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If $A=\{1,2,3,4\},~B=\{4,5\}$ and $C=\{5,6\}$ find:

  • $A\times (B-C)$
  • $(A\times B)-(A\times C)$
  • $(A\times B)\cap (A\times C)$

My answers that I came up with were:

  • $\{(1,4),(2,4),(3,4),(4,4),(1,5),(1,5),(3,5),(4,5)\}$
  • $\{(1,4),(2,4),(3,4),(4,4),(1,6),(2,6),(36),(4,6)\}$
  • $\{(1,5),(1,5),(3,5),(4,5)\}$

Please correct me if I am wrong in any of these answers.

Editor's note: originals found here https://i.stack.imgur.com/sPMvp.png https://i.stack.imgur.com/POZon.png

JMoravitz
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Harmaan
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  • will you please take the pain of properly posting the question instead of just posting an image and expecting to get an answer? – Devansh Kamra Mar 05 '21 at 17:01
  • Welcome to Math.ME! Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – md2perpe Mar 05 '21 at 17:01
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    Take a look at how I retyped your question. I left in the obvious typos that you made, like $(36)$ being an element and having multiple $(1,5)$'s. Feel free to correct those yourself. – JMoravitz Mar 05 '21 at 17:02
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    They are all wrong. – fleablood Mar 05 '21 at 17:04
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    Before you answer any of these. First calculate what $B-C$ is; and what $A\times B$ and $A\times C$ are – fleablood Mar 05 '21 at 17:05
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    As for your answers, apart from the obvious typos, you do have a number of errors. Specifically, your answers for the first two parts are both incorrect. You seem to be misunderstanding what set difference is. Note, for instance, that ${4,5}-{5,6}={4}$... so the first problem is asking for ${1,2,3,4}\times {4}$. All three of these problems will have answers who contain just four pairs, not eight. – JMoravitz Mar 05 '21 at 17:05

1 Answers1

1

$A \times (B -C)$:

First $B-C = \{4,\color{red}5\} - \{\color{red}5, \color{purple}6\} = \{4\}$

So $A \times (B-C) = \{1,2,3,4\} \times \{4\}= \{(1,4),(2,4),(3,4),(4,4)\}$

Why did you include any with $5$ in the second term?

$(A\times B) - (A\times C)$:

This means we can have any $(x,y)$ so that $x\in A;y\in B$ so long as we do not have both $x\in A$ and $y\in C$. As $C$ contains $5$ and $6$ we can't have any $x\in A; y = 5$ or $y= 6$. The only $y \in B$ that are not $5$ or $6$ are $y=4$ so we can only have $(x,y)$ there $x \in A$ and $y = 4$. And we can have all of those.

..... or more directly

$$A \times B = \{(1,4),(2,4),(3,4),(4,4),\\ (1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5)\}$$

but we must exclude $$A\times C = \{(1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5),\\ (1,\color{purple} 6),(2,\color{purple} 6),(3,\color{purple} 6),(4,\color{purple} 6)\}$$

So that leaves $A\times B - A\times C = \{(1,4),(2,4),(3,4),(4,4)\}$

$(A\times B)\cap (A\times C)$:

Again $$A \times B = \{(1,4),(2,4),(3,4),(4,4),\\ (1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5)\}$$

and $$A\times C = \{(1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5),\\ (1,\color{purple} 6),(2,\color{purple} 6),(3,\color{purple} 6),(4,\color{purple} 6)\}$$

so we must take what the have in common:

$(A\times B)\cap (A\times C)=\{(1,\color{red} 5),(2,\color{red} 5),(3,\color{red} 5),(4,\color{red} 5)\}$

fleablood
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