Mixing
Prove that $(A triangle C) cap (B triangle C) subseteq (A cap B) triangle C$
$begingroup$
This is a exercise from How To Prove It by Velleman. Exercise 3.5.24 to be exact. The solutions I can find online involve using 4 cases. My proof involves 2 so I was wondering if I was skipping some steps.
Proof: Let x be an arbitrary element of $(A triangle C) cap (B triangle C)$. It follows that $x in A triangle C$ and $x in B triangle C$.
We will consider 2 cases.
Case 1: $x in C$. Since $x in A triangle C$, it follows that $x notin A$. Also since $x in B triangle C$, it follows that $x notin B$. This means $x in C setminus (A cap B)$ thus $x in (A cap B) triangle C$.
Case 2: $x notin C$. Since $x in A triangle C$, it follows that $x in A$. Also since $x in B triangle C$, it follows that $x in B$. This means $x in (A cap B) setminus C$ thus $x in (A cap B) triangle C$.
Both cases gave us $x in (A cap B) triangle C$ and since x was arbitrary $(A triangle C) cap (B triangle C) subseteq (A cap B) triangle C$
Is this proof a valid one? Any feedback would be greatly appreciated.
Thanks in advance.
proof-verification
asked Jan 28 at 3:59
fesodesfesodes
262
$endgroup$
add a comment |
$begingroup$
This is a exercise from How To Prove It by Velleman. Exercise 3.5.24 to be exact. The solutions I can find online involve using 4 cases. My proof involves 2 so I was wondering if I was skipping some steps.
Proof: Let x be an arbitrary element of $(A triangle C) cap (B triangle C)$. It follows that $x in A triangle C$ and $x in B triangle C$.
We will consider 2 cases.
Case 1: $x in C$. Since $x in A triangle C$, it follows that $x notin A$. Also since $x in B triangle C$, it follows that $x notin B$. This means $x in C setminus (A cap B)$ thus $x in (A cap B) triangle C$.
Case 2: $x notin C$. Since $x in A triangle C$, it follows that $x in A$. Also since $x in B triangle C$, it follows that $x in B$. This means $x in (A cap B) setminus C$ thus $x in (A cap B) triangle C$.
Both cases gave us $x in (A cap B) triangle C$ and since x was arbitrary $(A triangle C) cap (B triangle C) subseteq (A cap B) triangle C$
Is this proof a valid one? Any feedback would be greatly appreciated.
Thanks in advance.
proof-verification
asked Jan 28 at 3:59
fesodesfesodes
262
$endgroup$
$begingroup$
Looks good to me...
$endgroup$
– Arturo Magidin
Jan 28 at 4:01
$begingroup$
The proof is fine but in Case 1 the second sentence is not necessary.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 5:56
add a comment |
$begingroup$
This is a exercise from How To Prove It by Velleman. Exercise 3.5.24 to be exact. The solutions I can find online involve using 4 cases. My proof involves 2 so I was wondering if I was skipping some steps.
Proof: Let x be an arbitrary element of $(A triangle C) cap (B triangle C)$. It follows that $x in A triangle C$ and $x in B triangle C$.
We will consider 2 cases.
Case 1: $x in C$. Since $x in A triangle C$, it follows that $x notin A$. Also since $x in B triangle C$, it follows that $x notin B$. This means $x in C setminus (A cap B)$ thus $x in (A cap B) triangle C$.
Case 2: $x notin C$. Since $x in A triangle C$, it follows that $x in A$. Also since $x in B triangle C$, it follows that $x in B$. This means $x in (A cap B) setminus C$ thus $x in (A cap B) triangle C$.
Both cases gave us $x in (A cap B) triangle C$ and since x was arbitrary $(A triangle C) cap (B triangle C) subseteq (A cap B) triangle C$
Is this proof a valid one? Any feedback would be greatly appreciated.
Thanks in advance.
proof-verification
asked Jan 28 at 3:59
fesodesfesodes
262
$endgroup$
This is a exercise from How To Prove It by Velleman. Exercise 3.5.24 to be exact. The solutions I can find online involve using 4 cases. My proof involves 2 so I was wondering if I was skipping some steps.
Proof: Let x be an arbitrary element of $(A triangle C) cap (B triangle C)$. It follows that $x in A triangle C$ and $x in B triangle C$.
We will consider 2 cases.
Case 1: $x in C$. Since $x in A triangle C$, it follows that $x notin A$. Also since $x in B triangle C$, it follows that $x notin B$. This means $x in C setminus (A cap B)$ thus $x in (A cap B) triangle C$.
Case 2: $x notin C$. Since $x in A triangle C$, it follows that $x in A$. Also since $x in B triangle C$, it follows that $x in B$. This means $x in (A cap B) setminus C$ thus $x in (A cap B) triangle C$.
Both cases gave us $x in (A cap B) triangle C$ and since x was arbitrary $(A triangle C) cap (B triangle C) subseteq (A cap B) triangle C$
Is this proof a valid one? Any feedback would be greatly appreciated.
Thanks in advance.
proof-verification
proof-verification
asked Jan 28 at 3:59
fesodesfesodes
262
asked Jan 28 at 3:59
fesodesfesodes
262
asked Jan 28 at 3:59
fesodesfesodes
262
asked Jan 28 at 3:59
fesodesfesodes
262
asked Jan 28 at 3:59
fesodesfesodes
262
262
$begingroup$
Looks good to me...
$endgroup$
– Arturo Magidin
Jan 28 at 4:01
$begingroup$
The proof is fine but in Case 1 the second sentence is not necessary.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 5:56
add a comment |
$begingroup$
Looks good to me...
$endgroup$
– Arturo Magidin
Jan 28 at 4:01
$begingroup$
The proof is fine but in Case 1 the second sentence is not necessary.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 5:56
$begingroup$
Looks good to me...
$endgroup$
– Arturo Magidin
Jan 28 at 4:01
$begingroup$
Looks good to me...
$endgroup$
– Arturo Magidin
Jan 28 at 4:01
$begingroup$
The proof is fine but in Case 1 the second sentence is not necessary.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 5:56
$begingroup$
The proof is fine but in Case 1 the second sentence is not necessary.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 5:56
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090460%2fprove-that-a-triangle-c-cap-b-triangle-c-subseteq-a-cap-b-triangle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090460%2fprove-that-a-triangle-c-cap-b-triangle-c-subseteq-a-cap-b-triangle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Looks good to me...
$endgroup$
– Arturo Magidin
Jan 28 at 4:01
$begingroup$
The proof is fine but in Case 1 the second sentence is not necessary.
$endgroup$
– Kavi Rama Murthy
Jan 28 at 5:56