is zero a natural number 0∈N?


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Question Number 114807 by Eric002 last updated on 21/Sep/20

$i s z e r o a n a t u r a l n u m b e r 0 \in N ?$
Commented by MJS_new last updated on 21/Sep/20
$there seem to be two different definitions of N, both are in use... confusion... old definition N={1, 2, 3, ...}; N∪{0}=N_0 new definition N={0, 1, 2, ...}; N\{0}=N^★ so the answer depends on in which country and on which school you are when I was in school (a zillion years ago) we leaned these things one after the other: (1) N: as you count (nobody would start to count by 0). (1.1) addition is always possible in N (1.2) multiplication is always possible in N but n×1=n (1.3) substraction only as a test of addition a+b=c ⇒ c−b=a∧c−a=b (1.4) substraction a−b possible only if a>b? ⇒ (2) Z=N∪{...−2, −1, 0} these are magical numbers we need to substract without limits a−b=x always solveable in Z (2.1) 0 neutral element of addition (2.2) 1 neutral element of multiplication then Q (a÷b=x solveable), then R (all non− periodic numbers like (√2), π; later e...), finally C$
$there seem to be two different definitions$ $of N, both are in use . . . confusion . . .$ $old definition N = {1, 2, 3, . . .}; N \cup {0} = N_{0}$ $new definition N = {0, 1, 2, . . .}; N ∖ {0} = N^{★}$ $so the answer depends on in which country$ $and on which school you are$ $when I was in school (a zillion years ago) we$ $leaned these things one after the other :$ $(1) N : a s y o u c o u n t (nobody would start to$ $count by 0) .$ $(1.1) addition is always possible in N$ $(1.2) multiplication is always possible in N$ $but n \times 1 = n$ $(1.3) substraction only as a test of addition$ $a + b = c \Rightarrow c - b = a \land c - a = b$ $(1.4) substraction a - b possible only if a > b ?$ $\Rightarrow$ $(2) Z = N \cup {. . . - 2, - 1, 0} these are magical$ $numbers we need to substract without limits$ $a - b = x always solveable in Z$ $(2.1) 0 n e u t r a l e l e m e n t o f a d d i t i o n$ $(2.2) 1 n e u t r a l e l e m e n t o f m u l t i p l i c a t i o n$ $then Q (a \div b = x solveable), then R (all non -$ $periodic numbers like \sqrt{2}, π; later e . . .),$ $finally C$
Commented by Eric002 last updated on 21/Sep/20

$t h a n k y o u f o r e x p l a n i n g b u t w h a t i s t h e$ $d i f f e r e n c e b e t w e e n N a n d N^{*}$
Commented by MJS_new last updated on 21/Sep/20
$the old N is the same as the new N^★ ={1, 2, 3, ...} the old N_0 is the same as the new N ={0, 1, 2, ...}$
$the old N is the same as the new N^{★}$ $= {1, 2, 3, . . .}$ $the old N_{0} is the same as the new N$ $= {0, 1, 2, . . .}$
Commented by Eric002 last updated on 21/Sep/20

$t h a n k y o u s i r$
Commented by Rasheed.Sindhi last updated on 21/Sep/20

$W h y {i s n}^{'} t t h e r e a n y i n t e r n a t i o n a l$ $i n s t i t u t e w h i c h p l a y r o l l i n t h i s$ $c o n n e c t i o n ? W h i c h m a k e$ $s t a n d a r d s s o t h a t s u c h$ $c o n f u s i o n s b e k i l l e d ?$
Commented by MJS_new last updated on 21/Sep/20

$standards are no longer accepted in the age$ $of internet; {everybody}^{'} s own opinion counts$ $more than any standard$
Commented by Rasheed.Sindhi last updated on 21/Sep/20

$T h a n k S s S i r!$
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