can you crack my code?

[mod-man]

ive always been interested in encrypted messages, so i decided to make my own
3.0/1.1/2.7
               
100.75/10.5/21.1

3.1/20.2/1.0/121.11

13.1/100.75

0.3/75.5/2.2/10.5
can you solve it?

[Ad Bot]

[Aide33]

arg

[ThatRandomGuy]

"420 blaze it" I win

[Rigel]

"420 blaze it" I win

I'd imagine that it's like a math problem where you have to show what you did so other people can get how you got the answer.

[SWAT One]

Could you arrange it in a table as a visual aid, perhaps.

[c]

solve what

[joe411]

arg

[Sheepocalypse]

3.0/1.1/2.7
              
100.75/10.5/21.1

3.1/20.2/1.0/121.11

13.1/100.75

0.3/75.5/2.2/10.5

3.0/1.1/2.7

3.0/

3

3

3

Your forgetin' code.

[Yosher]

if you divide the numbers in the first row, you get a binary messages which says UP

that's all i know so far

[Blockzillahead]

hey this reminds me of my dusty project a while back
i took someones broken masking algorithm online because i needed it for my RSA and revised it to actually work
then developed my own demasking algorithm because the guy never made one. he asked someone to help develop one and i was going to reply back with my solution but the post was like 2 years old

its like a caesar cipher on drugs
it uses a key and its theoretically uncrackable without knowing the key. i think its called "one time padding" but idk
i wouldnt call it encryption exactly. its more like a way to parse the data so it can be interpreted correctly. text -> padded to fit bit length -> hex -> mask -> rsa both mask and key to mask

caesar cipher:
"aaaaa" -> "ddddd"
me:
"aaaaa" -> key: 5 -> "fLrXD"

i enjoy cryptography too op

[Inap]

try to find out my code
01000010 01001100 01001111 01000111 01001100 01000001 01001110 01000100 00100000 01000110 01001111 01010010 01010101 01001101 01010011

[ThatRandomGuy]

try to find out my code
01000010 01001100 01001111 01000111 01001100 01000001 01001110 01000100 00100000 01000110 01001111 01010010 01010101 01001101 01010011

01001110 01001001 01000011 01000101 00100000 01001111 01000010 01010011 01000101 01010010 01010110 01000001 01010100 01001001 01001111 01001110

[Inap]

01001110 01001001 01000011 01000101 00100000 01001111 01000010 01010011 01000101 01010010 01010110 01000001 01010100 01001001 01001111 01001110

NICE OBSERVATION is what urs says

[The Resonte!]

to answer op's question

no

[[]----[]]

if you divide the numbers in the first row, you get a binary messages which says UP

that's all i know so far

http://www.wolframalpha.com/input/?i=3.0%2F1.1%2F2.7
?????????????????????????

two repeated numbers:

3.0/1.1/2.7
               
100.75/10.5/21.1

3.1/20.2/1.0/121.11

13.1/100.75

0.3/75.5/2.2/10.5

my guess is that it's simpler than a cryptogram, and that the ciphertext consists of 5 words of the form

ABC DEF GHIJ KD LMNE

but i can't get any farther than that without discovering the substitution algorithm (if there is one)