Why do spaces between guitar frets get smaller as the notes get higher?
This is more clear when you think of it in terms of the length of the string. Ignoring for a second that strings need to be a minimum length in order to vibrate, we can produce a full octave on the first half of any length of string. Open for the "base" note, midpoint for the octave, one third of the way up for the fifth, etc.
So on a guitar you have the first half of the string to make up an octave. Then on the second half, you have another half of that (a quarter) to make up the next octave. Then another half of a half of a half (eighth) to make the next octave. Each time, you only have half the space for the octave. Thus, each fret has to be half the size of the fret an octave lower.
This mirrors my comment above: While maintaining frequency ratios, the absolute frequency differences changes by a factor of 2, and that is represented in the physical frets as well.
Because frequency with regard to pitch is exponential, not linear.
Exponential (frequency doubles for each octave -- each higher octave fits perfectly into the lower one with the least possible interference -- 2:1 ratio):
A3: 220hz A4: 440hz A5: 880hz
Linear (same value added to each successive frequency, making the top note a 3:2 ratio with the previous -- much more interference than an octave):
A3 : 220hz A4 : 440hz 'A5': 660hz -- **nope**
In reality, this 660hz would be a just intonation fifth above A4, or E5.
The frets are getting arithmetically closer together (their absolute distance decreases). But they are not getting geometrically closer, in terms of their distance from the bridge.
If you take the distance from the bridge to any fret (call that X) and also the distance from the bridge to the next higher lower (call that Y), then the ratio X/Y is the same whether or not X and Y are frets 2 or 1, or frets 20 and 19.
This ratio produces the frequency ratio that corresponds to the (equal temperament) semitone. It is 21/12, or about 1.0595. (The two denotes that frequency doubles per octave, and the 1/12 indicates one half-step out of a potential twelve in an octave.)
Fret 19 is 1.0595 times farther away from the bridge than fret 20.
Fret 1 is 1.0595 times farther away from the bridge than fret 2.
This 1.0595 ratio is also the frequency ratio. If you know the frequency of some given note, like A = 440 Hz, you can figure out what is the frequency of A#, one semitone higher. Just multiply 440 x 1.0595 = 466.18. The A# above the 440A has a frequency of about 466.2.
There are 12 semitones in an octave. If we multiply a number by 1.0595 and do that 11 more times, we will get about twice the original number. You can try this on your calculator. Type 1 X 1.0595. Then hit the = key 12 times. You should get a number very close to 2.
For anyone interested the common ratio between the frets would be the twelfth root of 2.
Interestingly, all the answers are completely missing the point. Everyone brings up the exponential increase of frequency as the reason ("you double the frequency for each octave"), but that is a red herring.
Even if the frequencies of consecutive pitches increased linearly, the higher note frets would still be closer together.
The actual reason why this happens is because the length of the vibrating string is inversely proportional to the frequency of the sound it produces. This very simple physical explanation is the answer to the question.
Given two "consecutive" frequencies f1 and f2, the distance between the two frets is proportional to (1/f1-1/f2) or (f2-f1)/(f1*f2).
So even if (f2-f1) were constant (i.e., frequencies increase linearly) or outright increasing, the denominator (f1*f2) still increases very fast as f1 and f2 go higher, which means that no matter what formula you pick for the frequencies, the distance between frets will get smaller.