2 votes 2 votes What is the largest mantissa we can store in floating-point format if the size of the mantissa field is m-bit and exponent field is e-bit? The mantissa is normalized and has an implied $1$ in the left of the point. Normalized form of mantissa is 1.M Digital Logic gateforum-test-series digital-logic floating-point-representation + – Gupta731 asked Dec 24, 2018 edited Mar 12, 2019 by ajaysoni1924 Gupta731 853 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Shobhit Joshi commented Dec 24, 2018 reply Follow Share A ? 0 votes 0 votes Gupta731 commented Dec 24, 2018 reply Follow Share Answer is $A$ 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes Mantissa = $1.m$ $m=\underbrace{111..111111}_{m\,1^{'s}}=2^{m}-1$ $0.m = (2^m-1)*2^{-m}=1-2^{-m}$ Mantissa = $1.m=1+(1-2^{-m})=(2^{m+1}-1)\times2^{-m}$ So, answer should be $(A)$ Correct me if i'm wrong Shobhit Joshi answered Dec 24, 2018 Shobhit Joshi comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Gupta731 commented Dec 24, 2018 reply Follow Share Thanks, got it 0 votes 0 votes shashank joshi commented May 6, 2020 i edited by shashank joshi May 6, 2020 reply Follow Share please clear my doubt in second step $0.m=(2^m -1 ) *2^{-m} =1-2^{-m}$ $2^{-m}$ comes from ? if we shift ' . ' to left of m then it should be $2^{m}$ ... 0 votes 0 votes reboot commented Aug 15, 2020 reply Follow Share because you can write 0.111 as 111 * $2^{-3}$ 1 votes 1 votes Please log in or register to add a comment.