How is rearranging 56 x 100 ÷ 8 into 56 ÷8 x100 allowed by the commutative property?
$begingroup$
So according to the commutative property for multiplication:
$a * b = b * a$
However this does not hold for division
$a ÷ b$ $!= b ÷a$
Why is it that in the following case:
$56 * 100 ÷ 8 = 56 ÷ 8* 100$
It seems like division is breaking the rule. There is something I am misunderstanding here.
Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?
If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?
arithmetic
asked 45 mins ago
SphygmomanometerSphygmomanometer
717
$endgroup$
add a comment |
$begingroup$
So according to the commutative property for multiplication:
$a * b = b * a$
However this does not hold for division
$a ÷ b$ $!= b ÷a$
Why is it that in the following case:
$56 * 100 ÷ 8 = 56 ÷ 8* 100$
It seems like division is breaking the rule. There is something I am misunderstanding here.
Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?
If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?
arithmetic
asked 45 mins ago
SphygmomanometerSphygmomanometer
717
$endgroup$
add a comment |
$begingroup$
So according to the commutative property for multiplication:
$a * b = b * a$
However this does not hold for division
$a ÷ b$ $!= b ÷a$
Why is it that in the following case:
$56 * 100 ÷ 8 = 56 ÷ 8* 100$
It seems like division is breaking the rule. There is something I am misunderstanding here.
Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?
If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?
arithmetic
asked 45 mins ago
SphygmomanometerSphygmomanometer
717
$endgroup$
So according to the commutative property for multiplication:
$a * b = b * a$
However this does not hold for division
$a ÷ b$ $!= b ÷a$
Why is it that in the following case:
$56 * 100 ÷ 8 = 56 ÷ 8* 100$
It seems like division is breaking the rule. There is something I am misunderstanding here.
Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?
If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?
arithmetic
arithmetic
asked 45 mins ago
SphygmomanometerSphygmomanometer
717
asked 45 mins ago
SphygmomanometerSphygmomanometer
717
edited 43 mins ago
Sphygmomanometer
asked 45 mins ago
SphygmomanometerSphygmomanometer
717
asked 45 mins ago
SphygmomanometerSphygmomanometer
717
asked 45 mins ago
SphygmomanometerSphygmomanometer
717
717
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 36 mins ago
AndreiAndrei
12.1k21127
$endgroup$
add a comment |
$begingroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 36 mins ago
Angela RichardsonAngela Richardson
5,28911733
$endgroup$
add a comment |
$begingroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 31 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 31 mins ago
Rhys HughesRhys Hughes
6,7101530
$endgroup$
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%2f3117260%2fhow-is-rearranging-56-x-100-%25c3%25b7-8-into-56-%25c3%25b78-x100-allowed-by-the-commutative-prope%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 36 mins ago
AndreiAndrei
12.1k21127
$endgroup$
add a comment |
$begingroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 36 mins ago
AndreiAndrei
12.1k21127
$endgroup$
add a comment |
$begingroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 36 mins ago
AndreiAndrei
12.1k21127
$endgroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 36 mins ago
AndreiAndrei
12.1k21127
answered 36 mins ago
AndreiAndrei
12.1k21127
answered 36 mins ago
AndreiAndrei
12.1k21127
answered 36 mins ago
AndreiAndrei
12.1k21127
12.1k21127
add a comment |
add a comment |
$begingroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 36 mins ago
Angela RichardsonAngela Richardson
5,28911733
$endgroup$
add a comment |
$begingroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 36 mins ago
Angela RichardsonAngela Richardson
5,28911733
$endgroup$
add a comment |
$begingroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 36 mins ago
Angela RichardsonAngela Richardson
5,28911733
$endgroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 36 mins ago
Angela RichardsonAngela Richardson
5,28911733
answered 36 mins ago
Angela RichardsonAngela Richardson
5,28911733
answered 36 mins ago
Angela RichardsonAngela Richardson
5,28911733
answered 36 mins ago
Angela RichardsonAngela Richardson
5,28911733
5,28911733
add a comment |
add a comment |
$begingroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 31 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 31 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 31 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 31 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 31 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 31 mins ago
AdityaAditya
112
answered 31 mins ago
AdityaAditya
112
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 31 mins ago
Rhys HughesRhys Hughes
6,7101530
$endgroup$
add a comment |
$begingroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 31 mins ago
Rhys HughesRhys Hughes
6,7101530
$endgroup$
add a comment |
$begingroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 31 mins ago
Rhys HughesRhys Hughes
6,7101530
$endgroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 31 mins ago
Rhys HughesRhys Hughes
6,7101530
answered 31 mins ago
Rhys HughesRhys Hughes
6,7101530
answered 31 mins ago
Rhys HughesRhys Hughes
6,7101530
answered 31 mins ago
Rhys HughesRhys Hughes
6,7101530
6,7101530
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.
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%2f3117260%2fhow-is-rearranging-56-x-100-%25c3%25b7-8-into-56-%25c3%25b78-x100-allowed-by-the-commutative-prope%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
