Hierarchy of the Numbers

If

  • N \mathbb{N} is the set of natural numbers,
  • Q \mathbb{Q} is the set of rational numbers,
  • R \mathbb{R} is the set of real numbers,
  • and Z \mathbb{Z} is the set of integers,

then what is the correct ordering of the containment of these sets in each other?

R Q Z N \mathbb{R} \subset \mathbb{Q} \subset \mathbb{Z} \subset \mathbb{N} Z N Q R \mathbb{Z} \subset \mathbb{N} \subset \mathbb{Q} \subset \mathbb{R} Q N Z R \mathbb{Q} \subset \mathbb{N} \subset \mathbb{Z} \subset \mathbb{R} N Z Q R \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Ricky Ramcharitar
Jan 18, 2016

N \mathbb{N} \rightarrow Natural Numbers: 1,2,3,4,5, ...

Z \mathbb{Z} \rightarrow Integers: ...-3,-2,-1,0,1,2,3, ...

Contains the Set of Natural Numbers (i.e. N \mathbb{N} )

Q \mathbb{Q} \rightarrow Rational Numbers: Any number that can be written as a fraction --> m n \dfrac{m}{n}

Contains the Sets of Natural Numbers and Integers (i.e. Z \mathbb{Z} and N \mathbb{N} )

R \mathbb{R} \rightarrow Real Numbers: Any number that is not complex (i.e. all Rational and Irrational Numbers) --> π , e , 1 3 , 0 , 2 , 4 7 \pi, e, -\dfrac{1}{3}, \sqrt{0}, \sqrt{2}, \sqrt{\dfrac{4}{7}} , etc...

Contains the Sets of Natural Numbers, Integers, and Rational Numbers (i.e. N \mathbb{N} , Z \mathbb{Z} , and Q \mathbb{Q} )

Therefore:

N Z Q R \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}

Specifically where m and n are integers. Also, some reals can't be expressed with square roots. Pi, for example, and e, and the Euler-Mascheroni constant. In fact, I'd say "most" reals can't be expressed with square roots. You need to fix your definition.

Kieran Kaempen - 5 years, 4 months ago

Log in to reply

I also wrote that the real numbers contain all rational and irrational numbers, not just positive square roots.

I fixed it up a bit to highlight all rational and irrational at the beginning.

Ricky Ramcharitar - 5 years, 4 months ago
Fahim Abid
Jan 16, 2016

As N, Z, Q, R contains natural, integers, rational and real numbers respectively, that is why N ⊂ Z ⊂ Q ⊂ R

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...