Mixing
![Find a $x$ such that $2^{2015}xequiv 1 pmod{13}$ [on hold]](https://lh5.googleusercontent.com/-aB5MTzWhkNE/AAAAAAAAAAI/AAAAAAAAAAA/ABtNlbAd1seH-eyBHj6GysayZAN2T3zsDw/s72-c-mo/photo.jpg?sz=32)
3D spaces in $mathbb{R}^4$
$begingroup$
Let $V$ be a 3-dimensional subspace in $mathbb{R}^4$. Find such $V_1, V_2, V_3, V_4$ that $$V_1 cap V_2 cap V_3 cap V_4 = {0,0,0,0}$$.
I thought about such $V_i$ that
$V_1 = left{ begin{bmatrix}
1\
0\
0\
0\
end{bmatrix},
begin{bmatrix}
0\
1\
0\
0\
end{bmatrix},
begin{bmatrix}
0\
0\
1\
0\
end{bmatrix} right}
$
And $V_2$ is the same, but the bottom line changes to 1, in $V_3$ it changes to 2 and in $V_4$ to 3. Am I thinking right?
linear-algebra
asked Feb 1 at 22:23
aaaaaa
133
$endgroup$
add a comment |
$begingroup$
Let $V$ be a 3-dimensional subspace in $mathbb{R}^4$. Find such $V_1, V_2, V_3, V_4$ that $$V_1 cap V_2 cap V_3 cap V_4 = {0,0,0,0}$$.
I thought about such $V_i$ that
$V_1 = left{ begin{bmatrix}
1\
0\
0\
0\
end{bmatrix},
begin{bmatrix}
0\
1\
0\
0\
end{bmatrix},
begin{bmatrix}
0\
0\
1\
0\
end{bmatrix} right}
$
And $V_2$ is the same, but the bottom line changes to 1, in $V_3$ it changes to 2 and in $V_4$ to 3. Am I thinking right?
linear-algebra
asked Feb 1 at 22:23
aaaaaa
133
$endgroup$
add a comment |
$begingroup$
Let $V$ be a 3-dimensional subspace in $mathbb{R}^4$. Find such $V_1, V_2, V_3, V_4$ that $$V_1 cap V_2 cap V_3 cap V_4 = {0,0,0,0}$$.
I thought about such $V_i$ that
$V_1 = left{ begin{bmatrix}
1\
0\
0\
0\
end{bmatrix},
begin{bmatrix}
0\
1\
0\
0\
end{bmatrix},
begin{bmatrix}
0\
0\
1\
0\
end{bmatrix} right}
$
And $V_2$ is the same, but the bottom line changes to 1, in $V_3$ it changes to 2 and in $V_4$ to 3. Am I thinking right?
linear-algebra
asked Feb 1 at 22:23
aaaaaa
133
$endgroup$
Let $V$ be a 3-dimensional subspace in $mathbb{R}^4$. Find such $V_1, V_2, V_3, V_4$ that $$V_1 cap V_2 cap V_3 cap V_4 = {0,0,0,0}$$.
I thought about such $V_i$ that
$V_1 = left{ begin{bmatrix}
1\
0\
0\
0\
end{bmatrix},
begin{bmatrix}
0\
1\
0\
0\
end{bmatrix},
begin{bmatrix}
0\
0\
1\
0\
end{bmatrix} right}
$
And $V_2$ is the same, but the bottom line changes to 1, in $V_3$ it changes to 2 and in $V_4$ to 3. Am I thinking right?
linear-algebra
linear-algebra
asked Feb 1 at 22:23
aaaaaa
133
asked Feb 1 at 22:23
aaaaaa
133
asked Feb 1 at 22:23
aaaaaa
133
asked Feb 1 at 22:23
aaaaaa
133
asked Feb 1 at 22:23
aaaaaa
133
133
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are correct assuming I'm interpreting what you've said in the right way. Essentially you're letting $e_i$ be the $i$-th standard unit basis vector, i.e.
$$e_1 = begin{pmatrix}1&0&0&0 end{pmatrix}, e_2 = begin{pmatrix}0&1&0&0 end{pmatrix} ,..., e_4 = begin{pmatrix} 0&0&0&1end{pmatrix}$$
and then letting
$$V_1 = text{span} {e_1, e_2, e_3 }, V_2 = text{span}{e_2,e_3,e_4}, V_3 = text{span}{e_1,e_3,e_4}, V_4= {e_1,e_2,e_4} $$
As such, for $v_1$ to be in $V_1$, its fourth coordinate must be $0$. Further, for $v_1$ to be in $V_1 cap V_2$, both it's fourth and first coordinate must be zero. This goes on until we see that for $v_1$ to be in $cap_{i=1}^4 V_i$ that all coordinates of $v_1$ must be $0$.
answered Feb 1 at 22:35
Theo C.Theo C.
44228
$endgroup$
$begingroup$
Thank you very much! Yes, what I've said doesn't seem very clear as I read it now. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}.
$endgroup$
– aaa
Feb 1 at 22:46
1
$begingroup$
@aaa That example you gave doesn't work: for example, $V_3 = { x in mathbb{R}^4 mid x_4 = 2(x_1 + x_2 + x_3) }$, and then by combining similar descriptions for the others, $(1, 1, -2, 0)$ is in the intersection. Overall, the intersection would be ${ x in mathbb{R}^4 mid x_1 + x_2 + x_3 = x_4 = 0 }$.
$endgroup$
– Daniel Schepler
Feb 1 at 23:10
add a comment |
$begingroup$
You're working with vector spaces?
If so, then the idea will work. Take a nice simple basis with four elements: ${e_1, e_2, e_3, e_4}$. Let $V_n$ be the subspace generated by this basis with $e_n$ removed.
answered Feb 1 at 22:38
badjohnbadjohn
4,4171720
$endgroup$
$begingroup$
Hello. Yes, these are vector spaces. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}. And thank you for your answer!
$endgroup$
– aaa
Feb 1 at 22:52
$begingroup$
I got your idea. I just thought that my slight variation was a little neater.
$endgroup$
– badjohn
Feb 1 at 22:56
$begingroup$
Yes, your is very elegant. Thank you.
$endgroup$
– aaa
Feb 1 at 22:57
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%2f3096798%2f3d-spaces-in-mathbbr4%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
$begingroup$
You are correct assuming I'm interpreting what you've said in the right way. Essentially you're letting $e_i$ be the $i$-th standard unit basis vector, i.e.
$$e_1 = begin{pmatrix}1&0&0&0 end{pmatrix}, e_2 = begin{pmatrix}0&1&0&0 end{pmatrix} ,..., e_4 = begin{pmatrix} 0&0&0&1end{pmatrix}$$
and then letting
$$V_1 = text{span} {e_1, e_2, e_3 }, V_2 = text{span}{e_2,e_3,e_4}, V_3 = text{span}{e_1,e_3,e_4}, V_4= {e_1,e_2,e_4} $$
As such, for $v_1$ to be in $V_1$, its fourth coordinate must be $0$. Further, for $v_1$ to be in $V_1 cap V_2$, both it's fourth and first coordinate must be zero. This goes on until we see that for $v_1$ to be in $cap_{i=1}^4 V_i$ that all coordinates of $v_1$ must be $0$.
answered Feb 1 at 22:35
Theo C.Theo C.
44228
$endgroup$
$begingroup$
Thank you very much! Yes, what I've said doesn't seem very clear as I read it now. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}.
$endgroup$
– aaa
Feb 1 at 22:46
1
$begingroup$
@aaa That example you gave doesn't work: for example, $V_3 = { x in mathbb{R}^4 mid x_4 = 2(x_1 + x_2 + x_3) }$, and then by combining similar descriptions for the others, $(1, 1, -2, 0)$ is in the intersection. Overall, the intersection would be ${ x in mathbb{R}^4 mid x_1 + x_2 + x_3 = x_4 = 0 }$.
$endgroup$
– Daniel Schepler
Feb 1 at 23:10
add a comment |
$begingroup$
You are correct assuming I'm interpreting what you've said in the right way. Essentially you're letting $e_i$ be the $i$-th standard unit basis vector, i.e.
$$e_1 = begin{pmatrix}1&0&0&0 end{pmatrix}, e_2 = begin{pmatrix}0&1&0&0 end{pmatrix} ,..., e_4 = begin{pmatrix} 0&0&0&1end{pmatrix}$$
and then letting
$$V_1 = text{span} {e_1, e_2, e_3 }, V_2 = text{span}{e_2,e_3,e_4}, V_3 = text{span}{e_1,e_3,e_4}, V_4= {e_1,e_2,e_4} $$
As such, for $v_1$ to be in $V_1$, its fourth coordinate must be $0$. Further, for $v_1$ to be in $V_1 cap V_2$, both it's fourth and first coordinate must be zero. This goes on until we see that for $v_1$ to be in $cap_{i=1}^4 V_i$ that all coordinates of $v_1$ must be $0$.
answered Feb 1 at 22:35
Theo C.Theo C.
44228
$endgroup$
$begingroup$
Thank you very much! Yes, what I've said doesn't seem very clear as I read it now. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}.
$endgroup$
– aaa
Feb 1 at 22:46
1
$begingroup$
@aaa That example you gave doesn't work: for example, $V_3 = { x in mathbb{R}^4 mid x_4 = 2(x_1 + x_2 + x_3) }$, and then by combining similar descriptions for the others, $(1, 1, -2, 0)$ is in the intersection. Overall, the intersection would be ${ x in mathbb{R}^4 mid x_1 + x_2 + x_3 = x_4 = 0 }$.
$endgroup$
– Daniel Schepler
Feb 1 at 23:10
add a comment |
$begingroup$
You are correct assuming I'm interpreting what you've said in the right way. Essentially you're letting $e_i$ be the $i$-th standard unit basis vector, i.e.
$$e_1 = begin{pmatrix}1&0&0&0 end{pmatrix}, e_2 = begin{pmatrix}0&1&0&0 end{pmatrix} ,..., e_4 = begin{pmatrix} 0&0&0&1end{pmatrix}$$
and then letting
$$V_1 = text{span} {e_1, e_2, e_3 }, V_2 = text{span}{e_2,e_3,e_4}, V_3 = text{span}{e_1,e_3,e_4}, V_4= {e_1,e_2,e_4} $$
As such, for $v_1$ to be in $V_1$, its fourth coordinate must be $0$. Further, for $v_1$ to be in $V_1 cap V_2$, both it's fourth and first coordinate must be zero. This goes on until we see that for $v_1$ to be in $cap_{i=1}^4 V_i$ that all coordinates of $v_1$ must be $0$.
answered Feb 1 at 22:35
Theo C.Theo C.
44228
$endgroup$
You are correct assuming I'm interpreting what you've said in the right way. Essentially you're letting $e_i$ be the $i$-th standard unit basis vector, i.e.
$$e_1 = begin{pmatrix}1&0&0&0 end{pmatrix}, e_2 = begin{pmatrix}0&1&0&0 end{pmatrix} ,..., e_4 = begin{pmatrix} 0&0&0&1end{pmatrix}$$
and then letting
$$V_1 = text{span} {e_1, e_2, e_3 }, V_2 = text{span}{e_2,e_3,e_4}, V_3 = text{span}{e_1,e_3,e_4}, V_4= {e_1,e_2,e_4} $$
As such, for $v_1$ to be in $V_1$, its fourth coordinate must be $0$. Further, for $v_1$ to be in $V_1 cap V_2$, both it's fourth and first coordinate must be zero. This goes on until we see that for $v_1$ to be in $cap_{i=1}^4 V_i$ that all coordinates of $v_1$ must be $0$.
answered Feb 1 at 22:35
Theo C.Theo C.
44228
answered Feb 1 at 22:35
Theo C.Theo C.
44228
answered Feb 1 at 22:35
Theo C.Theo C.
44228
answered Feb 1 at 22:35
Theo C.Theo C.
44228
44228
$begingroup$
Thank you very much! Yes, what I've said doesn't seem very clear as I read it now. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}.
$endgroup$
– aaa
Feb 1 at 22:46
1
$begingroup$
@aaa That example you gave doesn't work: for example, $V_3 = { x in mathbb{R}^4 mid x_4 = 2(x_1 + x_2 + x_3) }$, and then by combining similar descriptions for the others, $(1, 1, -2, 0)$ is in the intersection. Overall, the intersection would be ${ x in mathbb{R}^4 mid x_1 + x_2 + x_3 = x_4 = 0 }$.
$endgroup$
– Daniel Schepler
Feb 1 at 23:10
add a comment |
$begingroup$
Thank you very much! Yes, what I've said doesn't seem very clear as I read it now. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}.
$endgroup$
– aaa
Feb 1 at 22:46
1
$begingroup$
@aaa That example you gave doesn't work: for example, $V_3 = { x in mathbb{R}^4 mid x_4 = 2(x_1 + x_2 + x_3) }$, and then by combining similar descriptions for the others, $(1, 1, -2, 0)$ is in the intersection. Overall, the intersection would be ${ x in mathbb{R}^4 mid x_1 + x_2 + x_3 = x_4 = 0 }$.
$endgroup$
– Daniel Schepler
Feb 1 at 23:10
$begingroup$
Thank you very much! Yes, what I've said doesn't seem very clear as I read it now. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}.
$endgroup$
– aaa
Feb 1 at 22:46
$begingroup$
Thank you very much! Yes, what I've said doesn't seem very clear as I read it now. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}.
$endgroup$
– aaa
Feb 1 at 22:46
1
1
$begingroup$
@aaa That example you gave doesn't work: for example, $V_3 = { x in mathbb{R}^4 mid x_4 = 2(x_1 + x_2 + x_3) }$, and then by combining similar descriptions for the others, $(1, 1, -2, 0)$ is in the intersection. Overall, the intersection would be ${ x in mathbb{R}^4 mid x_1 + x_2 + x_3 = x_4 = 0 }$.
$endgroup$
– Daniel Schepler
Feb 1 at 23:10
$begingroup$
@aaa That example you gave doesn't work: for example, $V_3 = { x in mathbb{R}^4 mid x_4 = 2(x_1 + x_2 + x_3) }$, and then by combining similar descriptions for the others, $(1, 1, -2, 0)$ is in the intersection. Overall, the intersection would be ${ x in mathbb{R}^4 mid x_1 + x_2 + x_3 = x_4 = 0 }$.
$endgroup$
– Daniel Schepler
Feb 1 at 23:10
add a comment |
$begingroup$
You're working with vector spaces?
If so, then the idea will work. Take a nice simple basis with four elements: ${e_1, e_2, e_3, e_4}$. Let $V_n$ be the subspace generated by this basis with $e_n$ removed.
answered Feb 1 at 22:38
badjohnbadjohn
4,4171720
$endgroup$
$begingroup$
Hello. Yes, these are vector spaces. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}. And thank you for your answer!
$endgroup$
– aaa
Feb 1 at 22:52
$begingroup$
I got your idea. I just thought that my slight variation was a little neater.
$endgroup$
– badjohn
Feb 1 at 22:56
$begingroup$
Yes, your is very elegant. Thank you.
$endgroup$
– aaa
Feb 1 at 22:57
add a comment |
$begingroup$
You're working with vector spaces?
If so, then the idea will work. Take a nice simple basis with four elements: ${e_1, e_2, e_3, e_4}$. Let $V_n$ be the subspace generated by this basis with $e_n$ removed.
answered Feb 1 at 22:38
badjohnbadjohn
4,4171720
$endgroup$
$begingroup$
Hello. Yes, these are vector spaces. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}. And thank you for your answer!
$endgroup$
– aaa
Feb 1 at 22:52
$begingroup$
I got your idea. I just thought that my slight variation was a little neater.
$endgroup$
– badjohn
Feb 1 at 22:56
$begingroup$
Yes, your is very elegant. Thank you.
$endgroup$
– aaa
Feb 1 at 22:57
add a comment |
$begingroup$
You're working with vector spaces?
If so, then the idea will work. Take a nice simple basis with four elements: ${e_1, e_2, e_3, e_4}$. Let $V_n$ be the subspace generated by this basis with $e_n$ removed.
answered Feb 1 at 22:38
badjohnbadjohn
4,4171720
$endgroup$
You're working with vector spaces?
If so, then the idea will work. Take a nice simple basis with four elements: ${e_1, e_2, e_3, e_4}$. Let $V_n$ be the subspace generated by this basis with $e_n$ removed.
answered Feb 1 at 22:38
badjohnbadjohn
4,4171720
answered Feb 1 at 22:38
badjohnbadjohn
4,4171720
answered Feb 1 at 22:38
badjohnbadjohn
4,4171720
answered Feb 1 at 22:38
badjohnbadjohn
4,4171720
4,4171720
$begingroup$
Hello. Yes, these are vector spaces. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}. And thank you for your answer!
$endgroup$
– aaa
Feb 1 at 22:52
$begingroup$
I got your idea. I just thought that my slight variation was a little neater.
$endgroup$
– badjohn
Feb 1 at 22:56
$begingroup$
Yes, your is very elegant. Thank you.
$endgroup$
– aaa
Feb 1 at 22:57
add a comment |
$begingroup$
Hello. Yes, these are vector spaces. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}. And thank you for your answer!
$endgroup$
– aaa
Feb 1 at 22:52
$begingroup$
I got your idea. I just thought that my slight variation was a little neater.
$endgroup$
– badjohn
Feb 1 at 22:56
$begingroup$
Yes, your is very elegant. Thank you.
$endgroup$
– aaa
Feb 1 at 22:57
$begingroup$
Hello. Yes, these are vector spaces. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}. And thank you for your answer!
$endgroup$
– aaa
Feb 1 at 22:52
$begingroup$
Hello. Yes, these are vector spaces. Actually I meant something like that: V1 = {(1,0,0,0), (0,1,0,0), (0,0,1,0)}, V2 = {(1,0,0,1), (0,1,0,1), (0,0,1,1)}, V3 = {(1,0,0,2), (0,1,0,2), (0,0,1,2)}, V4 = {(1,0,0,3), (0,1,0,3), (0,0,1,3)}. And thank you for your answer!
$endgroup$
– aaa
Feb 1 at 22:52
$begingroup$
I got your idea. I just thought that my slight variation was a little neater.
$endgroup$
– badjohn
Feb 1 at 22:56
$begingroup$
I got your idea. I just thought that my slight variation was a little neater.
$endgroup$
– badjohn
Feb 1 at 22:56
$begingroup$
Yes, your is very elegant. Thank you.
$endgroup$
– aaa
Feb 1 at 22:57
$begingroup$
Yes, your is very elegant. Thank you.
$endgroup$
– aaa
Feb 1 at 22:57
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%2f3096798%2f3d-spaces-in-mathbbr4%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
