 For this example, I'm going to be converting a number from its floating point representation back into its real number representation. To do this, I'll start by placing all of the bits from my number into the floating point format so I can tell which bits are associated with the sign, the exponent, and the mantissa. So I have 8, 0, 0, c, and then four more blocks of 0s that won't affect the value of my number. I have a 1 in the sign bit, so I know my number is negative. For the rest of the problem, there's two different ways that I can approach this. One way is to say, okay, this is a denormalized number. That means I don't know what that first bit in my number actually is, so I can't just assume it's a 1 the way I did before. I'll have to use this first bit to tell me what the first bit in my number is. So I could write down 0.0011 times 2 negative, then take my exponent, negative 127. Alternatively, I could say, okay, I don't know what the leading bit in my number is, but I just want to leave everything as mantissa. I don't want to try putting one bit in front of the binary point. Then I'd have 0.00011 times 2 to the negative 126. So in this case, my exponent is one larger, but I don't place the first bit in front of the binary point. Both of these are in fact equal, and see that I can get from here to here just by moving my binary point over one place. Then I'd have 2 to the minus 127, and I'd have a zero in front of it, two zeros, one, one afterwards. So both of these methods work. They will both get me to the same place. So now I have to figure out what this number actually is. So I need to move my binary point over three more places to figure out what that first bit is. So that's negative 2 to the minus 130th, 128, 129, 130. I also have a 2 to the minus 131st. That goes same over here as well. So this is the value of my number. That's probably not terribly helpful for us. If you put that into a calculator, it will tell you that this is approximately equal to negative 1.102 times 10 to the minus 39. So this is a really small number, but it's smaller than what we could have gotten without using a denormalized number. We would have been limited by the 2 to the minus 127th that we would have had otherwise, and it would have been forced to assume the first bit here in this place was actually a 1 instead, but it's not. We were able to move it over a couple more places by using the denormalized notation instead. So we got a really, really small number.