Denormal Numbers - More Floating Point Madness!

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We have previously seen floating point numbers, and the logic that goes into how they can represent numbers with a fractional portion. However, there's one edge case I missed out in my original video, that merits a whole video of its own!

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Finally a good explanation for denormal FP number online! :)

nehavedam
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Amazing Video! I was working on a problem for an hour but then in literally 30 seconds you answered everything!

emiliomartineziii
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This is a very important (and interesting) topic for me and you have explained it exceedenly well. Now I need to go back to study for my upcoming exam :S - I thank you for this video; I owe you a lot!

black-t-shirt
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Nice video! :) It's worth mentioning another reason why denormal numbers are useful. Not only is it a way to cram in a few more numbers around zero; it also induces a useful property on the set of floating-point numbers: If the difference between two numbers, say x and y, is 0, then the numbers are equal (x-y = 0 -> x = y). This property fails on normalized-only floating-point numbers because the difference between the least and second-least positive normalized number is precisely the smallest denormal number, which would underflow to zero. But with denormals, it will never underflow.

gubgubbubbub
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Cheers man - sometimes you hear about something in a paper and just need a quick explainer. Glad to have found your video when I needed it. Also, are you from Singapore/Malaysia?

shashvatshukla
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Thank you for the video. For those who may be wondering how does this matter in real life (as in the working world ouside of academia and other schooling) I have encountered a problem with commercial real time programs where RTUs are reading a very tiny value close to but not zero. When the real time program polls the RTU and sees this, it causes the program to flag that value and ignore it based on how the product was developed. :( These tiny tiny tiny close to but not quite zero RTU values are not encountered often so when it is encountered people have to remember what is going on and how the program functions. Typically this involves a trouble call to the vendor and research on their part and ta-da De-normalized IEEE values... Too bad the RTUs are not programmed to round those tiny tiney tiny not quite zero values to zero so these problems don't pop up in the program :D Thank you once again :D

OStuxbird
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Though topic.. but explained clearly. thank you

studytimewithjency
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Wow awesome!!
I just saw the first video of floating point and entered to your channel and I saw this one also !
Can you please make a video about the things you told at the end of the video ?
Thanks!! :P

AlmogYosef
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So what if the number instead of 2^130 is 2^250 and it exceeds the bit limit for the mantissa?

Averageyoutubeenjoyer-pypl
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I'm confused. Why does the value -126 make the exponent go all zero? If the exponent of 2 is -126 then the offset exponent in the representation should be -126 + 127 = 1, so the binary representation should be shouldn't it? What's so special about -126? I love your video but the nature of this topic is itself pretty confusing I guess, so it would be great if you could suggest some reading material on this. 💖

amkhrjee
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Hi, thanks for sharing this topic.
Can you explain more detail why we have to make offset for the exponent part?
As I read somewhere, there is some error when we convert to binary but I don't know what it is

phucnguyenkimthanh
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Thank you for your video but I am so confused. 1.0 x 2^2 is equals to 0.1x2^3. and these two numbers can be representable but differently, how can I know which representation will be used?.

ahmetkarakartal
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Cool video! I was able to understand the binary to float but not the float to binary, At the extracting the mantissa you have your multiplication operations on both sides of the equal sign which to me from my knowledge of algebra says that both operations come to the same answer. to my understanding 2^-1 x 2 = 0.4 am i doing this wrong?

xzs
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Hi, i went to your website and realised that a Binary of actually is 0 in decimals. but since there is always a 1 imply before the mantissa, how can in Binary actually be 0 in decimal? Since 0(sign) * 2^[0-127] (Exponent) * [1+ ( mantissa), it will be 2^-127 * 1

ngkhai
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Still waiting for the third episode.. XD

htxdy
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What if you have a denormalized number with a fraction : 000101100011..1
How you deal with that ?

dorifourer
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