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#151 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 05:47:13
Ah ok, yes xD Ok so p1p2pn is even
#152 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 05:44:53
Well, I don't see why it couldn't be odd.
#153 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 05:41:18
mmmmh.....not following.
#154 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 05:35:22
Just in case you didn't see my second post..
#155 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 05:33:10
Oh ok, could you explain me where the +1 comes from? It seems rather mysterious...
#156 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 05:30:59
Yes, I'm following you. Ok, I'll go re-read the proof to see if there's anything I want to ask you. (I love understanding deeply something, hope it won't bother you.)
Thank you for your help
#157 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 05:19:03
Ok, but now m/q work yes or no ??? If it is outside the list, than it should divide m, because every number is divisible.(And we don't have that remainder of 1)
#158 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 05:14:03
Ok, but it's not part of our list p1,p2,etc (the q) right ?
#159 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 05:07:25
Ok, just another thing :
Now suppose m is a product of primes. Let q be one of the primes in this product. Then m is divisible by q. But weve already seen that m is not divisible by any of the numbers in the list p1, p2, . . . , pn,
Why is he saying that m/q. I'm guessing that he's taking a prime "q" not in this list, because like we said, none of the primes in our lists divide m. And knowing that any number must be divided, he divides m/q and it works. Is it correct ?
#160 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 04:57:26
Yes, for sure it would leave a remainder. Then, if m is a product of primes, what are those primes ???
#161 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 04:53:47
I don't understand, what do you mean by : But m is not a product of any of the primes in our list,yes?
I say yes, it is a product of p1,p2,pn.
#162 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 04:49:01
Yes, because if we assumed m to be prime, then it would mean that our list wasn't complete and there was a new prime that juste showed up, which wasn't in it(the complete list)
#163 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 04:46:04
yeah
#164 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 04:37:56
Yes !
#165 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 04:35:09
Yes, I agree
#166 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 04:31:10
Yup again
#167 Re: Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 04:28:39
Yes also
#169 Re: Help Me ! » mathematical symbol » 2013-08-09 04:04:56
It's ok, I won't bite you. I'll just return to it when I know more about those things.
#170 Re: Help Me ! » mathematical symbol » 2013-08-09 04:00:07
Ahh, never seen this topic ... Anyway, Ill return to it another time. By the way, could you go see my topic in exercices please ??? thanks
#171 Re: Help Me ! » mathematical symbol » 2013-08-09 03:52:32
Thank you, even if I have no idea what It means xD But thanks again
#172 Exercises » Euclid's proof of infinitude of primes. » 2013-08-09 03:49:23
- Al-Allo
- Replies: 45
Hi, I'm having trouble with euclid's proof :
http://imageshack.us/photo/my-images/703/uhvj.jpg/
I'm having a problem with the last paragraph... If m is a product of primes, and q is one of those primes taken randomly, then we should be able to divide m by q to obtain the rest of the primes(p1,p2,etc.). But we established earlier that we CANNOT divide evenly, because we will have a remainder of one (By the way, I wonder where it comes from...)So now, I understand that this is a contradiction, but I'm not sure of following the ''we have a contradiction with the assumption that this list included all prime numbers." I'm not having the same conclusion about it, how does this prove that the list doesn't have ALL primes, even if we can't divide evenly? If it really contained all the primes, should we be supposed to divide it without a remainder, is it because primes are missing that it won't divide ??? Thank you for your help !
Can somebody help me ?? Thank you
#173 Help Me ! » mathematical symbol » 2013-08-09 03:36:27
- Al-Allo
- Replies: 7
Can somebody tell me what is the meaning of this symbol :
Is it approximation ?
#174 Exercises » Proof question. » 2013-08-02 06:44:22
- Al-Allo
- Replies: 1
I have a question concerning a proof that two negatives makes a positive :
Here is the outline of this proof: Let us prove first that 3 . (-5)= -15. What is -15? It is a number opposite to 15, that is, a number that produces zero when added to 15. So we must prove that 3 . (-5) + 15=0
Indeed, 3 . (-5) + 15= 3 . (-5) +3 .5 = 3. (-5+5)=3.0=0
(When taking 3 out of the parentheses we use the law ab+ac=a(b+c) for a=3, b=-5, c=5;we assume that it is true for all numbers, including negative ones.) So 3.(-5)=-15
....
One thing which I do not follow is, why the need to prove that -15 is really -15 ? We know that, (-5)+(-5)+(-5)=-15
Logically, I don't see what else it could be 
But in the above text, are we really trying to show that the number -15 is really -15 by adding it to positive 15 ?So if it really gives 0, we are sure to be in presence of negative fifteen(3.-5=-15) ? Or is there anything that I've missed about it ? ANy help would appreciated ! Thank you.
#175 Re: Exercises » Ver.. » 2013-07-13 09:02:09
hi Al-Allo
You want the complete reason for this. I suggest you start by looking at the fractions that terminate. What property causes them to terminate? then you can state which fractions will produce recurring digits.
How many in a recurring cycle? Well consider how many possible remainders there can be. 1/7ths are a good example of this. If there is a limit to the number of remainders then what happens when you 'run out' of 'new' remainders?
Bob
I just wanted to answer that last question, before answering to the one with 1/7,2/7,etc.
We know that there is a direct equality between the number of remainders and the number of digits in your quotient (The cycle)
For example : 1/7
7*0,1=0,7 >> 1-0,7=0,3 1
7*0,04=0,28 >> 0,3-0,28=0,02
7*0,002=0,014 >> 0,02-0,014=0,006 3
7*0,0008=0,0056 >> 0,006-0,0056=0,0004 4
7*0,00005=0.00035 >> 0,0004-0.00035=0,00005 5
7*0,000007=0.000049 >> 0,00005-0.000049=0,000001 6
Which gives us : 0,142857... ( we have exactly 6 digits in one cycle)
As we can see, there is always remainders(The quotient going to infinity), but there is a limit to the kinds of remainders. We always have new old ones the moment we hit a certain point (This certain point in this example corresponds to 0,000001, but I'm not sure if it always needs to be of one unit of difference to have a new cycle, I would need to check it) and this is when the cycle continues again, being periodic
Anyway, any opinion on what I said ? If you think something is missing, please tell me. The only thing that is irritating me is, I haven't discoverred why certain numbers have longer periods than others, I think that by knowing this, it will help me a lot. If you know the answer, please don't tell me, I want to be able to answer it myself. Thank you again!