted July 3, 2004 Posted July 3, 2004 [math]10^{10^{34}} = \underbrace{10 \cdot 10 \cdots 10}_{\displaystyle 10^{34} \mbox{ times}}[/math] [math]10^{10^{10^{34}}} = \underbrace{10 \cdot 10 \cdots 10}_{\displaystyle 10^{10^{34}} \mbox{ times}}[/math]
jordan Posted July 3, 2004 Author Posted July 3, 2004 [math']10^{10^{34}} = \underbrace{10 \cdot 10 \cdots 10}_{\displaystyle 10^{34} \mbox{ times}}[/math] Is that what I said or not?
Martin Posted July 3, 2004 Posted July 3, 2004 Is that what I said or not? Hi jordan, I didnt look to see I assume it was what you said----especially if you assure me of that. I was just playing around with this LaTex notation it is so wonderful to be able to share formulas easily can you do the basics with it? just, like do "quote" on a Dave post and then dont bother to carry through and reply just look at how he does the type-setting
jordan Posted July 3, 2004 Author Posted July 3, 2004 Thanks. I do have LaTeX down. I've used it extensively in this is the learning calculus threads. Anyway, I started the questioni in post #6 and rephrased in post #9. I was asking if you said what I was saying or not. I can't see the difference right now, but I am extremely tired of looking at the problem. Thanks.
bloodhound Posted July 3, 2004 Posted July 3, 2004 i hate stacked powers as well. i find it better to understand if the author puts brackets eg. (10^10)^34 is 10^10 34 times which is equal to 10^340. if its 10^(10^34) then its a different matter
Dave Posted July 4, 2004 Posted July 4, 2004 that I doubt the trouble comes when you have to do a lot of copy/paste operations and forget to erase things---LaTex is a marvel but can be a pain in the butt as well Wasn't a particularly complex equation, just got my head in a muddle.
lqg Posted July 5, 2004 Posted July 5, 2004 0^0 is undefined. like division by 0 is undefined because they are the same: 0/0=0^0
JaKiri Posted July 5, 2004 Posted July 5, 2004 because they are the same:0/0=0^0 They most definitely are not. However, 0^0 should be strictly speaking undefined, as the proof by notation that works for x^0 where x =! 0 doesn't hold if x = 0.
jordan Posted July 5, 2004 Author Posted July 5, 2004 I don't remember where now, but it seems I remember reading something saying that it changes depending on what feild you're working in. I can't recall the specifics, but in one case, it's appropriate to say it equals 1, in another 0 and in a third it can be undefined.
Freeman Posted July 5, 2004 Posted July 5, 2004 Isn't it [math]0^0[/math], which means "how many 0s are there in 0" = 1.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now