Question

- from ,,fundamental theorem of arithmetic" the below formula can be considered proven :
2^a *(2b +1) = N ,where ,,a" and ,,b" are natural numbers or zero , ,,N" is the set of natural numbers
- true or false ?

Answer 1

for example we know that

2^0 *(2*0 +1) = 1
2^1 *(2*0 +1) = 2
2^0 *(2*1 +1) = 3
...

Answer 2

induction?

Answer 3

yes i have thought it same but from this fundamental theorem of aritmethic not result directly so what was proven by Euclid ?

Answer 4

I am sorry, I don't know what you just said.

Answer 5

I don't see why this is surprising.\[N = 2^a 3^{x_1}5^{x_2}\cdots\]Now the \(3^{x_1}5^{x_2}\cdots\) part represents an odd number.

Answer 6

In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1[note 1] either is prime itself or is the product of prime numbers, and that this product is unique, up to the order of the factors.[3][4][5] For example,

1200 = 24 × 31 × 52 = 3 × 2 × 2 × 2 × 2 × 5 × 5 = 5 × 2 × 3 × 2 × 5 × 2 × 2 = etc.

Answer 7

this is from wikipedia.org

Answer 8

yes exactly. by that theorem, all natural numbers are of the form\[N = 2^a 3^{x_1}5^{x_2}\cdots\]Now the \(3^{x_1}5^{x_2}\cdots\) part represents an odd number so we can write that as \(2b+1\) for some natural number \(b\).\[N = 2^a (2b+1)\]thus we've shown that all natural numbers can be expressed in this form

Answer 9

,,ParthKohli" so this mean that your answer wann being ,,TRUE" on this question - yes ?

Answer 10

I think so

Answer 11

thank you very much ,,ParthKohli" !

Answer 12

No problem :)

Answer 13

so i will please you cooperate with me in a proof of a conjecture where i think that can be used this formula easy - ok. ?

Answer 14

Oh, which conjecture?

Answer 15

I remember you working on that 3n+1 Collatz conjecture.

Answer 16

yes you remember it right

Answer 17

are you there?

Answer 18

sorry now i have got your words - but you can write me than you like on my email address too ,on ,,gbrndrs1968@yahoo.com" - thank you very much - and happy weekend night
jhonyy9 as. Andrew

Answer 19

- using these knowledge in case of Collatz's conjecture

- so from these all result that because we know that 2^a / 2 - for indifferent value of a>0 at the end will get 2/2 = 1
- so in case of second part of these formula so the (2b+1) about this we know that always this mean an odd number inclusively for b=0 - so this mean that always will need multiplie
by 3 and adding 1 ,so
(2b+1)*3+1=6b+3 +1 =6b+4 = 2(3b+2) so divide by 2 because is even will get 3b+2 what is odd for b=1 than multiplie by 3 and add 1 and we get (3b+2)*3 +1 = 9b+6+1 =9b+7 so what will be for b=1 equal 16=2^4 so what is a power of 2 so what mean that at the end will get 2/2 =1
for b=k we get (2b+1)=(2k+1)suppose is true too
- so than we need to prove it for k=k+1 for what we get
(2k+1)=(2(k+1)+1)=(2k+2+1)=2k+3 for k=1 this is odd than multiplie it by 3 and add 1 we get (2k+3)*3+1=6k+9+1=6k+10 so what in case of k=1 will be 6*1+10=6+10=16 so what is 2^4 so a power of 2 what at the end will result 2/2 =1
- with these above the Collatz's conjecture is proven ?
@ganeshie8
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@mathstudent55
@dan815
@whpalmer4
@Compassionate
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@saifoo.khan

Answer 20

I don't understand what you've written, but I have gone down pretty similar alleyways in the past.

Here's what I think you're saying, and I'll show you how this ends up not leading anywhere:

Start with a number of the form

\(2^a*(2b+1)\)

Divide by 2 until you get:

\(2b+1\)

this is odd, so you get:
\(3(2b+1)+1 = 6b+4\)

This number is even:

\(6b+4 = 2(3b+2) \)

So we have to divide to get:

\(3b+2\)

3b+2 could be even or odd and since

\(3b+2 > 2b+1\)

We haven't shown that the number will decrease and, thus inevitably converge to 1. It could so be that \(3b+2\) is another number of the form \(2^a*(2b+1)\) which then leads to yet another and another infinitely.

Answer 21

@dan815 do you have same opinion please ?

Answer 22

This isn't an opinion, I either gave a counterexample to your thing or you should correct me and tell me what you're actually saying.

Answer 23

ok. so i understand you but than you check there above what i have wrote so there has proven the case with 3b+2 that on the end result always power of 2 so on the end 1 - right ?

Answer 24

check it please there

Answer 25

What I'm saying is I don't understand what you're saying so I can't read it. How is the 3b+2 case proven? 3b+2 is not always a power of 2.

Answer 26

so using induction same above
(3b+2)*3+1=9b+7 what for b=1 is 2^4 so power of 2 too

Answer 27

b=1? What about b=9353?

Answer 28

so do you know the method of proof by complete induction - so you need just one case like in this case b=1 - and in general so make the proof with -

Answer 29

Yes, but I don't see where you do that proof.

Answer 30

@Empty so just for you

(3b+2)*3+1=9b+7 so what for b=1 result 9+7=16 what is equal 2^4 so power of two
- for b=2
(3*2+2)=(6+2)=8 =2^3 what is power of two
(3*3+2)=11 so is odd than 11*3+1=34/2=17*3+1=52/2=26/2=13*3+1=40/8=
=5*3+1=16 = 2^4
- for b=k suppose is true
- for b=k+1 will get
(3(k+1)+2)=(3k+3+2)=(3k+5) what for k=1 will result 3*1+5=8 =2^3

- so do you accept this proof now ?

ty.

Answer 31

I don't understand how this proves anything, there's not really a logical argument that I can discern, I'm sorry.

Answer 32

true

Answer 33

@workareject ty. but do you understand the importantly think of this subject in case of Collatz"s conjecture ? this is the way where is like to go using this above wrote

Answer 34

Yes I understand