diff --git a/lab2-recursive-arithmetic.pl b/lab2-recursive-arithmetic.pl
new file mode 100644
index 0000000..60d3c54
--- /dev/null
+++ b/lab2-recursive-arithmetic.pl
@@ -0,0 +1,95 @@
+% Lab 2 - Recursive Arithmetic in Prolog
+% LABSHEET:
+% https://teaching.csse.uwa.edu.au/units/CITS3005/labs/lab02-arithmetic.pdf
+
+% Build a Prolog program to do recursive arithmetic. It should have
+
+% 1. a constant, zero, to represent zero.
+zero.
+
+% 2. a function, next(X), that represents the next number (so next(zero) represents one).
+next(X) :- X.
+
+% 3. a predicate, sum(X,Y,Z), which is true if X + Y = Z.
+% sum(zero,zero,Z) :- Z is zero.
+% sum(zero,Y,Z) :- Z is Y.
+% sum(next(zero),Y,Z) :- Z is next(Y).
+sum(zero,zero,zero).                    % CASE 1 : 0 + 0 == 0
+sum(zero,X,X).                          % CASE 2 : 0 + X == X
+sum(next(X),Y,next(Z)) :- sum(X,Y,Z).   % CASE 3 : (X+1) + Y == (Z+1) (Works based on previous two cases???)
+% TEST 1:
+% sum(next(zero),next(zero),next(next(zero)))
+%=sum(zero,next(zero),next(zero))   [ TRUE ]
+% TEST 2:
+% sum(next(next(zero)),zero,next(zero))
+%=sum(next(zero),zero,zero)         [ FALSE ]
+
+% 4. a predicate, mult(X,Y,Z) which is true if X × Y = Z.
+mult(zero,zero,zero).
+mult(zero,_,zero).
+mult(next(zero),X,X).
+mult(next(X),Y,Z) :- mult(X,Y,Zp), sum(Y,Zp,Z). % (x+1)*y == x*y+y
+
+% 5. a predicate, equals(X,Y) which is true if X = Y .
+equals(X,X).
+
+% 6. a predicate, lessThan(X,Y) which is true if X < Y ¿
+lessThan(zero,next(_)).
+lessThan(next(X),next(Y)) :- lessThan(X,Y).     % X+1 < Y+1 => X < Y. As X->0, Y->N => 0 < N.
+
+% binary_(zero,0).
+% binary_(next(X),IntValue) :- binary_(X,PrevIntValue), write(PrevIntValue), IntValue is PrevIntValue+1.
+% binary(X) :- binary_(X,Buffer), write(Buffer).
+
+
+% Test your program and consider the efficinecy of the implementations. Is there a way to get faster annswers?
+% Next, add additional functions:
+% 1. Write a program binary(X) that will print the binary representation of X out, so for example, binary(next(next(zero)
+% will print out 10.
+% NOTE: Binary array is in reverse!
+increment_b([],[]).
+increment_b([ODigit | ORest],[Digit | Rest]) :- 
+    ODigit =:= 0 -> Digit is 1,
+    equals(Rest,ORest);
+    ODigit =:= 1 -> Digit is 0,
+    (
+        equals(ORest,[]) -> equals(Rest,[1]);
+        increment_b(ORest,Rest)
+    ).
+
+% NOTE: Binary array is in reverse!
+binary_arr(zero,[0]).
+binary_arr(next(X),NextBinary) :- 
+    binary_arr(X,Binary),
+    increment_b(Binary,NextBinary).
+
+binary_arr_print([]).
+binary_arr_print([Digit|Rest]) :- write(Digit), binary_arr_print(Rest).
+
+binary(X) :-
+    binary_arr(X,BinaryArrRev),
+    reverse(BinaryArrRev, BinaryArr),
+    binary_arr_print(BinaryArr).
+
+% 2. Implement predicates for odd, even and prime numbers.
+even(zero).
+even(next(X)) :- \+even(X).
+
+odd(X) :- \+even(X).
+
+% Cheating a bit here ... would be harder doing the whole primality test in the above notation, since I don't have a modulo predicate defined
+to_integer(zero,0).
+to_integer(next(X),Integer) :- to_integer(X,OInteger), Integer is OInteger + 1.
+
+not_prime_i(I,Comp) :- 
+    Comp*Comp < I+1,
+    (I mod Comp =:= 0; I mod (Comp + 2) =:= 0;
+    not_prime_i(I,Comp+6)).
+
+prime_i_entry(I) :- 
+    I > 1, (I =:= 2; I =:= 3;
+    I mod 2 =\= 0, I mod 3 =\= 0,
+    \+not_prime_i(I, 5)).
+
+
+prime(X) :- to_integer(X,I), prime_i_entry(I).
\ No newline at end of file