Mixing
Proof tasks math student need help
Can we add congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a+c≡b+d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
Can we subtract congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a−c≡b−d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
proof-verification proof-explanation
asked Nov 20 '18 at 13:49
S. Totland
22
add a comment |
Can we add congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a+c≡b+d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
Can we subtract congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a−c≡b−d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
proof-verification proof-explanation
asked Nov 20 '18 at 13:49
S. Totland
22
1
Welcome to Math.SE, S. Totland! You’ll find that you get a better response here if you give context to your question (where did it come from? Why are you asking about it?) and talk about your thoughts on the problem (what have you tried to solve it? What are the definitions involved?). We try not to answer questions that look like homework problems, or that simply state a question and expect a solution.
– Santana Afton
Nov 20 '18 at 14:03
1
You'll notice that your question already has three downvotes and vote to close. That's almost certainly because it appears to be a direct transcription of a homework problem, and the folks who contribute to this site generally don't like doing people's homework. You can show us what you've tried so far by clicking "edit" beneath your question, and you might get more positive results.
– John Hughes
Nov 20 '18 at 14:04
add a comment |
Can we add congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a+c≡b+d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
Can we subtract congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a−c≡b−d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
proof-verification proof-explanation
asked Nov 20 '18 at 13:49
S. Totland
22
Can we add congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a+c≡b+d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
Can we subtract congruences? If a≡b mod m and c≡d mod m, is it necessarily true that a−c≡b−d mod m? If so, why? If not, provide an example that illustrates why not. To get started on this question, do some numerical examples.
proof-verification proof-explanation
proof-verification proof-explanation
asked Nov 20 '18 at 13:49
S. Totland
22
asked Nov 20 '18 at 13:49
S. Totland
22
edited Nov 20 '18 at 14:02
asked Nov 20 '18 at 13:49
S. Totland
22
asked Nov 20 '18 at 13:49
S. Totland
22
asked Nov 20 '18 at 13:49
S. Totland
22
22
1
Welcome to Math.SE, S. Totland! You’ll find that you get a better response here if you give context to your question (where did it come from? Why are you asking about it?) and talk about your thoughts on the problem (what have you tried to solve it? What are the definitions involved?). We try not to answer questions that look like homework problems, or that simply state a question and expect a solution.
– Santana Afton
Nov 20 '18 at 14:03
1
You'll notice that your question already has three downvotes and vote to close. That's almost certainly because it appears to be a direct transcription of a homework problem, and the folks who contribute to this site generally don't like doing people's homework. You can show us what you've tried so far by clicking "edit" beneath your question, and you might get more positive results.
– John Hughes
Nov 20 '18 at 14:04
add a comment |
1
Welcome to Math.SE, S. Totland! You’ll find that you get a better response here if you give context to your question (where did it come from? Why are you asking about it?) and talk about your thoughts on the problem (what have you tried to solve it? What are the definitions involved?). We try not to answer questions that look like homework problems, or that simply state a question and expect a solution.
– Santana Afton
Nov 20 '18 at 14:03
1
You'll notice that your question already has three downvotes and vote to close. That's almost certainly because it appears to be a direct transcription of a homework problem, and the folks who contribute to this site generally don't like doing people's homework. You can show us what you've tried so far by clicking "edit" beneath your question, and you might get more positive results.
– John Hughes
Nov 20 '18 at 14:04
1
1
Welcome to Math.SE, S. Totland! You’ll find that you get a better response here if you give context to your question (where did it come from? Why are you asking about it?) and talk about your thoughts on the problem (what have you tried to solve it? What are the definitions involved?). We try not to answer questions that look like homework problems, or that simply state a question and expect a solution.
– Santana Afton
Nov 20 '18 at 14:03
Welcome to Math.SE, S. Totland! You’ll find that you get a better response here if you give context to your question (where did it come from? Why are you asking about it?) and talk about your thoughts on the problem (what have you tried to solve it? What are the definitions involved?). We try not to answer questions that look like homework problems, or that simply state a question and expect a solution.
– Santana Afton
Nov 20 '18 at 14:03
1
1
You'll notice that your question already has three downvotes and vote to close. That's almost certainly because it appears to be a direct transcription of a homework problem, and the folks who contribute to this site generally don't like doing people's homework. You can show us what you've tried so far by clicking "edit" beneath your question, and you might get more positive results.
– John Hughes
Nov 20 '18 at 14:04
You'll notice that your question already has three downvotes and vote to close. That's almost certainly because it appears to be a direct transcription of a homework problem, and the folks who contribute to this site generally don't like doing people's homework. You can show us what you've tried so far by clicking "edit" beneath your question, and you might get more positive results.
– John Hughes
Nov 20 '18 at 14:04
add a comment |
2 Answers
2
active
oldest
votes
Usually we talk about adding congruence classes. For example, if $a=b$ mod(m), we usually say the congruence classes of a and b are equal, that is $[a]=[b]$, and we can add/subtract/multiply two congruence classes.
I'll show the first question and you can do the second one.
By definition, $a=b$ mod(m) $iff$ $a-b=0$ mod(m) $iff$ $a-b=tm$ for some $tinmathbb{Z}$.
Using this fact, we can see that $$a-b+c-d=tm+sm=(t+s)m$$ for some $t,sinmathbb{Z}$. If we add $b+d$ to both sides, we obtain $$a+c=b+d+(t+s)m.$$ Taking mod(m) of both sides, we obtain our result.
answered Nov 20 '18 at 14:04
Blake Jackson
765
1
+1 This is correct and well written. That said, I'd rather you waited to answer until the OP showed some effort of his or her own - note the comments on the questions.
– Ethan Bolker
Nov 20 '18 at 14:45
@EthanBolker Noted. Thank you for the input!
– Blake Jackson
Nov 20 '18 at 18:46
add a comment |
Consider: $$a≡b mod(m) hspace{0.5cm} and hspace{0.5cm} c≡d mod(m)$$
Then, $$m mid (a-b) hspace{0.5cm} and hspace{0.5cm} m mid (c-d)$$
So, $$a-b=km hspace{0.5cm} and hspace{0.5cm} c-d=k'm$$
Adding, $$(a+c)-(b+d)=(k+k')m$$
Hence, $$ m mid (a+c)-(b+d)$$
We get: $$(a+c)≡(b+d) mod m$$
Similarly, proceed for: $$(a-c)≡(b-d) mod m$$
You wanted numerical examples:
Take $$7≡5 mod(2) hspace{0.5cm} and hspace{0.5cm} 10≡4 mod(2)$$
$$7+10≡5+4 mod(2)$$ or $$10≡4 mod(2)$$ $$17≡9 mod(2)$$
Similarly,
$$7-10≡5-4 mod(2)$$ or $$-3≡1 mod(2)$$
answered Nov 20 '18 at 14:15
idea
2,15441025
add a comment |
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%2f3006323%2fproof-tasks-math-student-need-help%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Usually we talk about adding congruence classes. For example, if $a=b$ mod(m), we usually say the congruence classes of a and b are equal, that is $[a]=[b]$, and we can add/subtract/multiply two congruence classes.
I'll show the first question and you can do the second one.
By definition, $a=b$ mod(m) $iff$ $a-b=0$ mod(m) $iff$ $a-b=tm$ for some $tinmathbb{Z}$.
Using this fact, we can see that $$a-b+c-d=tm+sm=(t+s)m$$ for some $t,sinmathbb{Z}$. If we add $b+d$ to both sides, we obtain $$a+c=b+d+(t+s)m.$$ Taking mod(m) of both sides, we obtain our result.
answered Nov 20 '18 at 14:04
Blake Jackson
765
1
+1 This is correct and well written. That said, I'd rather you waited to answer until the OP showed some effort of his or her own - note the comments on the questions.
– Ethan Bolker
Nov 20 '18 at 14:45
@EthanBolker Noted. Thank you for the input!
– Blake Jackson
Nov 20 '18 at 18:46
add a comment |
Usually we talk about adding congruence classes. For example, if $a=b$ mod(m), we usually say the congruence classes of a and b are equal, that is $[a]=[b]$, and we can add/subtract/multiply two congruence classes.
I'll show the first question and you can do the second one.
By definition, $a=b$ mod(m) $iff$ $a-b=0$ mod(m) $iff$ $a-b=tm$ for some $tinmathbb{Z}$.
Using this fact, we can see that $$a-b+c-d=tm+sm=(t+s)m$$ for some $t,sinmathbb{Z}$. If we add $b+d$ to both sides, we obtain $$a+c=b+d+(t+s)m.$$ Taking mod(m) of both sides, we obtain our result.
answered Nov 20 '18 at 14:04
Blake Jackson
765
1
+1 This is correct and well written. That said, I'd rather you waited to answer until the OP showed some effort of his or her own - note the comments on the questions.
– Ethan Bolker
Nov 20 '18 at 14:45
@EthanBolker Noted. Thank you for the input!
– Blake Jackson
Nov 20 '18 at 18:46
add a comment |
Usually we talk about adding congruence classes. For example, if $a=b$ mod(m), we usually say the congruence classes of a and b are equal, that is $[a]=[b]$, and we can add/subtract/multiply two congruence classes.
I'll show the first question and you can do the second one.
By definition, $a=b$ mod(m) $iff$ $a-b=0$ mod(m) $iff$ $a-b=tm$ for some $tinmathbb{Z}$.
Using this fact, we can see that $$a-b+c-d=tm+sm=(t+s)m$$ for some $t,sinmathbb{Z}$. If we add $b+d$ to both sides, we obtain $$a+c=b+d+(t+s)m.$$ Taking mod(m) of both sides, we obtain our result.
answered Nov 20 '18 at 14:04
Blake Jackson
765
Usually we talk about adding congruence classes. For example, if $a=b$ mod(m), we usually say the congruence classes of a and b are equal, that is $[a]=[b]$, and we can add/subtract/multiply two congruence classes.
I'll show the first question and you can do the second one.
By definition, $a=b$ mod(m) $iff$ $a-b=0$ mod(m) $iff$ $a-b=tm$ for some $tinmathbb{Z}$.
Using this fact, we can see that $$a-b+c-d=tm+sm=(t+s)m$$ for some $t,sinmathbb{Z}$. If we add $b+d$ to both sides, we obtain $$a+c=b+d+(t+s)m.$$ Taking mod(m) of both sides, we obtain our result.
answered Nov 20 '18 at 14:04
Blake Jackson
765
answered Nov 20 '18 at 14:04
Blake Jackson
765
answered Nov 20 '18 at 14:04
Blake Jackson
765
answered Nov 20 '18 at 14:04
Blake Jackson
765
765
1
+1 This is correct and well written. That said, I'd rather you waited to answer until the OP showed some effort of his or her own - note the comments on the questions.
– Ethan Bolker
Nov 20 '18 at 14:45
@EthanBolker Noted. Thank you for the input!
– Blake Jackson
Nov 20 '18 at 18:46
add a comment |
1
+1 This is correct and well written. That said, I'd rather you waited to answer until the OP showed some effort of his or her own - note the comments on the questions.
– Ethan Bolker
Nov 20 '18 at 14:45
@EthanBolker Noted. Thank you for the input!
– Blake Jackson
Nov 20 '18 at 18:46
1
1
+1 This is correct and well written. That said, I'd rather you waited to answer until the OP showed some effort of his or her own - note the comments on the questions.
– Ethan Bolker
Nov 20 '18 at 14:45
+1 This is correct and well written. That said, I'd rather you waited to answer until the OP showed some effort of his or her own - note the comments on the questions.
– Ethan Bolker
Nov 20 '18 at 14:45
@EthanBolker Noted. Thank you for the input!
– Blake Jackson
Nov 20 '18 at 18:46
@EthanBolker Noted. Thank you for the input!
– Blake Jackson
Nov 20 '18 at 18:46
add a comment |
Consider: $$a≡b mod(m) hspace{0.5cm} and hspace{0.5cm} c≡d mod(m)$$
Then, $$m mid (a-b) hspace{0.5cm} and hspace{0.5cm} m mid (c-d)$$
So, $$a-b=km hspace{0.5cm} and hspace{0.5cm} c-d=k'm$$
Adding, $$(a+c)-(b+d)=(k+k')m$$
Hence, $$ m mid (a+c)-(b+d)$$
We get: $$(a+c)≡(b+d) mod m$$
Similarly, proceed for: $$(a-c)≡(b-d) mod m$$
You wanted numerical examples:
Take $$7≡5 mod(2) hspace{0.5cm} and hspace{0.5cm} 10≡4 mod(2)$$
$$7+10≡5+4 mod(2)$$ or $$10≡4 mod(2)$$ $$17≡9 mod(2)$$
Similarly,
$$7-10≡5-4 mod(2)$$ or $$-3≡1 mod(2)$$
answered Nov 20 '18 at 14:15
idea
2,15441025
add a comment |
Consider: $$a≡b mod(m) hspace{0.5cm} and hspace{0.5cm} c≡d mod(m)$$
Then, $$m mid (a-b) hspace{0.5cm} and hspace{0.5cm} m mid (c-d)$$
So, $$a-b=km hspace{0.5cm} and hspace{0.5cm} c-d=k'm$$
Adding, $$(a+c)-(b+d)=(k+k')m$$
Hence, $$ m mid (a+c)-(b+d)$$
We get: $$(a+c)≡(b+d) mod m$$
Similarly, proceed for: $$(a-c)≡(b-d) mod m$$
You wanted numerical examples:
Take $$7≡5 mod(2) hspace{0.5cm} and hspace{0.5cm} 10≡4 mod(2)$$
$$7+10≡5+4 mod(2)$$ or $$10≡4 mod(2)$$ $$17≡9 mod(2)$$
Similarly,
$$7-10≡5-4 mod(2)$$ or $$-3≡1 mod(2)$$
answered Nov 20 '18 at 14:15
idea
2,15441025
add a comment |
Consider: $$a≡b mod(m) hspace{0.5cm} and hspace{0.5cm} c≡d mod(m)$$
Then, $$m mid (a-b) hspace{0.5cm} and hspace{0.5cm} m mid (c-d)$$
So, $$a-b=km hspace{0.5cm} and hspace{0.5cm} c-d=k'm$$
Adding, $$(a+c)-(b+d)=(k+k')m$$
Hence, $$ m mid (a+c)-(b+d)$$
We get: $$(a+c)≡(b+d) mod m$$
Similarly, proceed for: $$(a-c)≡(b-d) mod m$$
You wanted numerical examples:
Take $$7≡5 mod(2) hspace{0.5cm} and hspace{0.5cm} 10≡4 mod(2)$$
$$7+10≡5+4 mod(2)$$ or $$10≡4 mod(2)$$ $$17≡9 mod(2)$$
Similarly,
$$7-10≡5-4 mod(2)$$ or $$-3≡1 mod(2)$$
answered Nov 20 '18 at 14:15
idea
2,15441025
Consider: $$a≡b mod(m) hspace{0.5cm} and hspace{0.5cm} c≡d mod(m)$$
Then, $$m mid (a-b) hspace{0.5cm} and hspace{0.5cm} m mid (c-d)$$
So, $$a-b=km hspace{0.5cm} and hspace{0.5cm} c-d=k'm$$
Adding, $$(a+c)-(b+d)=(k+k')m$$
Hence, $$ m mid (a+c)-(b+d)$$
We get: $$(a+c)≡(b+d) mod m$$
Similarly, proceed for: $$(a-c)≡(b-d) mod m$$
You wanted numerical examples:
Take $$7≡5 mod(2) hspace{0.5cm} and hspace{0.5cm} 10≡4 mod(2)$$
$$7+10≡5+4 mod(2)$$ or $$10≡4 mod(2)$$ $$17≡9 mod(2)$$
Similarly,
$$7-10≡5-4 mod(2)$$ or $$-3≡1 mod(2)$$
answered Nov 20 '18 at 14:15
idea
2,15441025
edited Nov 20 '18 at 14:24
answered Nov 20 '18 at 14:15
idea
2,15441025
answered Nov 20 '18 at 14:15
idea
2,15441025
answered Nov 20 '18 at 14:15
idea
2,15441025
2,15441025
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3006323%2fproof-tasks-math-student-need-help%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
1
Welcome to Math.SE, S. Totland! You’ll find that you get a better response here if you give context to your question (where did it come from? Why are you asking about it?) and talk about your thoughts on the problem (what have you tried to solve it? What are the definitions involved?). We try not to answer questions that look like homework problems, or that simply state a question and expect a solution.
– Santana Afton
Nov 20 '18 at 14:03
1
You'll notice that your question already has three downvotes and vote to close. That's almost certainly because it appears to be a direct transcription of a homework problem, and the folks who contribute to this site generally don't like doing people's homework. You can show us what you've tried so far by clicking "edit" beneath your question, and you might get more positive results.
– John Hughes
Nov 20 '18 at 14:04