Write the shortest code, in number of bytes, to display, return, or evaluate to the golden ratio (that is, the positive root of the quadratic equation: \$x^2-x-1=0\$, approximately 1.618033988749895), to at least 15 significant figures. No input will be given to your program.
Sample in Stutsk programming language:
1 100 { 1 + 1 swp / } repeat print 
# (commonmark migration) \$\endgroup\$ φ Builtins ftw ¯\_(ツ)_/¯
Without builtins it's 6 bytes:
51α√½Σ Explanation:
φ # Push golden ratio builtin 1.618033988749895 # (output the entire stack joined together implicitly as result) 5 # Push 5 1 # Push 1 α # Wrap the last two values into a list: [5,1] √ # Take the square-root of each value: [2.23606797749979,1.0] ½ # Halve each: [1.118033988749895,0.5] Σ # And sum this list: 1.618033988749895 # (after which the entire stack joined together is output implicitly as result) 
X5mq+2/ Just one more byte than 05AB1E! Pretty simple stack-based translation of the equation on the Wikipedia page:
X5 Push 1 and 5 on to the stack mq Square root the top number + Add the top two numbers on the stack 2/ Divide by two (implicit) output the stack 
w 5**.5+1/2 Output: 1.618033988749894849
This highlights a quick about MUMPS: order of operations is evaluated left to right. Something like .5+5**.5/2 would give us (5.5**.5)/2 (1.172603939955857389).
5t>; 5t>; # full program ; # divide... t # square root of... 5 # literal... > # plus 1... ; # by 2 # implicit output 
S1n; This answer feels like cheating because S1 simply pushes the golden ratio onto the stack.
This answer doesn't work with TIO, but it does work with this interpreter: Try it online!
bc forbids non-integer exponents, so:
.5+sqrt(5/4) (incl. the newline) plus -l for the math library (which also conveniently bumps the scale); therefore:
$ echo '.5+sqrt(5)/2' | bc -l 1.61803398874989484820 Not the shortest possible J-uby solution but more interesting. Same idea as coltim's K solution and others.
-:p^(:/&1.0|:+&1!~1) -:p ^ (:/ & 1.0 | :+ & 1 !~ 1) :/ & 1.0 | :+ & 1 is equivalent to ->n{ 1+1.0/n }. !~ applies that function to an initial value (here 1) until a fixed point. -:p prints it.
1¥ᵑ{ŗ→ Nekomata uses fractions instead of floats, so a Diophantine approximation is used here. The output differs from the true value of \$\phi\$ by \$\sim8.5\times10^{-108}\$.
1¥ᵑ{ŗ→ | Full program -------+---------------------------- 1 | Starting with 1, ¥ᵑ{ | do the following 256 times: ŗ | Reciprocal → | Add 1 
ratio=(1+5**(1/2))/2 print(ratio) Output:
1.618033988749895 
(1+5**(1/2))/2 is by itself is a snippet so wouldn't be a valid answer, and ratio=(1+5**(1/2))/2<newline>print(ratio) is sufficiently long it may not be a serious contender for the winning criteria of the challenge. \$\endgroup\$ Very basic solution, calculates exact value & prints.
main(){printf("%.15f",.5+sqrt(5)/2);} 
