Determine if a number is less than another [closed]

Question

Introduction

Your job is to write a function or program that detects if a number a is less than a number b.

But... You are not allowed to use built-in alternatives for equals, less than, greater than, less than or equal to, greater than or equal to, or not equal to operators. (Writing your own implementation is allowed.) Similar alternative methods/alternatives in other languages that are designed to replace those operations are also restricted.

You are allowed to use operators that are not banned, e.g. bitwise operators and the not operator (but it would be better if your answer did not have the not operator).

Challenge

Simply return a truthy or falsy value (or something that evaluates in the language to true or false like 0 or 1) to indicate whether a is less than b. The winner of the challenge can be determined by the shortest code. External sources are not allowed, but anonymous functions are. Explaining your answer is not mandatory, but encouraged.

Example using Javascript:

(a,b)=>!!Math.max(0,b-a) //remove the !! for truthy and falsy numbers
/*Explanation
(a,b)                    Parameters a and b
     =>                  Function is
       !!                returns true if truthy and false if falsy
         Math.max(0,b-a) returns the greater number, 0 or b-a
*/

Examples of Input and Output

Input | Output
------|-------
1, 3  | true
2, 2  | false
3, 2  | false

Answer 1

APL (Dyalog Unicode), 2 bytes^SBCS

Anonymous tacit prefix function. Requires ⎕IO←0

⊃⍒

Try it online!

⍒ gives the "grade". The indices into the argument that would stably sort it in descending order

• If the first element is less than the second, it gives [1,0]

• If the the elements are equal, it gives [0,1]

• If the second element is less than the second, it gives [0,1]

⊃ picks the first element of that

• If the first element is less than the second, it gives 1

• If the the elements are equal, it gives 0

• If the second element is less than the second, it gives 0