 This time, I'm going to convert 6.25 into the floating point format. I have this helpful box here that's divided up into each of the three parts that I'm going to need to fill in. So I'm going to figure out what those parts are, and then I will just be able to fill them into the box, and there will be my floating point representation. So I will start by converting my number into binary. So 5.625 is 5 point, and this is a half plus a 1 eighth. So I've got a half bit, no quarter bit, but a 1 eighth bit. So there's 5.625 in binary. Next I need to represent this in normalized scientific notation. So I'm going to want to move my binary point two places to the left. This will give me 1.01101 times 2 squared. I moved it two places to the left, so my exponent is a 2. Now I get to fill in all of these fields. So 5.65 is a positive number, so my sine bit is a 0. Next I need an exponent. My exponent is 2, but I have a bias of 127. So 2 plus 127 is 129, and in binary that will look like 128 plus 1. So I will copy those 8 bits into my exponent. And last I need my mantissa. For my mantissa, I will just copy these 5 bits in. Again, I don't need my leading 1 because my number is in normalized scientific notation, so I know that I always have a leading 1. Since I already know what this number is, there's no reason to write it down. Now I need a 32 bit number, so I'm going to need to fill in the right-hand side with 0s until I've got 32 bits. I have 23 bits for my mantissa. I've used 5 of them so far, which means I have 18 more to go. So I will just add 18 0s to the right-hand side. And there is 5.65 in the floating point format. Now I might want to also convert this to a hexadecimal representation because that might make it easier to read, to talk about, in which case I'm going to look for blocks of 4 and just kind of read those off. So here I have 4, 0, 8 plus 3 is 11, which is B, and then 4, and then 4 blocks of 0s. And I'll put a 0x at the beginning to indicate that this is a hexadecimal representation. So either of these would be an acceptable form of the floating point representation for 5.625.