• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Modular arithmetic Stereotomy Junior Fellow Posts: 2 Threads: 1 Joined: Apr 2010 04/02/2010, 07:16 PM I'll just preface this by saying I'm just a physics undergrad, so this might be a bit beyond my understanding, and I may well be missing something obvious or making a stupid mistake, but while playing around I noticed that it seems to be true that $^{n}a\text{ mod }b= {}^{m}a\text{ mod }b$ $\text{For }^{n}a,{}^{m}a > b,\text{ }a,b,n,m \in \mathbb{N}$ Is this actually true? And if so is there a proof of it I'll be able to wrap my mind around? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/02/2010, 10:57 PM (04/02/2010, 07:16 PM)Stereotomy Wrote: $^{n}a\text{ mod }b= {}^{m}a\text{ mod }b$ $\text{For }^{n}a,{}^{m}a > b,\text{ }a,b,n,m \in \mathbb{N}$ Is this actually true? And if so is there a proof of it I'll be able to wrap my mind around? I dont think it is true. For example: $7^7 \text{mod} 5 = 3\neq 2 = 7 \text{mod} 5$ Stereotomy Junior Fellow Posts: 2 Threads: 1 Joined: Apr 2010 04/03/2010, 12:18 AM (This post was last modified: 04/03/2010, 12:18 AM by Stereotomy.) (04/02/2010, 10:57 PM)bo198214 Wrote: (04/02/2010, 07:16 PM)Stereotomy Wrote: $^{n}a\text{ mod }b= {}^{m}a\text{ mod }b$ $\text{For }^{n}a,{}^{m}a > b,\text{ }a,b,n,m \in \mathbb{N}$ Is this actually true? And if so is there a proof of it I'll be able to wrap my mind around? I dont think it is true. For example: $7^7 \text{mod} 5 = 3\neq 2 = 7 \text{mod} 5$ Ah, good point, though $7^{7^{7}} \text{mod} 5 = 3$ Is true as well. In fact, thinking about it, the numbers I tried out with this all had b>a. Perhaps that's an additional condition that either b > a or m, n > 1? Just quickly tried this for a few low examples, a = 8, 9, 10, 11, and it seems to hold. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 04/03/2010, 12:00 PM (04/03/2010, 12:18 AM)Stereotomy Wrote: $7^{7^{7}} \text{mod} 5 = 3$ Is true as well. In fact, thinking about it, the numbers I tried out with this all had b>a. Perhaps that's an additional condition that either b > a or m, n > 1? There is an article which proves that ${^n a}$ (which is $a{\uparrow}^2 n$ in Knuth's arrow notation) finally will be constant for $n\to\infty$ mod any $M$, see this thread. « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post [rule 30] Is it possible to easily rewrite rule 30 in terms of modular arithmetic ? tommy1729 0 1,939 07/24/2014, 11:09 PM Last Post: tommy1729 Generalized arithmetic operator hixidom 16 16,491 06/11/2014, 05:10 PM Last Post: hixidom Tetration and modular arithmetic. tommy1729 0 2,518 01/12/2014, 05:07 AM Last Post: tommy1729 modular tetration tommy1729 0 2,860 12/26/2010, 10:11 PM Last Post: tommy1729 Arithmetic in the height-parameter (sums, series) Gottfried 7 12,343 02/06/2010, 12:52 AM Last Post: bo198214

Users browsing this thread: 1 Guest(s)