How do you center multiple equations that have multiple steps?












2















I am writing a paper that deals with Gaussian integers and I have a section where I am talking about the norm of the units in the Gaussian integers. Since there are only four units in the Gaussian integers, I'd like to have the calculations of their norms appear neatly in four columns that are centered. Basically so that you can easily look down each of the four columns and see the calculations for each unit. What I have right now sort of does that but its very sloppy and I'm sure there is a much cleaner way to do it. I'm still new to Latex so I realize this code is probably painfully inefficient but I would appreciate any suggestions of how to fix it and what packages I'd need to use to implement those changes.



begin{center}  
$N(1)=N(1+0i)$
hspace{.025cm}
$N(-1)=N(-1+0i)$
hspace{.025cm}
$N(i)=N(0+1i)$
hspace{.025cm}
$N(-i)=N(0-1i)$
end{center}

begin{center}
$N(1)=1^2+0$
hspace{.1cm}
$N(-1)=(-1)^2+0^2$
hspace{.1cm}
$N(i)=0^2+1^2$
hspace{.1cm}
$N(-i)=0^2+(-1)^2$
end{center}

begin{center}
$N(1)=1$
hspace{.5cm}
$N(-1)=1$
hspace{.5cm}
$N(i)=1$
hspace{.5cm}
$N(-i)=1$
end{center}









share|improve this question

























  • Maybe use an array environment?

    – leandriis
    1 hour ago













  • The only problem is that I don't know anything about arrays in Latex. I can look them up and see what I can find but are there any other suggestions that come to mind?

    – Alex Adinolfi
    1 hour ago
















2















I am writing a paper that deals with Gaussian integers and I have a section where I am talking about the norm of the units in the Gaussian integers. Since there are only four units in the Gaussian integers, I'd like to have the calculations of their norms appear neatly in four columns that are centered. Basically so that you can easily look down each of the four columns and see the calculations for each unit. What I have right now sort of does that but its very sloppy and I'm sure there is a much cleaner way to do it. I'm still new to Latex so I realize this code is probably painfully inefficient but I would appreciate any suggestions of how to fix it and what packages I'd need to use to implement those changes.



begin{center}  
$N(1)=N(1+0i)$
hspace{.025cm}
$N(-1)=N(-1+0i)$
hspace{.025cm}
$N(i)=N(0+1i)$
hspace{.025cm}
$N(-i)=N(0-1i)$
end{center}

begin{center}
$N(1)=1^2+0$
hspace{.1cm}
$N(-1)=(-1)^2+0^2$
hspace{.1cm}
$N(i)=0^2+1^2$
hspace{.1cm}
$N(-i)=0^2+(-1)^2$
end{center}

begin{center}
$N(1)=1$
hspace{.5cm}
$N(-1)=1$
hspace{.5cm}
$N(i)=1$
hspace{.5cm}
$N(-i)=1$
end{center}









share|improve this question

























  • Maybe use an array environment?

    – leandriis
    1 hour ago













  • The only problem is that I don't know anything about arrays in Latex. I can look them up and see what I can find but are there any other suggestions that come to mind?

    – Alex Adinolfi
    1 hour ago














2












2








2








I am writing a paper that deals with Gaussian integers and I have a section where I am talking about the norm of the units in the Gaussian integers. Since there are only four units in the Gaussian integers, I'd like to have the calculations of their norms appear neatly in four columns that are centered. Basically so that you can easily look down each of the four columns and see the calculations for each unit. What I have right now sort of does that but its very sloppy and I'm sure there is a much cleaner way to do it. I'm still new to Latex so I realize this code is probably painfully inefficient but I would appreciate any suggestions of how to fix it and what packages I'd need to use to implement those changes.



begin{center}  
$N(1)=N(1+0i)$
hspace{.025cm}
$N(-1)=N(-1+0i)$
hspace{.025cm}
$N(i)=N(0+1i)$
hspace{.025cm}
$N(-i)=N(0-1i)$
end{center}

begin{center}
$N(1)=1^2+0$
hspace{.1cm}
$N(-1)=(-1)^2+0^2$
hspace{.1cm}
$N(i)=0^2+1^2$
hspace{.1cm}
$N(-i)=0^2+(-1)^2$
end{center}

begin{center}
$N(1)=1$
hspace{.5cm}
$N(-1)=1$
hspace{.5cm}
$N(i)=1$
hspace{.5cm}
$N(-i)=1$
end{center}









share|improve this question
















I am writing a paper that deals with Gaussian integers and I have a section where I am talking about the norm of the units in the Gaussian integers. Since there are only four units in the Gaussian integers, I'd like to have the calculations of their norms appear neatly in four columns that are centered. Basically so that you can easily look down each of the four columns and see the calculations for each unit. What I have right now sort of does that but its very sloppy and I'm sure there is a much cleaner way to do it. I'm still new to Latex so I realize this code is probably painfully inefficient but I would appreciate any suggestions of how to fix it and what packages I'd need to use to implement those changes.



begin{center}  
$N(1)=N(1+0i)$
hspace{.025cm}
$N(-1)=N(-1+0i)$
hspace{.025cm}
$N(i)=N(0+1i)$
hspace{.025cm}
$N(-i)=N(0-1i)$
end{center}

begin{center}
$N(1)=1^2+0$
hspace{.1cm}
$N(-1)=(-1)^2+0^2$
hspace{.1cm}
$N(i)=0^2+1^2$
hspace{.1cm}
$N(-i)=0^2+(-1)^2$
end{center}

begin{center}
$N(1)=1$
hspace{.5cm}
$N(-1)=1$
hspace{.5cm}
$N(i)=1$
hspace{.5cm}
$N(-i)=1$
end{center}






horizontal-alignment equations vertical-alignment






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









Mico

289k32395783




289k32395783










asked 1 hour ago









Alex AdinolfiAlex Adinolfi

484




484













  • Maybe use an array environment?

    – leandriis
    1 hour ago













  • The only problem is that I don't know anything about arrays in Latex. I can look them up and see what I can find but are there any other suggestions that come to mind?

    – Alex Adinolfi
    1 hour ago



















  • Maybe use an array environment?

    – leandriis
    1 hour ago













  • The only problem is that I don't know anything about arrays in Latex. I can look them up and see what I can find but are there any other suggestions that come to mind?

    – Alex Adinolfi
    1 hour ago

















Maybe use an array environment?

– leandriis
1 hour ago







Maybe use an array environment?

– leandriis
1 hour ago















The only problem is that I don't know anything about arrays in Latex. I can look them up and see what I can find but are there any other suggestions that come to mind?

– Alex Adinolfi
1 hour ago





The only problem is that I don't know anything about arrays in Latex. I can look them up and see what I can find but are there any other suggestions that come to mind?

– Alex Adinolfi
1 hour ago










3 Answers
3






active

oldest

votes


















3














Here are three different suggestions using either array or alignedat:



enter image description here



documentclass{article}
usepackage{geometry}
usepackage{amsmath}

begin{document}

[
begin{array}{llll}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]

[
begin{array}{cccc}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]


begin{alignat*}{4}
N(1)&=N(1+0i) &quad N(-1)&=N(-1+0i) &quad N(i)&=N(0+1i) &quad N(-i)&=N(0-1i) \
N(1)&=1^2+0 & N(-1)&=(-1)^2+0^2 & N(i)&=0^2+1^2 & N(-i)&=0^2+(-1)^2 \
N(1)&=1 & N(-1)&=1 & N(i)&=1 & N(-i)&=1
end{alignat*}

end{document}





share|improve this answer
























  • Thank you very much! This does exactly what I was looking to do.

    – Alex Adinolfi
    1 hour ago






  • 1





    @AlexAdinolfi: Glad I helped you. If you like my answer and it was helpful, please consider upvoting (by clicking on the arrows next to the score) and/or marking it as the accepted answer (by clicking on the checkmark ✓). This also applies to all your other questions to which you already recieved answers.

    – leandriis
    58 mins ago





















1














For example I would have used the tables in sequence (but it's just a personal taste) to give a touch of vitality.
In this case I have used the booktabs package to have toprule and bottomrule.



enter image description here



documentclass[a4paper,12pt]{article}
usepackage{array}
usepackage{booktabs}

begin{document}
[
setlengtharraycolsep{0pt}
renewcommandarraystretch{1.25}
begin{array}{r @{{}={}} l}
toprule
N(1) &N(1+0i) \
N(-1)&N(-1+0i) \
N(i) &N(0+1i) \
N(-i)&N(1+0i) \
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1^2+0 \
N(-1)&(-1)^2+0^2\
N(i) &0^2+1^2 \
N(-i)&0^2+(-1)^2\
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1 \
N(-1)&1 \
N(i) &1 \
N(-i)&1 \
bottomrule
end{array}
]
end{document}





share|improve this answer


























  • @Mico Correct observation. Obviously I just wanted to give another interpretation based simply on aesthetics. If you don't need it, I'll remove it.

    – Sebastiano
    43 mins ago



















1














I propose this layout:



documentclass{article}
usepackage[utf8]{inputenc}%
usepackage{geometry}
usepackage[table, svgnames]{xcolor}
usepackage{mathtools}
colorlet{shadecolor}{Gainsboro!50! Lavender}
newcommand*shadebox[1]{%
colorbox{shadecolor}{hspace{1em}$displaystyle #1 $hspace{1em}}}
begin{document}


begin{equation}
begin{gathered}
begin{aligned}
N(1) & =N(1+0i)\
& =1^2+0^2
end{aligned} \[1.5ex]
shadebox{N(1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-1) & =N(-1+0i) \
& =(-1)^2+0^2
end{aligned} \[1ex]
shadebox{N(-1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(i) & =N(0+1i) \
& =0^2+1^2
end{aligned} \[1ex]
shadebox{ N(i) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-i) & =N(0-1i) \
& =0^2+(-1)^2
end{aligned} \[1ex]
shadebox{N(-i) = 1}
end{gathered}
end{equation}

end{document}


enter image description here






share|improve this answer



















  • 1





    Very nice. The parfum of the lavander is fantastic and also the color :-)

    – Sebastiano
    14 mins ago






  • 1





    I like it because I feel it adds some freshness to the Gainsboro grey (b.t.w., is there a Leonardo grey? ;o)

    – Bernard
    8 mins ago












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "85"
};
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f488304%2fhow-do-you-center-multiple-equations-that-have-multiple-steps%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














Here are three different suggestions using either array or alignedat:



enter image description here



documentclass{article}
usepackage{geometry}
usepackage{amsmath}

begin{document}

[
begin{array}{llll}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]

[
begin{array}{cccc}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]


begin{alignat*}{4}
N(1)&=N(1+0i) &quad N(-1)&=N(-1+0i) &quad N(i)&=N(0+1i) &quad N(-i)&=N(0-1i) \
N(1)&=1^2+0 & N(-1)&=(-1)^2+0^2 & N(i)&=0^2+1^2 & N(-i)&=0^2+(-1)^2 \
N(1)&=1 & N(-1)&=1 & N(i)&=1 & N(-i)&=1
end{alignat*}

end{document}





share|improve this answer
























  • Thank you very much! This does exactly what I was looking to do.

    – Alex Adinolfi
    1 hour ago






  • 1





    @AlexAdinolfi: Glad I helped you. If you like my answer and it was helpful, please consider upvoting (by clicking on the arrows next to the score) and/or marking it as the accepted answer (by clicking on the checkmark ✓). This also applies to all your other questions to which you already recieved answers.

    – leandriis
    58 mins ago


















3














Here are three different suggestions using either array or alignedat:



enter image description here



documentclass{article}
usepackage{geometry}
usepackage{amsmath}

begin{document}

[
begin{array}{llll}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]

[
begin{array}{cccc}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]


begin{alignat*}{4}
N(1)&=N(1+0i) &quad N(-1)&=N(-1+0i) &quad N(i)&=N(0+1i) &quad N(-i)&=N(0-1i) \
N(1)&=1^2+0 & N(-1)&=(-1)^2+0^2 & N(i)&=0^2+1^2 & N(-i)&=0^2+(-1)^2 \
N(1)&=1 & N(-1)&=1 & N(i)&=1 & N(-i)&=1
end{alignat*}

end{document}





share|improve this answer
























  • Thank you very much! This does exactly what I was looking to do.

    – Alex Adinolfi
    1 hour ago






  • 1





    @AlexAdinolfi: Glad I helped you. If you like my answer and it was helpful, please consider upvoting (by clicking on the arrows next to the score) and/or marking it as the accepted answer (by clicking on the checkmark ✓). This also applies to all your other questions to which you already recieved answers.

    – leandriis
    58 mins ago
















3












3








3







Here are three different suggestions using either array or alignedat:



enter image description here



documentclass{article}
usepackage{geometry}
usepackage{amsmath}

begin{document}

[
begin{array}{llll}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]

[
begin{array}{cccc}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]


begin{alignat*}{4}
N(1)&=N(1+0i) &quad N(-1)&=N(-1+0i) &quad N(i)&=N(0+1i) &quad N(-i)&=N(0-1i) \
N(1)&=1^2+0 & N(-1)&=(-1)^2+0^2 & N(i)&=0^2+1^2 & N(-i)&=0^2+(-1)^2 \
N(1)&=1 & N(-1)&=1 & N(i)&=1 & N(-i)&=1
end{alignat*}

end{document}





share|improve this answer













Here are three different suggestions using either array or alignedat:



enter image description here



documentclass{article}
usepackage{geometry}
usepackage{amsmath}

begin{document}

[
begin{array}{llll}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]

[
begin{array}{cccc}
N(1)=N(1+0i) & N(-1)=N(-1+0i) & N(i)=N(0+1i) & N(-i)=N(0-1i) \
N(1)=1^2+0 & N(-1)=(-1)^2+0^2 & N(i)=0^2+1^2 & N(-i)=0^2+(-1)^2 \
N(1)=1 & N(-1)=1 & N(i)=1 & N(-i)=1
end{array}
]


begin{alignat*}{4}
N(1)&=N(1+0i) &quad N(-1)&=N(-1+0i) &quad N(i)&=N(0+1i) &quad N(-i)&=N(0-1i) \
N(1)&=1^2+0 & N(-1)&=(-1)^2+0^2 & N(i)&=0^2+1^2 & N(-i)&=0^2+(-1)^2 \
N(1)&=1 & N(-1)&=1 & N(i)&=1 & N(-i)&=1
end{alignat*}

end{document}






share|improve this answer












share|improve this answer



share|improve this answer










answered 1 hour ago









leandriisleandriis

12.3k1833




12.3k1833













  • Thank you very much! This does exactly what I was looking to do.

    – Alex Adinolfi
    1 hour ago






  • 1





    @AlexAdinolfi: Glad I helped you. If you like my answer and it was helpful, please consider upvoting (by clicking on the arrows next to the score) and/or marking it as the accepted answer (by clicking on the checkmark ✓). This also applies to all your other questions to which you already recieved answers.

    – leandriis
    58 mins ago





















  • Thank you very much! This does exactly what I was looking to do.

    – Alex Adinolfi
    1 hour ago






  • 1





    @AlexAdinolfi: Glad I helped you. If you like my answer and it was helpful, please consider upvoting (by clicking on the arrows next to the score) and/or marking it as the accepted answer (by clicking on the checkmark ✓). This also applies to all your other questions to which you already recieved answers.

    – leandriis
    58 mins ago



















Thank you very much! This does exactly what I was looking to do.

– Alex Adinolfi
1 hour ago





Thank you very much! This does exactly what I was looking to do.

– Alex Adinolfi
1 hour ago




1




1





@AlexAdinolfi: Glad I helped you. If you like my answer and it was helpful, please consider upvoting (by clicking on the arrows next to the score) and/or marking it as the accepted answer (by clicking on the checkmark ✓). This also applies to all your other questions to which you already recieved answers.

– leandriis
58 mins ago







@AlexAdinolfi: Glad I helped you. If you like my answer and it was helpful, please consider upvoting (by clicking on the arrows next to the score) and/or marking it as the accepted answer (by clicking on the checkmark ✓). This also applies to all your other questions to which you already recieved answers.

– leandriis
58 mins ago













1














For example I would have used the tables in sequence (but it's just a personal taste) to give a touch of vitality.
In this case I have used the booktabs package to have toprule and bottomrule.



enter image description here



documentclass[a4paper,12pt]{article}
usepackage{array}
usepackage{booktabs}

begin{document}
[
setlengtharraycolsep{0pt}
renewcommandarraystretch{1.25}
begin{array}{r @{{}={}} l}
toprule
N(1) &N(1+0i) \
N(-1)&N(-1+0i) \
N(i) &N(0+1i) \
N(-i)&N(1+0i) \
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1^2+0 \
N(-1)&(-1)^2+0^2\
N(i) &0^2+1^2 \
N(-i)&0^2+(-1)^2\
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1 \
N(-1)&1 \
N(i) &1 \
N(-i)&1 \
bottomrule
end{array}
]
end{document}





share|improve this answer


























  • @Mico Correct observation. Obviously I just wanted to give another interpretation based simply on aesthetics. If you don't need it, I'll remove it.

    – Sebastiano
    43 mins ago
















1














For example I would have used the tables in sequence (but it's just a personal taste) to give a touch of vitality.
In this case I have used the booktabs package to have toprule and bottomrule.



enter image description here



documentclass[a4paper,12pt]{article}
usepackage{array}
usepackage{booktabs}

begin{document}
[
setlengtharraycolsep{0pt}
renewcommandarraystretch{1.25}
begin{array}{r @{{}={}} l}
toprule
N(1) &N(1+0i) \
N(-1)&N(-1+0i) \
N(i) &N(0+1i) \
N(-i)&N(1+0i) \
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1^2+0 \
N(-1)&(-1)^2+0^2\
N(i) &0^2+1^2 \
N(-i)&0^2+(-1)^2\
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1 \
N(-1)&1 \
N(i) &1 \
N(-i)&1 \
bottomrule
end{array}
]
end{document}





share|improve this answer


























  • @Mico Correct observation. Obviously I just wanted to give another interpretation based simply on aesthetics. If you don't need it, I'll remove it.

    – Sebastiano
    43 mins ago














1












1








1







For example I would have used the tables in sequence (but it's just a personal taste) to give a touch of vitality.
In this case I have used the booktabs package to have toprule and bottomrule.



enter image description here



documentclass[a4paper,12pt]{article}
usepackage{array}
usepackage{booktabs}

begin{document}
[
setlengtharraycolsep{0pt}
renewcommandarraystretch{1.25}
begin{array}{r @{{}={}} l}
toprule
N(1) &N(1+0i) \
N(-1)&N(-1+0i) \
N(i) &N(0+1i) \
N(-i)&N(1+0i) \
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1^2+0 \
N(-1)&(-1)^2+0^2\
N(i) &0^2+1^2 \
N(-i)&0^2+(-1)^2\
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1 \
N(-1)&1 \
N(i) &1 \
N(-i)&1 \
bottomrule
end{array}
]
end{document}





share|improve this answer















For example I would have used the tables in sequence (but it's just a personal taste) to give a touch of vitality.
In this case I have used the booktabs package to have toprule and bottomrule.



enter image description here



documentclass[a4paper,12pt]{article}
usepackage{array}
usepackage{booktabs}

begin{document}
[
setlengtharraycolsep{0pt}
renewcommandarraystretch{1.25}
begin{array}{r @{{}={}} l}
toprule
N(1) &N(1+0i) \
N(-1)&N(-1+0i) \
N(i) &N(0+1i) \
N(-i)&N(1+0i) \
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1^2+0 \
N(-1)&(-1)^2+0^2\
N(i) &0^2+1^2 \
N(-i)&0^2+(-1)^2\
bottomrule
end{array}
quadRightarrowquad
begin{array}{r @{{}={}} l}
toprule
N(1) &1 \
N(-1)&1 \
N(i) &1 \
N(-i)&1 \
bottomrule
end{array}
]
end{document}






share|improve this answer














share|improve this answer



share|improve this answer








edited 21 mins ago









Mico

289k32395783




289k32395783










answered 52 mins ago









SebastianoSebastiano

11.8k42467




11.8k42467













  • @Mico Correct observation. Obviously I just wanted to give another interpretation based simply on aesthetics. If you don't need it, I'll remove it.

    – Sebastiano
    43 mins ago



















  • @Mico Correct observation. Obviously I just wanted to give another interpretation based simply on aesthetics. If you don't need it, I'll remove it.

    – Sebastiano
    43 mins ago

















@Mico Correct observation. Obviously I just wanted to give another interpretation based simply on aesthetics. If you don't need it, I'll remove it.

– Sebastiano
43 mins ago





@Mico Correct observation. Obviously I just wanted to give another interpretation based simply on aesthetics. If you don't need it, I'll remove it.

– Sebastiano
43 mins ago











1














I propose this layout:



documentclass{article}
usepackage[utf8]{inputenc}%
usepackage{geometry}
usepackage[table, svgnames]{xcolor}
usepackage{mathtools}
colorlet{shadecolor}{Gainsboro!50! Lavender}
newcommand*shadebox[1]{%
colorbox{shadecolor}{hspace{1em}$displaystyle #1 $hspace{1em}}}
begin{document}


begin{equation}
begin{gathered}
begin{aligned}
N(1) & =N(1+0i)\
& =1^2+0^2
end{aligned} \[1.5ex]
shadebox{N(1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-1) & =N(-1+0i) \
& =(-1)^2+0^2
end{aligned} \[1ex]
shadebox{N(-1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(i) & =N(0+1i) \
& =0^2+1^2
end{aligned} \[1ex]
shadebox{ N(i) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-i) & =N(0-1i) \
& =0^2+(-1)^2
end{aligned} \[1ex]
shadebox{N(-i) = 1}
end{gathered}
end{equation}

end{document}


enter image description here






share|improve this answer



















  • 1





    Very nice. The parfum of the lavander is fantastic and also the color :-)

    – Sebastiano
    14 mins ago






  • 1





    I like it because I feel it adds some freshness to the Gainsboro grey (b.t.w., is there a Leonardo grey? ;o)

    – Bernard
    8 mins ago
















1














I propose this layout:



documentclass{article}
usepackage[utf8]{inputenc}%
usepackage{geometry}
usepackage[table, svgnames]{xcolor}
usepackage{mathtools}
colorlet{shadecolor}{Gainsboro!50! Lavender}
newcommand*shadebox[1]{%
colorbox{shadecolor}{hspace{1em}$displaystyle #1 $hspace{1em}}}
begin{document}


begin{equation}
begin{gathered}
begin{aligned}
N(1) & =N(1+0i)\
& =1^2+0^2
end{aligned} \[1.5ex]
shadebox{N(1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-1) & =N(-1+0i) \
& =(-1)^2+0^2
end{aligned} \[1ex]
shadebox{N(-1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(i) & =N(0+1i) \
& =0^2+1^2
end{aligned} \[1ex]
shadebox{ N(i) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-i) & =N(0-1i) \
& =0^2+(-1)^2
end{aligned} \[1ex]
shadebox{N(-i) = 1}
end{gathered}
end{equation}

end{document}


enter image description here






share|improve this answer



















  • 1





    Very nice. The parfum of the lavander is fantastic and also the color :-)

    – Sebastiano
    14 mins ago






  • 1





    I like it because I feel it adds some freshness to the Gainsboro grey (b.t.w., is there a Leonardo grey? ;o)

    – Bernard
    8 mins ago














1












1








1







I propose this layout:



documentclass{article}
usepackage[utf8]{inputenc}%
usepackage{geometry}
usepackage[table, svgnames]{xcolor}
usepackage{mathtools}
colorlet{shadecolor}{Gainsboro!50! Lavender}
newcommand*shadebox[1]{%
colorbox{shadecolor}{hspace{1em}$displaystyle #1 $hspace{1em}}}
begin{document}


begin{equation}
begin{gathered}
begin{aligned}
N(1) & =N(1+0i)\
& =1^2+0^2
end{aligned} \[1.5ex]
shadebox{N(1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-1) & =N(-1+0i) \
& =(-1)^2+0^2
end{aligned} \[1ex]
shadebox{N(-1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(i) & =N(0+1i) \
& =0^2+1^2
end{aligned} \[1ex]
shadebox{ N(i) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-i) & =N(0-1i) \
& =0^2+(-1)^2
end{aligned} \[1ex]
shadebox{N(-i) = 1}
end{gathered}
end{equation}

end{document}


enter image description here






share|improve this answer













I propose this layout:



documentclass{article}
usepackage[utf8]{inputenc}%
usepackage{geometry}
usepackage[table, svgnames]{xcolor}
usepackage{mathtools}
colorlet{shadecolor}{Gainsboro!50! Lavender}
newcommand*shadebox[1]{%
colorbox{shadecolor}{hspace{1em}$displaystyle #1 $hspace{1em}}}
begin{document}


begin{equation}
begin{gathered}
begin{aligned}
N(1) & =N(1+0i)\
& =1^2+0^2
end{aligned} \[1.5ex]
shadebox{N(1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-1) & =N(-1+0i) \
& =(-1)^2+0^2
end{aligned} \[1ex]
shadebox{N(-1) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(i) & =N(0+1i) \
& =0^2+1^2
end{aligned} \[1ex]
shadebox{ N(i) = 1}
end{gathered}
qquad
begin{gathered}
begin{aligned}
N(-i) & =N(0-1i) \
& =0^2+(-1)^2
end{aligned} \[1ex]
shadebox{N(-i) = 1}
end{gathered}
end{equation}

end{document}


enter image description here







share|improve this answer












share|improve this answer



share|improve this answer










answered 19 mins ago









BernardBernard

178k779211




178k779211








  • 1





    Very nice. The parfum of the lavander is fantastic and also the color :-)

    – Sebastiano
    14 mins ago






  • 1





    I like it because I feel it adds some freshness to the Gainsboro grey (b.t.w., is there a Leonardo grey? ;o)

    – Bernard
    8 mins ago














  • 1





    Very nice. The parfum of the lavander is fantastic and also the color :-)

    – Sebastiano
    14 mins ago






  • 1





    I like it because I feel it adds some freshness to the Gainsboro grey (b.t.w., is there a Leonardo grey? ;o)

    – Bernard
    8 mins ago








1




1





Very nice. The parfum of the lavander is fantastic and also the color :-)

– Sebastiano
14 mins ago





Very nice. The parfum of the lavander is fantastic and also the color :-)

– Sebastiano
14 mins ago




1




1





I like it because I feel it adds some freshness to the Gainsboro grey (b.t.w., is there a Leonardo grey? ;o)

– Bernard
8 mins ago





I like it because I feel it adds some freshness to the Gainsboro grey (b.t.w., is there a Leonardo grey? ;o)

– Bernard
8 mins ago


















draft saved

draft discarded




















































Thanks for contributing an answer to TeX - LaTeX 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f488304%2fhow-do-you-center-multiple-equations-that-have-multiple-steps%23new-answer', 'question_page');
}
);

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







Popular posts from this blog

Alcázar de San Juan

Griza ansero

Heinkel He 51