 So, to recap from the last lecture, the consonances correspond to ratios of small whole numbers – 2 to 1, 3 to 2, 4 to 3. Western music also has octave equivalents. Two notes corresponding to the 2 to 1 ratio are regarded as the same note. This leads to the tuning problem. Construct a scale by starting at a note and ascending by consonances, then scale all notes back into a single octave. Let's see how that works out. So, let's construct a scale. Suppose we start with a string of length 16, and that's purely for convenience. In the following, we'll treat the length as the note. And so we can say the string of length 16 produces the note 16. So, if we use the interval of an octave and produce additional lengths in a ratio of 2 to 1, our notes are going to be 16, then 8, which will be in a 2 to 1 ratio. Then 4, then 2, and we could keep going, but let's stop here for now. Now, since we want all the notes to be within one octave, we'll scale all the lengths to be between 16 and 8. So, because of octave equivalents, multiplying by 2 gives us the same note. And if we do that, our notes become the following. The note 8, scaled down an octave, multiplied by 2 in other words, becomes the note 16. The note 4, scaled down an octave, becomes the note 8, but I already scaled 8 down an octave back to 16. So, this note 4 also becomes 16. The note 2, scaled down an octave to 4, scaled down another octave to 8, scaled down another octave to 16. And what this means is that if we scale these higher notes down into the same octave, they all turn out to be the same note, the one we started with. And our scale is literally monotonous. Well, let's try and produce another scale. Let's use the 3 to 2 ratio. So again, for convenience, we'll start with 81, because we're going to be taking 2 thirds of a number. So, if we start with 81, our notes will be 2 thirds of 81, that's 54, then 2 thirds of 54, or 36, then 2 thirds of 36, that's 24, 2 thirds of 24, that's 16, and we'll stop there for now. Now, this octave will run from 81 to 81 divided by 2, that's 40 and a half. So, we can scale the notes into this interval. So, some of the notes are already in the interval 81 and 54, 36 isn't, so we'll double it to lower than an octave. So, the note 36 can be lower than an octave to 72, 24 can be lower than an octave to 48, 16 can be lower than an octave to 32, but our interval runs from 81 to 40 and a half, so we'll scale it down another octave to 64. And that gives us five distinct notes. And this leads to an important problem. We can continue to produce additional notes, then scaling them back into the octave, but do we ever stop? And this is known as the problem of the circle of fifths. Start to get a note. Can you ascend by fifths to get a note that is the same up to octave equivalence as your starting note? Well, let's try it. Since we need to find 2 thirds of a value, let's start with the power of 3, say 3 to the eighth or 243, and we'll find our successor of notes. So, 2 thirds of 243 is 162, 2 thirds of 162 is 108, and we can find a few more notes. So, our octave is going to run from 243 to 122 and a half, so let's scale all of our notes into the octave. So that 108, I can scale that down into the octave by doubling its length to 216. I'll scale 72 into the octave, and so on. And remember, sometimes we may have to scale more than one octave down, and this gives us our set of five distinct notes, 243 to 16, 192, 144, 128. And it's worth noting that the first and last notes are almost in a 2 to 1 ratio. That ratio of 243 to 128 is almost 2 to 1. And so we almost span an octave. It's close, but not quite. So let's apply some mathematics to this problem. Suppose our strength has length k. If we go up by fifths, our notes will be k times sum power of 2 thirds. We want this to be m octaves above our starting note, so that's going to be k times sum power of one half. And so that means we want k times sum power of two thirds to be k times sum power of one half. And we can do a little bit of math. And what we find is that solving this problem requires that we just need to find a power of two that's equal to a power of three. But there's a problem. The powers of two are all even numbers, and the powers of three are all odd numbers. So we need to find an even number that's equal to an odd number, but that's impossible. And this leads to the following. All of the proceeding was well within the mathematical knowledge and musical inclinations of the Pythagoreans. Moreover, it would have been consistent with the Pythagorean interest in number. Knowledge or interest in geometry would not be required. And so if we want to consider how the Pythagoreans first discovered the existence of irrational quantities, it seems likely that this tuning problem may have played an important role.