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Mathematics 15 Online
OpenStudy (anonymous):

zero cant be classified as wich type of number?

OpenStudy (angela210793):

Real no

OpenStudy (anonymous):

Imaginary

OpenStudy (angela210793):

not natural

OpenStudy (anonymous):

Ishaan94, whole numbers include 0, natural numbers do not.

OpenStudy (anonymous):

how many numbers are between 1 and 2? how would you classify the numbers?

OpenStudy (angela210793):

yea.....but u said natural not Ishaan

OpenStudy (angela210793):

thr r infinite no.s between 1 and 2

OpenStudy (anonymous):

i thought he said zero can be classified sorry reading error it is can't

OpenStudy (anonymous):

0 is an integer

OpenStudy (anonymous):

pk is right

OpenStudy (anonymous):

Im a girl thank u

OpenStudy (anonymous):

lol whats to do with girl

OpenStudy (anonymous):

maths is maths

OpenStudy (angela210793):

okokokokok....so is 0 in natural no.s or not? I know it is not..

OpenStudy (anonymous):

is every whole number an integer? is every integer a whole number?

OpenStudy (anonymous):

it's not natural number you right angela

OpenStudy (angela210793):

kk thnx :)

OpenStudy (anonymous):

but the question is "zero can't be classified as wich type of number"...

OpenStudy (anonymous):

can anyone answer me question

OpenStudy (angela210793):

:O:O:O:O:O:O:O OMG!!! oops :S :$:$:$

OpenStudy (anonymous):

pk did that right what kinda answer your expecting

OpenStudy (anonymous):

Integers are whole numbers, so yes @carrie

OpenStudy (anonymous):

natural for sure

OpenStudy (anonymous):

I asked is every whole number an integer? is every integer a whole number?

OpenStudy (anonymous):

Thax

OpenStudy (anonymous):

whole numbers are always positive,1,2,3,4.....etc but integers can be positive or negative ......-1,-2,-3,0,1,2,3....

OpenStudy (anonymous):

if the number is negative, which of the following statement is false? 1,The opposite number is to the left of zero 2. The number is to the left of zer 3, the absolute value of the number is to the right of zero.

OpenStudy (anonymous):

so the answer of ur 1st question is YES,2nd is NO

OpenStudy (anonymous):

it just depends on how you define these sets of numbers. I consider a whole number to be a number that doesnt have a fractional part. So in that sense, -5, -3, etc are whole numbers. If you were taught they had to be positive as well, then I see why you would disagree, and thats cool. @carrie if the number os negative, then it is to the left of 0: ...-3, -2, -1, 0, 1, 2, 3...

OpenStudy (anonymous):

1

OpenStudy (anonymous):

List all the number sets that contain the number 15

OpenStudy (anonymous):

is it true that all numbers belong to the set of complex numbers?

OpenStudy (anonymous):

no.real numbers are not definitely.

OpenStudy (anonymous):

yes, real numbers are also complex numbers. They are complex numbers in the form a+bi with b = 0.

OpenStudy (anonymous):

you have to allow real numbers to be complex numbers so the set of complex numbers can remain closed under multiplication and addition. If a set isnt closed under those opperations, its not a Field, and it cant have inverses. Which is disastrous!

OpenStudy (anonymous):

some more explanation please joemath.

OpenStudy (anonymous):

yeah me too

OpenStudy (anonymous):

So, a Field is a set of numbers (or objects) that satisfy a couple of properties. One of the properties is called "Closure under Addition/Multiplication" A set is closed under addition if you can pick any two elements in the set, add them, and you still remain in the set. For example: The set of Natural Numbers is closed under addition {1,2,3,4....} If i add any two natural numbers, i will still have a natural number. The set {-1,0,1} is not closed under addition. If i pick 1 + 1 i get 2, which isnt in the set. Closure under multiplication means the same thing, only with multiplication. (still more to come lol)

OpenStudy (anonymous):

Now, if you say that Real numbers arent complex, then that would force complex numbers not to be closed under addition. a+bi is complex, a-bi is complex, if you add them you get 2a, which (if real numbers weren't complex as well) would not be complex. Also, we know that: \[(a+bi)(a-bi) = a^2+b^2\] which also wouldnt be complex If there wasnt closure under multiplication, there wouldnt be solutions to: \[cx = d\] in the complex plane.

OpenStudy (anonymous):

i understand.thanks

OpenStudy (anonymous):

ahh I didb;t knew things go like this...this is interesting ...thanks joe

OpenStudy (anonymous):

no prob :) i learned the formal definitions last year lol. It was a lot to grasp, but it helps.

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