Question

let formula 2^n *(2m +1) with n and m natural numbers plus/(with)* zero
* - because zero not accepted natural

- using this formula can be writing ALL numbers ,like for example : 2^0 *(2*0 +1) = 1
2^1 *(2*0 +1) = 2
2^0 *(2*1 +1) = 3
2^2 *(2*0 +1) = 4
.....

can being this proven so how ?

Answer 1

`n=0` gives all odd natural numbers since every odd number can be expressed as `2m+1` by division algorithm

Answer 2

yes

thank you

so but in case of even numbers ?

Answer 3

2^1 *(2*0 +1)= 2
2^2 *(2*0 +1)= 4
2^3 *(2*0 +1)= 8
....

- so we get every power of 2

Answer 4

what are you trying to prove?

Answer 5

that using this formula we can getting every natural numbers

Answer 6

you are trying to prove
can any natural number be written as the product of a power of 2 and an odd number

Answer 7

You want to prove that for any integer k, there exist m, n such that
2^n *(2m +1)= k

Answer 8

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

Answer 9

every odd integer can be written as 2m+1

Answer 10

so we only have to look at even numbers

Answer 11

show that for any even number a=2t , a = 2^n ( 2m + 1)

Answer 12

if a = 2t, then t is either even or odd

Answer 13

case 1: t is even, then t = 2h for some integer h
2 (2h) = 2^n (2m+1)

Answer 14

at some point h as to be odd

Answer 15

2^n *(2m +1) from D.A 2m+1 is the formula of all odd numbers , thus
2^0 *(2m +1)=2m+1 gives all odd
u cant represent odd numbers if \(n\neq 0\)

now let n be any number n>0
2^n *(2m +1) = 2 (2^n-1 *(2m +1) = 2k all even numbers :)

Answer 16

here 1 case to get even numbers :-
let n=1 , m>=0
2^n *(2m +1)

2(1)=2
2(3)=6
2(4)=8
2(5)=10
.....
and let n=2 , m=0 to get 4
4(1)=0
...

Answer 17

clever !

Answer 18

@ikram002p you have wrote 2(1)=2 this mean m=0 and n=1 so on next step 2(3)=6 so in this case n bot equal 3 and m=0 ?

on the second case 2(3)= 6 not will be the right result 8 because 2^3 =8 ?

how you think it please ?

Answer 19

thats a good catch

Answer 20

2(3)
n=0,m=1 :O
2(5)
n=0 , m=2
... and go on with n=0 , m>=0 condition

Answer 21

ok but you have wrote in the first case 2(1)=2

Answer 22

im i missing something in the question ?
i can chose m and n as i want , right ?

Answer 23

yes but how do u construct the even number 8 using n=1 and m>= 0 ?

Answer 24

oh sorry my bad xD
i made a typo for 8

Answer 25

ok then 3 cases for numbers :-
1- n =0 , and m>=0 we get all odds
2- m=0 , and n>0 we get numbers n s.t n=2^k
3- n=1 , and m>=0 we get all even number n s.t n= 2k , and k has an odd factor

Answer 26

BINGO!

Answer 27

:3

Answer 28

ty @ganeshie8 and @ikram002p ! so van being this proven by mathinduction or any ideas ?

Answer 29

can being

sorry

Answer 30

sure, ikram u want to try proving it by induction?

Answer 31

hmm we used proof by division algorithm :) we give cases

Answer 32

sure its trivial :)

Answer 33

i feel proving it by contradiction is more natural here though

Answer 34

as a start let
k=2^n *(2m +1)
then show that
k+1 can be written as 2^j*(2i+1)

Answer 35

all works

Answer 36

@ganeshie8 above you have wrote ba contradiction mean by reductio ad absurdum ?

Answer 37

by contradiction

Answer 38

Exactly! we have already established that odd numbers can be constructed by division alogirthm.

For even numbers, we can use contradiction

Answer 39

suppose there exists some even number which cannot be represented in form : \(2^n(2m+1)\)

Answer 40

by definition an even number is divisible by 2.
that means an even number can be divided repeateadly by 2 till you get an odd number.

Answer 41

you stop dividing once you reach an odd number. after repeated divisions the original even number number can be represented as \(2^{n}(2m+1)\)

Answer 42

actually this argument looks nice for a direct proof :)

Answer 43

do u have a proof using induction @jhonyy9 ?

Answer 44

it would be interesing to see how induction works here xD

Answer 45

even for induction , we would also have to discuss 3 cases of the number ;)

Answer 46

when k is odd
k=2m+1 for some m in N
k+1 =2m+2=2(m+1)
m+1 might be even or odd,
... continue like this so u would conclude the other two cases

Answer 47

one idea for even numbers

2^n *(2m +1) = 2(2^(n-1) *(2m +1)

opinions please ?

Answer 48

so for n=N and m=0 will result all power of 2

Answer 49

so wat Nnesha :O