 This time, I've got a bigger, more complex number, 89.34375, but we're pretty much going to do the same thing. We'll convert this number into binary, convert that into normalized scientific notation, and then fill in all of the fields of our floating point format. So, first of all, I need to convert this number to binary. I'm going to try to do this in two parts. I'll start with the whole number part. So, 89 is 64 plus 25. So, I've got a 64 bit, no 32 bit, I'll have a 16 bit. This leaves me with nine more, so one zero's one. So, there's the 89. But you're probably looking at that fraction and thinking, I have no idea what to do with this. So, I'll use the multiplication method to figure out what those bits are. I will take this zero and put it up here and multiply this by two again. I will take this one and put it here. Now, since I didn't multiply my one, I've got a zero and I will put that in the third position. And I can see I've got a 0.75, that will just give me two more ones. So, I will just write those two down. That will be my binary representation of 89.34375. Next, I need to convert this to normalized scientific notation. In this case, my binary point needs to move over one, two, three, four, five, six places. So, I'll have an exponent of six. And this is a positive number. So, my sign bit will be a zero. My exponent is a six. So, six plus 127 is 133, which is the 128 bit plus five. So, I will copy that in for my exponent. And lastly, I need my mantissa. So, as usual, I will take everything that's after the binary point and copy that in for my mantissa. And then I will fill in the right-hand side until I've got my full 23 bits. I've used 11, so I'll need 12 more. And there's my binary floating point number. Again, I can convert this to hexadecimal by finding blocks of four and converting them. So, I've got four, two, B, two, B, zero, zero, zero. So, there's the hexadecimal equivalent of our binary floating point number.