# Divide Numbers by 0

We've all been told at some point in our lives that dividing by 0 is impossible. And for the most part, that statement is true. But what if there was a way to perform the forbidden operation? Welcome to my newest creation: b-numbers.

b-numbers are a little bit like imaginary numbers: the main pronumeral involved represents an expression that isn't mathematically impossible (i represents $$\\sqrt{-1}\$$). In this case $$\b\$$ will be said to represent the expression $$\\frac{1}{0}\$$. From here, it is easy to determine what $$\\frac{x}{0}\$$ would equal:

$$\frac{x}{0} = \frac{x}{1} \cdot \frac{1}{0} = xb$$

Given an expression involving a division by 0, output the simplified value in terms of $$\b\$$. Note that input will be in the form of n/0 where n is any rational number or any b-number in decimal form. Leading 0s and trailing 0s wont be included.

## Example Input

4/0
1/0
0/0
80/0
-8/0
1.5/0
2.03/0
-1/0
-3.14/0
b/0
3b/0
-b/0
121/0


## Example Output

4b
b
0
80b
-8b
1.5b
2.03b
-b
-3.14b
b
3b
-b
121b


# Score

This is code golf, so fewest bytes wins. Standard loopholes are forbidden.

Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.

# Language Name, N bytes


where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

# Ruby, <s>104</s> <s>101</s> 96 bytes


If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

# Perl, 43 + 2 (-p flag) = 45 bytes


You can also make the language name a link which will then show up in the leaderboard snippet:

# [><>](http://esolangs.org/wiki/Fish), 121 bytes


var QUESTION_ID=191101;
var OVERRIDE_USER=8478;
var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;function answersUrl(d){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+d+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(d,e){return"https://api.stackexchange.com/2.2/answers/"+e.join(";")+"/comments?page="+d+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){answers.push.apply(answers,d.items),answers_hash=[],answer_ids=[],d.items.forEach(function(e){e.comments=[];var f=+e.share_link.match(/\d+/);answer_ids.push(f),answers_hash[f]=e}),d.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(d){d.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),d.has_more?getComments():more_answers?getAnswers():process()}})}getAnswers();var SCORE_REG=function(){var d=String.rawh\d,e=String.raw\-?\d+\.?\d*,f=String.raw[^\n<>]*,g=String.raw<s>${f}</s>|<strike>${f}</strike>|<del>${f}</del>,h=String.raw[^\n\d<>]*,j=String.raw<[^\n<>]+>;return new RegExp(String.raw<${d}>+String.raw\s*([^\n,]*[^\s,]),.*?+String.raw(${e})+String.raw(?=+String.raw${h}+String.raw(?:(?:${g}|${j})${h})*+String.raw</${d}>+String.raw))}(),OVERRIDE_REG=/^Override\s*header:\s*/i;function getAuthorName(d){return d.owner.display_name}function process(){var d=[];answers.forEach(function(n){var o=n.body;n.comments.forEach(function(q){OVERRIDE_REG.test(q.body)&&(o="<h1>"+q.body.replace(OVERRIDE_REG,"")+"</h1>")});var p=o.match(SCORE_REG);p&&d.push({user:getAuthorName(n),size:+p[2],language:p[1],link:n.share_link})}),d.sort(function(n,o){var p=n.size,q=o.size;return p-q});var e={},f=1,g=null,h=1;d.forEach(function(n){n.size!=g&&(h=f),g=n.size,++f;var o=jQuery("#answer-template").html();o=o.replace("{{PLACE}}",h+".").replace("{{NAME}}",n.user).replace("{{LANGUAGE}}",n.language).replace("{{SIZE}}",n.size).replace("{{LINK}}",n.link),o=jQuery(o),jQuery("#answers").append(o);var p=n.language;p=jQuery("<i>"+n.language+"</i>").text().toLowerCase(),e[p]=e[p]||{lang:n.language,user:n.user,size:n.size,link:n.link,uniq:p}});var j=[];for(var k in e)e.hasOwnProperty(k)&&j.push(e[k]);j.sort(function(n,o){return n.uniq>o.uniq?1:n.uniq<o.uniq?-1:0});for(var l=0;l<j.length;++l){var m=jQuery("#language-template").html(),k=j[l];m=m.replace("{{LANGUAGE}}",k.lang).replace("{{NAME}}",k.user).replace("{{SIZE}}",k.size).replace("{{LINK}}",k.link),m=jQuery(m),jQuery("#languages").append(m)}}
body{text-align:left!important}#answer-list{padding:10px;float:left}#language-list{padding:10px;float:left}table thead{font-weight:700}table td{padding:5px}
 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="https://cdn.sstatic.net/Sites/codegolf/primary.css?v=f52df912b654"> <div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> 

• I suspect that I'm doing something wrong, but if b/0 = b then if I divide both parts by b then 1/0 = 1. Do I need c-numbers to divide like this? Commented Aug 31, 2019 at 13:16
• @Erik that way, b/b = 0 when it's normally (and I'm pretty sure it's easily proven from all the various axioms) expected to be 1 (otherwise, b's multiplicative inverse seems to be not its multiplicative inverse). I'm pretty sure you just can't loophole against division by zero by adding b=1/0 or anything similar. Commented Aug 31, 2019 at 17:51
• There's a reason division by zero is undefined... $b=1\cdot b=\frac11\cdot b=\frac33\cdot b=\frac{3\cdot1}{3\cdot0}=\frac30=3\cdot\frac10=3b$. So you should be able to simplify all the examples (excepting the third of 0) to just $b$ Commented Aug 31, 2019 at 20:40
• Shouldn't the 3rd example have output 0b rather than 0? If the two expressions were equivalent then the question would have no premise Commented Aug 31, 2019 at 20:43
• Suggested test case: 3.1b/0 Commented Sep 1, 2019 at 18:04

# Malbolge Unshackled (20-trit rotation variant), 3,62e6 bytes

Size of this answer exceeds maximum postable program size (eh), so the code is located in my GitHub repository (note: Don't copy the code using CTRL+A and CTRL+C, just rightclick and click "Save destination element as...").

## How to run this?

This might be a tricky part, because naive Haskell interpreter will take ages upon ages to run this. TIO has decent Malbogle Unshackled interpreter, but sadly I won't be able to use it (limitations).

The best one I could find is the fixed 20-trit rotation width variant, that performs very well, calculating (pretty much) instantly.

To make the interpreter a bit faster, I've removed all the checks from Matthias Lutter's Malbolge Unshackled interpreter.

#include <malloc.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

const char* translation = "5z]&gqtyfr$(we4{WP)H-Zn,[%\\3dL+Q;>U!pJS72Fh" "OA1CB6v^=I_0/8|jsb9m<.TVacuY*MK'X~xDl}REokN:#?G\"i@"; typedef struct Word { unsigned int area; unsigned int high; unsigned int low; } Word; void word2string(Word w, char* s, int min_length) { if (!s) return; if (min_length < 1) min_length = 1; if (min_length > 20) min_length = 20; s[0] = (w.area%3) + '0'; s[1] = 't'; char tmp[20]; int i; for (i=0;i<10;i++) { tmp[19-i] = (w.low % 3) + '0'; w.low /= 3; } for (i=0;i<10;i++) { tmp[9-i] = (w.high % 3) + '0'; w.high /= 3; } i = 0; while (tmp[i] == s[0] && i < 20 - min_length) i++; int j = 2; while (i < 20) { s[j] = tmp[i]; i++; j++; } s[j] = 0; } unsigned int crazy_low(unsigned int a, unsigned int d){ unsigned int crz[] = {1,0,0,1,0,2,2,2,1}; int position = 0; unsigned int output = 0; while (position < 10){ unsigned int i = a%3; unsigned int j = d%3; unsigned int out = crz[i+3*j]; unsigned int multiple = 1; int k; for (k=0;k<position;k++) multiple *= 3; output += multiple*out; a /= 3; d /= 3; position++; } return output; } Word zero() { Word result = {0, 0, 0}; return result; } Word increment(Word d) { d.low++; if (d.low >= 59049) { d.low = 0; d.high++; if (d.high >= 59049) { fprintf(stderr,"error: overflow\n"); exit(1); } } return d; } Word decrement(Word d) { if (d.low == 0) { d.low = 59048; d.high--; }else{ d.low--; } return d; } Word crazy(Word a, Word d){ Word output; unsigned int crz[] = {1,0,0,1,0,2,2,2,1}; output.area = crz[a.area+3*d.area]; output.high = crazy_low(a.high, d.high); output.low = crazy_low(a.low, d.low); return output; } Word rotate_r(Word d){ unsigned int carry_h = d.high%3; unsigned int carry_l = d.low%3; d.high = 19683 * carry_l + d.high / 3; d.low = 19683 * carry_h + d.low / 3; return d; } // last_initialized: if set, use to fill newly generated memory with preinitial values... Word* ptr_to(Word** mem[], Word d, unsigned int last_initialized) { if ((mem[d.area])[d.high]) { return &(((mem[d.area])[d.high])[d.low]); } (mem[d.area])[d.high] = (Word*)malloc(59049 * sizeof(Word)); if (!(mem[d.area])[d.high]) { fprintf(stderr,"error: out of memory.\n"); exit(1); } if (last_initialized) { Word repitition[6]; repitition[(last_initialized-1) % 6] = ((mem[0])[(last_initialized-1) / 59049]) [(last_initialized-1) % 59049]; repitition[(last_initialized) % 6] = ((mem[0])[last_initialized / 59049]) [last_initialized % 59049]; unsigned int i; for (i=0;i<6;i++) { repitition[(last_initialized+1+i) % 6] = crazy(repitition[(last_initialized+i) % 6], repitition[(last_initialized-1+i) % 6]); } unsigned int offset = (59049*d.high) % 6; i = 0; while (1){ ((mem[d.area])[d.high])[i] = repitition[(i+offset)%6]; if (i == 59048) { break; } i++; } } return &(((mem[d.area])[d.high])[d.low]); } unsigned int get_instruction(Word** mem[], Word c, unsigned int last_initialized, int ignore_invalid) { Word* instr = ptr_to(mem, c, last_initialized); unsigned int instruction = instr->low; instruction = (instruction+c.low + 59049 * c.high + (c.area==1?52:(c.area==2?10:0)))%94; return instruction; } int main(int argc, char* argv[]) { Word** memory[3]; int i,j; for (i=0; i<3; i++) { memory[i] = (Word**)malloc(59049 * sizeof(Word*)); if (!memory) { fprintf(stderr,"not enough memory.\n"); return 1; } for (j=0; j<59049; j++) { (memory[i])[j] = 0; } } Word a, c, d; unsigned int result; FILE* file; if (argc < 2) { // read program code from STDIN file = stdin; }else{ file = fopen(argv[1],"rb"); } if (file == NULL) { fprintf(stderr, "File not found: %s\n",argv[1]); return 1; } a = zero(); c = zero(); d = zero(); result = 0; while (!feof(file)){ unsigned int instr; Word* cell = ptr_to(memory, d, 0); (*cell) = zero(); result = fread(&cell->low,1,1,file); if (result > 1) return 1; if (result == 0 || cell->low == 0x1a || cell->low == 0x04) break; instr = (cell->low + d.low + 59049*d.high)%94; if (cell->low == ' ' || cell->low == '\t' || cell->low == '\r' || cell->low == '\n'); else if (cell->low >= 33 && cell->low < 127 && (instr == 4 || instr == 5 || instr == 23 || instr == 39 || instr == 40 || instr == 62 || instr == 68 || instr == 81)) { d = increment(d); } } if (file != stdin) { fclose(file); } unsigned int last_initialized = 0; while (1){ *ptr_to(memory, d, 0) = crazy(*ptr_to(memory, decrement(d), 0), *ptr_to(memory, decrement(decrement(d)), 0)); last_initialized = d.low + 59049*d.high; if (d.low == 59048) { break; } d = increment(d); } d = zero(); unsigned int step = 0; while (1) { unsigned int instruction = get_instruction(memory, c, last_initialized, 0); step++; switch (instruction){ case 4: c = *ptr_to(memory,d,last_initialized); break; case 5: if (!a.area) { printf("%c",(char)(a.low + 59049*a.high)); }else if (a.area == 2 && a.low == 59047 && a.high == 59048) { printf("\n"); } break; case 23: a = zero(); a.low = getchar(); if (a.low == EOF) { a.low = 59048; a.high = 59048; a.area = 2; }else if (a.low == '\n'){ a.low = 59047; a.high = 59048; a.area = 2; } break; case 39: a = (*ptr_to(memory,d,last_initialized) = rotate_r(*ptr_to(memory,d,last_initialized))); break; case 40: d = *ptr_to(memory,d,last_initialized); break; case 62: a = (*ptr_to(memory,d,last_initialized) = crazy(a, *ptr_to(memory,d,last_initialized))); break; case 81: return 0; case 68: default: break; } Word* mem_c = ptr_to(memory, c, last_initialized); mem_c->low = translation[mem_c->low - 33]; c = increment(c); d = increment(d); } return 0; }  ## It's working! • I hope you didn't type all that. Commented Aug 31, 2019 at 20:31 • When I opened your program, Chrome attempted to translate it from Polish Commented Sep 3, 2019 at 9:06 • How to generate this program? Commented Jan 22, 2022 at 1:00 # PHP, 656461 58 bytes -1 byte by using a b instead of '' (empty string). Since "b"s are trimmed, it will be same as an empty string in this specific case. -3 bytes by using substr instead of explode to get first part of input. -3 bytes by using better methods to detect 1 and -1. <?=($n=substr($argn,0,-2))?trim($n+1?$n-1?$n:b:'-',b).b:0;


Try it online!

Tests: Try it online!

If first part of input before "/" (we call it $n) is 0, prints 0. Else prints $n itself with any "b" at the end trimmed from it and special cases of -1 and 1 handled, so the "1" digit is not printed. And at the end appends a single "b". The trimming part is to make sure we don't get a double "b" at the end like "3bb".

• very nicely done! Commented Aug 31, 2019 at 6:39
• Replacing $n==-1 with $n>0 (-2 bytes) seems to work. You could try that. Commented Sep 3, 2019 at 9:42
• @IsmaelMiguel, that is not working, if you meant $n<0, that won't work too, since we have inputs like -8/0. Commented Sep 3, 2019 at 11:47 • @IsmaelMiguel, but you gave me an idea, replaced $n==-1?'-':$n with $n+1?$n:'-' to save 2 bytes! Commented Sep 3, 2019 at 11:50 • :/ When I tested, it seemed to work. but well, good thing that you found another way. Commented Sep 3, 2019 at 15:42 # Jelly, 18 bytes I ended up stealing Erik's ṾṖ$İƑ¡ for this one (otherwise I'd also have 19)...

ṖṖv0ḢṾṖ$İƑ¡,Ạ¡”boḢ  A full program which prints the result. Try it online! Or see the test-suite. ### How? ṖṖv0ḢṾṖ$İƑ¡,Ạ¡”boḢ - Main Link: list of characters S
Ṗ                  - discard right-most (of S)
0               - literal zero
v                - evaluate as Jelly code with right argument (0)
- ... b is covert-to-base, so "nb0" gives [n]
Ḣ              - head ([n]->n or n->n)
¡        - repeat...
Ƒ         - ...# of times: is invariant under:
İ          -   reciprocation (n->1/n)
$- ...action: last two links as a monad: Ṿ - un-evaluate (-1->"-1" or 1->"1") Ṗ - discard right-most ("-1"->"-" or "1"->"") ¡ - repeat... Ạ - ...# of times: all? , ”b - ...action: pair with a 'b' character o - logical OR with: Ḣ - head (S) (i.e. if we end with 0 use the 1st character of the input) - implicit print  • Ahhh, and I was thinking of ways I could abuse v... :D Commented Aug 31, 2019 at 19:57 # Perl 6, 32 bytes {~m/^0/||S/[(\-|^)1|b]?\/0/$0b/}


Try it online!

A couple of regexes, one to check if the input is 0/0, and the other to replace the trailing /0 with just b (and to remove the old b, 1, and/or -1)

### Explanation (old)

{                          }  # Anonymous codeblock
~m/^0/     # Return 0 if the input starts with 0
||   # Otherwise
S/             / /  # Substitute
\/0       # The /0
(        )?          # Optionally starting with
<wb>1               # 1 or -1
|b             # Or b
b   # With just b

• Won't this fail on input 0.5/0? Although, I guess that kind of decimal isn't included in the spec Commented Mar 22, 2020 at 19:19

ṖṖḟ”bȯ1VṾṖ$İƑ¡,Ạ¡”b  Try it online! Full program. # Retina, 28 24 bytes b?/0 b ^0b 0 (^|-)1b$1b


Try it online!

First try at using Retina, so there's probably considerable room for golfing.

• 18 bytes: Try it online! Commented Aug 31, 2019 at 8:09
• After hunting for what \b does (I'm that inexperienced with regex), I was a bit disappointed to find that it can't be shortened to the unprintable backspace character. Anyhow, thanks Commented Aug 31, 2019 at 8:31
• @UnrelatedString of course it can't be shortened to backspace, after all, \b is merely an ASCII representation of the backspace character in normal strings :P Commented Sep 5, 2019 at 2:40

# convey, 47 bytes

['b'}
>>~^]] '-'[
91+>##"@]
{?~""@@<}
]11-^}


Try it online!

I would like to put off the space betwen the number and 'b' in the output but i dont know how to do it :(

How it works (-3/0):

How it works (1/0):

• Oh, the output was a bit jagged, you couldn't have done it properly. Fixed it in the interpreter, so it's now 3b as one would expect it. :-)
– xash
Commented Jun 4, 2021 at 14:18

# Python 3.8 (pre-release), 69 66 bytes / 73 70 bytes

Python already had a solution here but it didn't work properly on -1/0. That solution also uses regex while this one is vanilla python.

## Lambda version (66 bytes)

lambda x:{"0":0,"1":"b","-1":"-b"}.get(a:=x[:-2].strip("b"),a+"b")


Try it online!

## Print version (70 bytes)

print({"0":0,"1":"b","-1":"-b"}.get(a:=input()[:-2].strip("b"),a+"b"))


Try it online!

## How it works :

starting from a string x:

• x[:-2] remove the /0 part
• .strip("b") remove any eventual b present in the string
• a:= store our new string into a

Then we check if our new sring is equal to any "0", "1" or "-1":

• if True, {...}.get() will replace it by the appropriate string
• else, we will print it inchanged adding "b" at the end of the string

Thanks to ovs for reminding me that str.strip existed saving 3 bytes on each program

• .strip("b") is 3 bytes shorter than .split("b")[0].
– ovs
Commented Jun 3, 2021 at 10:31
• @ovs my bad I totally forgot about that ^^' Commented Jun 3, 2021 at 10:58

## Keg, 18B

All credit is to Jono 2906.

__:b=;[b]^:\1=[_]^


## Explanation

__                 # Take implicit input and remove the "trash" (/0).
:b=              # Is the last character equal to b?
;             # Negate(decrement) this value.
[b]          # If the last character is not b, append b.
^         # Reverse the stack.
:\1=     # Is the first character equal to 1?
[_]  # If so, reduce the value.
^ # Reverse the stack back and implicit output.


TIO!

# Python 3, 68 bytes

import re
print(re.sub('^0b$','0',re.sub(r'(^1)?b?/0','b',input())))  Try it online! • Nice solution! But import re rises the bytecount to 64. Commented Sep 1, 2019 at 15:22 • @movatica good point, new here so didn't realize import statement was included (although of course it is). Edited. Commented Sep 2, 2019 at 7:18 • Welcome! :) You still can keep the shorter lambda version though! It does not need to be a full program. And the import statement can be placed after the lambda definition, so 64 bytes are possible! Commented Sep 2, 2019 at 15:44 • @movatica ah, nice! I didn't find a way to make it work with import and lambda. Thank you Commented Sep 2, 2019 at 20:29 • In the given exmple, -1/0 should return -b and not -1b. You could achieve that bay changing '(^1)?b?/0','b' by '(^(-)?1)?b?/0','\\2b' at the cost of 7 bytes :/ Also you don't need the $ in your first regex, neither you need the r in front of your string Commented Jun 3, 2021 at 9:10

# JavaScript (ES6), 45 bytes

s=>+(n=s.split/[0])?[n*n-1?n:'-'[~n]]+'b':n


Try it online!

### Commented

s =>                  // s = input: "numerator/0"
+(                  //
n = s.split/[0] // n = numerator, as a string
) ?                 // if n coerced to a Number is neither equal to 0 nor NaN:
[ n * n - 1 ?     //   if abs(n) is not equal to 1:
n             //     append the numerator
:               //   else:
'-'[~n]       //     append '-' if n = -1, or an empty string otherwise
] + 'b'           //   append 'b'
:                   // else:
n                 //   just output the numerator because it's either "0" or
//   an expression that already contains 'b'


# C (clang), 298263217 179 bytes

i;f(char*s){for(i=0;s[i]-47&&(*s-49|s[1]-98||!printf("b"));)write(1,s+i,s[i++]!=98);}main(){char*s;gets(s);printf(*s-98&&*s-49|s[1]-47?*s-48?*s-45|s[1]-49?f(s),"b":"-b":"0":"b");}


Try it online!

After a while, I actually managed to make this function the way you want. It's pretty big, I will say, I had to debug a lot.

Thanks to ceilingcat for golfing 35 bytes, another 46 bytes, and another 38 bytes.

• Very well done! I'm proud of you! Commented Jun 3, 2021 at 8:07
• @lyxal Thanks for the compliment. It took a long time to make this. Commented Jun 3, 2021 at 10:32
• 171 bytes Commented Jun 10, 2021 at 6:04

# Wolfram Language (Mathematica), 37 bytes

ToExpression[""<>#~Drop~-2]b/.b^_->b&


Try it online!

Takes a list of characters as input.

## C, 209203 137 bytes

-66 bytes thanks to ceilingcat

char a[9];main(f){gets(a);f=strlen(a)-3;a[f+1]=0;printf((*a==55&a[1]==49&f==1?a[1]=98:*a==49&!f?*a=98:a[f]==98|*a==48&!f)?"%s":"%sb",a);}


TIO

• Putting in -0/0 gives -0b, but it was never in the example input or test cases, so it's correct. Commented Sep 4, 2019 at 1:35
• 135 bytes Commented Mar 22, 2020 at 20:06

# naz, 64 bytes

6a8m1s2x1v2m4a2x2v1x1f1r3x1v2e3x2v3e1o1f0x1x2f2m4a1o0x1x3f1o0x1f


Explanation (with 0x commands removed)

6a8m1s2x1v             # Set variable 1 equal to 47 ("/")
2m4a2x2v               # Set variable 2 equal to 98 ("b")
1x1f                   # Function 1
1r                 # Read a byte of input
1o1f # Otherwise, output it and jump back to the start of the function
1x2f2m4a1o             # Function 2
# Set the register equal to 98 and output once
1x3f1o                 # Function 3
# Output once
1f                     # Call function 1


# Ruby-p, 37 bytes

sub /(^1|-\K1)?b?\/0/,?b
sub /^0b/,?0


Try it online!

# C++ (gcc), 96 bytes

#import<ios>
int f(int r,int i){i-=~r;i||printf("-");i--*~-i&&printf("%d",i?:r);i&&printf("b");}


Takes as input an integer r denoting a real number and the coefficient of a multiple of $$\b\$$ i.

Then "divides" by zero and does some hacking to avoid printing deviating results.

Try it online!

• 87 bytes Commented Dec 7, 2020 at 4:16

# C (gcc), 71 bytes

f(r,i){i-=~r;i||printf("-");i--*~-i&&printf("%d",i?:r);i&&printf("b");}


Same as my C++ solution but with less boilerplate.

Try it online!

• @ceilingcat Thanks. Commented Mar 22, 2020 at 23:02
• 65 bytes Commented Mar 23, 2020 at 5:25

# 05AB1E, 15 bytes

¨¨ÐĀiáм'b«Dþi1K


Explanation:

¨¨               # Remove the last two characters of the (implicit) input
Ð              # Then triplicate the string
Āi            # Pop one, and if it's NOT 0 (using Python-style truthify):
á           #  Pop again and only leave the letters (in case it contains a "b")
м          #  And remove those letters
'b«      '#  Then append a trailing "b"
D      #  Duplicate it
þ     #  Pop and only leave the digits
i    #  If this is exactly 1:
1K  #   Remove the 1
# (after which the top of the stack is output implicitly as result)


# Brainfuck, 25 bytes

>,[>,]<[-<+>]<+++[<]>[.>]


## Explanation

>,[>,]        read from stdin
<[-<+>]<+++   add last two cells and add three ( ascii('/') + ascii('0') + 3 = ascii('b')
[<]>          move pointer to first char to output
[.>]          output until cell w/ value 0

• b/0 expected b, got bb; 0/0 expected 0, got 0b; -1/0 expected -b, got -1b.
– user85052
Commented Sep 3, 2019 at 13:47
• Yeah, basically this only substitutes the /0 for b and doesn't take into account any of the cases for 0b, 1b, -1b or any inputs that already contain a b
– Jo King
Commented Sep 4, 2019 at 1:09

# Symja, 85 bytes

d(x_):=If((e=StringReplace(x,"/0"->""))!="0",If(StringContainsQ(e,"b"),e,e<>"b"),"0")
`

Try it Online!

I finally answered my own question!