Cartesian Product!

We know that given two non - void sets, say AA and BB, their Cartesian Product\textbf{Cartesian Product} is defined as A×B={(a,b):aA,bB}A\,\times\,B\,=\,\{(a,b)\,:\,a\,\in\,A\,,\,b\,\in\,B\}. For eg. If A={1,3}A\,=\,\{1,3\} and B={2,4,9}B\,=\,\{2,4,9\}, then their Cartesian product is defined as {(1,2),(1,4),(1,9),(3,2),(3,4),(3,9)}\{(1,2),(1,4),(1,9),(3,2),(3,4),(3,9)\}.

I was wondering what if the set AA itself consists of ordered pairs, like what if A={(1,5),(1,9),(2,5)}A\,=\,\{(1,5),(1,9),(2,5)\} and say B={3,7}B\,=\,\{3,7\}, then how do we define the A×BA\,\times B?

One thing that I've observed is that ,in case, if the set that consists of ordered pairs ( n-tuples in general) can be expressed as Cartesian product of simple sets (sets that consists of numbers only), then the overall cartesian product can be found. What I mean is

Say, A={(3,4),(3,7),(2,4),(2,7)}A\,=\,\{(3,4),(3,7),(2,4),(2,7)\} and B={2,5}B\,=\,\{2,5\} and A×BA\times B is to be found, then one can proceed as follows :

A×B={(3,4),(3,7),(2,4),(2,7)}×{2,5}={3,2}×{4,7}×{2,5}={(2,3,4),(2,3,7),(2,2,4),(2,2,7),(5,3,4),(5,3,7),(5,2,4),(5,2,7)}A\times B\,=\,\{(3,4),(3,7),(2,4),(2,7)\}\,\times\,\{2,5\}\,\,=\,\{3,2\}\,\times\,\{4,7\}\,\times\,\{2,5\}\,\,=\,\{(2,3,4),(2,3,7),(2,2,4),(2,2,7),(5,3,4),(5,3,7),(5,2,4),(5,2,7)\}.

Note by Aditya Sky
4 years, 3 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

Do the same thing. A×B={(a,b)aA,bB} A \times B = \{ ( a, b) | a \in A, b \in B \} . It doesn't matter what the sets AA and BB are.

E.g. you can determine { elephant, zinc}×{12,(1,2)} \{ \text{ elephant, zinc} \} \times \{ \frac{1}{2} , ( 1, 2) \} .

Calvin Lin Staff - 4 years, 3 months ago

Log in to reply


Aditya Sky - 4 years, 3 months ago

Log in to reply

Infact, this is the method followed to generate n-ary relations.......

Aaghaz Mahajan - 2 years, 3 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...