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Can someone please help in highlighted part. Thanks

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0
what is problem there?
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Not getting 2^149 and 2^1074.
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Did you get this? I am also stuck on this. :(
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Yes. U still need explanation or got it resolved?
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Not yet understood :(
How those $2^{-126}$ and $2^{-149}$ came?
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smallest fractional form in 32 bit will be 0.00000.....1(23rd bit). Weight of this bit is 2^-23. Just used fractional form for denormalized case. You will get 2^-149
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I got

smallest fractional form in 32 bit will be 0.00000.....1(23rd bit). Weight of this bit is 2^-23. 

But. I did not get 

Just used fractional form for denormalized case. You will get 2^-149

I am interpreting it as you added -126 to -23 to get -149. But what is this -126?

  1. To get unbiased exponent in normalized numbers, we do $e-b$, where  $e$ is biased exponent and $b$ is bias. $b$ in IEEE 754 is 127. So it will be $e-127$ for normalized numbers. I know exponent in denormalized numbers is zero. So if we are to calculate unbiased exponent of denormalized numbers in similar way that in normalized numbers, it would have been $0-127=-127$, not $-126$.
  2. I just checked this link. In the table of summary section, it says value of positive denormalized real is $0.f\times 2^{(-b+1)}$, which will correctly give $2^{-149}$ . But, I didnt get from where that $-b\color{red}{+1}$ came. Especially that $+1$. 
  3. Also isnt denormalized numbers unbiased? Or they indeed are biased, just that their exponent is 0, more precisely a biased zero. 
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bro super thanks!!!!

Maaan you seems to be enlighted one!!!!  Did you just come up with that reasoning? Or you found it somewhere? Anyways thanks.
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That link only I put in earlier comment and asked doubt on it:

Just tell from where $-b\color{red}{+1}$ came on that page in summary table at the end

2
Denormalized numbers are used to fill the gap between 0 and smallest positive number in normalized form.

what is smallest positive number in normalized form? S=0 M =0 and E =1 so it will become 1.0 * 2 ^ -126

Now if number is smaller than this, it will be of form 0.M * 2 ^ -126

Hence when E=0 to make it -126 than +1 is added. i.e E- bias +1 = 0-127+1= -126.
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super thanks

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