anonymous
  • anonymous
State whether each of the following sets have closure for + - x and division. (Natural #s), (Rational #s) (irrational #s) (whole #s)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
do you know what "closure" means?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
ok then what about \(\mathbb{N}\) if you add two natural numbers, do you get a natural number?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Yes.
anonymous
  • anonymous
k then it is closed under \(+\) if you subtract one natural number from another, do you always get a natural number?
anonymous
  • anonymous
I sorta need these quick... I have 3 more pages of homework left o_O
anonymous
  • anonymous
then lets do them quick if you subtract, do you get a natural number?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
what about \(5-10\)?
anonymous
  • anonymous
Oh... I see what u did there. lol Then it's not closed.
anonymous
  • anonymous
right
anonymous
  • anonymous
Multi is Closed, right
anonymous
  • anonymous
if you multiply two natural numbers, do you get a natural number? yes, it is closed under \(\times\)
anonymous
  • anonymous
And division can't be a negative
anonymous
  • anonymous
division is \(2\div 7\) a natural number?
anonymous
  • anonymous
Nope... so it's not closed
anonymous
  • anonymous
k now since you want to be quick, rationals are closed under addition, subtraction \[\frac{a}{b}\pm\frac{c}{d}=\frac{ad\pm bc}{bd}\]
anonymous
  • anonymous
Thank you for this!!
anonymous
  • anonymous
also multiplication \[\frac{a}{b}\times \frac{c}{d}=\frac{ac}{bc}\]
anonymous
  • anonymous
yw
anonymous
  • anonymous
division is a bit trickier it is closed where the operation is defined, so i would say yes
anonymous
  • anonymous
in fact the answer is "yes" closed under division
anonymous
  • anonymous
as for irrationals, that is trickier still
anonymous
  • anonymous
you might think they were closed under addition, but they are not for example \[2+\sqrt{3}\] is irrational , but so is \(-\sqrt3\) and if you add them you get \(2\)
anonymous
  • anonymous
so irrationals not closed under addition, and of course therefore not closed under subtraction as for multiplication, it is clear that they are not closed take \(\sqrt{2}\times \sqrt{2}\) and get \(2\)
anonymous
  • anonymous
likewise for division
anonymous
  • anonymous
Thank you so much! I do have a couple other problems similar to this, they have various numbers in brackets. Not sure how to solve them. I will write a few
anonymous
  • anonymous
as for "whole numbers" i have never in my too long life understood the distinction between "natural" and "whole" there is some argument about what is what, and i stay out of it
anonymous
  • anonymous
go ahead and post if i know the answer i will help
anonymous
  • anonymous
{1} {0,1} {-1,0,1} {0,2,4,6....} {1,2,3} {1,3,5} {-1,1} that's all of them, each one with the brackets is a new problem and not part of the old one. Same directions at the top, I'm supposed to State whether the following sets have closure in each 4 ways
anonymous
  • anonymous
ok \(\{1\}\) is pretty clearly not closed under addition since \(1+1=2\) damn sometime math is not so hard
anonymous
  • anonymous
neither under subtraction since \(1-1=0\) however, it is closed under multiplication and division right? \[1\times 1=1,\frac{1}{1}=1\]
anonymous
  • anonymous
you got the hang of this?
anonymous
  • anonymous
Yep, it's just I don't understand with multiple numbers in the bracket what I'm supposed to do. Plus I have a ton more homework due tomorrow morning
anonymous
  • anonymous
\(\{-1,0,1\}\) not closed under addition right? since \(1+1=2\)
anonymous
  • anonymous
or subtraction but if you multiply any two of those numbers you get another one also if you divide (with the understanding that you cannot divide by 0)
anonymous
  • anonymous
Ok great so multiply and divide are closed
anonymous
  • anonymous
even whole numbers add any two you get another even whole number subtract, no because you get \(2-2=-2\)
anonymous
  • anonymous
multiply, yes, the product of two evens is even divide, no \(\frac{4}{4}=1\)
anonymous
  • anonymous
{1,2,3} i don't think this is closed under anything you can check though
anonymous
  • anonymous
likewise for {1,3,5}
anonymous
  • anonymous
as for {-1,1} i think it is closed wrt multiplication and division
anonymous
  • anonymous
Thank you!

Looking for something else?

Not the answer you are looking for? Search for more explanations.