64 bit double precision IEEE 754 binary floating point number 1 - 100 0000 0010 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 converted to decimal base ten (double)

How to convert 64 bit double precision IEEE 754 binary floating point:
1 - 100 0000 0010 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
to decimal system (base ten)

1. Identify the elements that make up the binary representation of the number:

First bit (the leftmost) indicates the sign,
1 = negative, 0 = positive.


The next 11 bits contain the exponent:
100 0000 0010


The last 52 bits contain the mantissa:
0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

2. Convert the exponent from binary (base 2) to decimal (base 10):

The exponent is allways a positive integer.

100 0000 0010(2) =


1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 0 × 23 + 0 × 22 + 1 × 21 + 0 × 20 =


1,024 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 2 + 0 =


1,024 + 2 =


1,026(10)

3. Adjust the exponent.

Subtract the excess bits: 2(11 - 1) - 1 = 1023, that is due to the 11 bit excess/bias notation:

Exponent adjusted = 1,026 - 1023 = 3


4. Convert the mantissa from binary (base 2) to decimal (base 10):

Mantissa represents the number's fractional part (the excess beyond the number's integer part, comma delimited)

0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000(2) =

0 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 0 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 0 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 2-15 + 0 × 2-16 + 0 × 2-17 + 0 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 0 × 2-23 + 0 × 2-24 + 0 × 2-25 + 0 × 2-26 + 0 × 2-27 + 0 × 2-28 + 0 × 2-29 + 0 × 2-30 + 0 × 2-31 + 0 × 2-32 + 0 × 2-33 + 0 × 2-34 + 0 × 2-35 + 0 × 2-36 + 0 × 2-37 + 0 × 2-38 + 0 × 2-39 + 0 × 2-40 + 0 × 2-41 + 0 × 2-42 + 0 × 2-43 + 0 × 2-44 + 0 × 2-45 + 0 × 2-46 + 0 × 2-47 + 0 × 2-48 + 0 × 2-49 + 0 × 2-50 + 0 × 2-51 + 0 × 2-52 =


0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0(10)

5. Put all the numbers into expression to calculate the double precision floating point decimal value:

(-1)Sign × (1 + Mantissa) × 2(Exponent adjusted) =


(-1)1 × (1 + 0) × 23 =


-1 × 23 =


-8

Conclusion:

1 - 100 0000 0010 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000
converted from
64 bit double precision IEEE 754 binary floating point
to
base ten decimal system (double) =

-8(10)

More operations of this kind:

1 - 100 0000 0010 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 = ?


Convert 64 bit double precision IEEE 754 floating point standard binary numbers to base ten decimal system (double)

64 bit double precision IEEE 754 binary floating point standard representation of numbers requires three building blocks: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

Latest 64 bit double precision IEEE 754 floating point binary standard numbers converted to decimal base ten (double)

1 - 100 0000 0010 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -8 Nov 29 10:29 UTC (GMT)
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 1000 0000 0010 1000 0100 0000 = 0 Nov 29 10:29 UTC (GMT)
0 - 011 1111 1111 - 0001 1100 0110 1010 0111 1110 1111 1001 1101 1011 0010 0010 0111 = 1.110 999 999 999 998 655 297 872 574 010 398 238 897 323 608 398 437 5 Nov 29 10:28 UTC (GMT)
0 - 100 0000 0001 - 0100 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0100 = 5.099 999 999 999 997 868 371 792 719 699 442 386 627 197 265 625 Nov 29 10:28 UTC (GMT)
0 - 011 1111 1100 - 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 = 0.149 999 999 999 999 994 448 884 876 874 217 297 881 841 659 545 898 437 5 Nov 29 10:28 UTC (GMT)
1 - 100 0000 0000 - 1010 1001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -3.320 312 5 Nov 29 10:27 UTC (GMT)
0 - 000 0000 0000 - 1100 1010 0010 1101 1101 0000 0000 0000 0000 0000 0000 0000 0000 = 0 Nov 29 10:24 UTC (GMT)
0 - 100 0000 0100 - 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 34 Nov 29 10:23 UTC (GMT)
1 - 010 0100 1101 - 1100 1100 0111 0001 1000 1011 1011 0011 0001 0011 0010 1010 0111 = -0 Nov 29 10:20 UTC (GMT)
0 - 100 0000 0000 - 1110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0111 0000 = 3.800 000 000 000 004 263 256 414 560 601 115 226 745 605 468 75 Nov 29 10:18 UTC (GMT)
1 - 000 0000 0000 - 0010 1000 0010 0010 0100 1001 0100 1010 1001 0101 0010 1010 1001 = -0 Nov 29 10:17 UTC (GMT)
0 - 011 1110 0111 - 0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 = 0.000 000 089 406 967 163 085 924 265 110 199 151 557 202 057 460 926 880 594 342 947 006 225 585 937 5 Nov 29 10:15 UTC (GMT)
1 - 100 0000 0011 - 1111 1010 0011 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -31.636 718 75 Nov 29 10:15 UTC (GMT)
All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

How to convert numbers from 64 bit double precision IEEE 754 binary floating point standard to decimal system in base 10

Follow the steps below to convert a number from 64 bit double precision IEEE 754 binary floating point representation to base 10 decimal system:

  • 1. Identify the elements that make up the binary representation of the number:
    First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
    The next 11 bits contain the exponent.
    The last 52 bits contain the mantissa.
  • 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10).
  • 3. Adjust the exponent, subtract the excess bits, 2(11 - 1) - 1 = 1,023, that is due to the 11 bit excess/bias notation.
  • 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10).
  • 5. Put all the numbers into expression to calculate the double precision floating point decimal value:
    (-1)Sign × (1 + Mantissa) × 2(Exponent adjusted)

Example: convert the number 1 - 100 0011 1101 - 1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000 from 64 bit double precision IEEE 754 binary floating point system to base ten decimal (double):

  • 1. Identify the elements that make up the binary representation of the number:
    First bit (leftmost) indicates the sign, 1 = negative, 0 = pozitive.
    The next 11 bits contain the exponent: 100 0011 1101
    The last 52 bits contain the mantissa:
    1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000
  • 2. Convert the exponent, that is allways a positive integer, from binary (base 2) to decimal (base 10):
    100 0011 1101(2) =
    1 × 210 + 0 × 29 + 0 × 28 + 0 × 27 + 0 × 26 + 1 × 25 + 1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 =
    1,024 + 0 + 0 + 0 + 0 + 32 + 16 + 8 + 4 + 0 + 1 =
    1,024 + 32 + 16 + 8 + 4 + 1 =
    1,085(10)
  • 3. Adjust the exponent, subtract the excess bits, 2(11 - 1) - 1 = 1,023, that is due to the 11 bit excess/bias notation:
    Exponent adjusted = 1,085 - 1,023 = 62
  • 4. Convert the mantissa, that represents the number's fractional part (the excess beyond the number's integer part, comma delimited), from binary (base 2) to decimal (base 10):
    1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000(2) =
    1 × 2-1 + 0 × 2-2 + 0 × 2-3 + 0 × 2-4 + 0 × 2-5 + 0 × 2-6 + 0 × 2-7 + 0 × 2-8 + 0 × 2-9 + 0 × 2-10 + 1 × 2-11 + 0 × 2-12 + 0 × 2-13 + 0 × 2-14 + 0 × 2-15 + 1 × 2-16 + 0 × 2-17 + 1 × 2-18 + 0 × 2-19 + 0 × 2-20 + 0 × 2-21 + 0 × 2-22 + 0 × 2-23 + 0 × 2-24 + 0 × 2-25 + 1 × 2-26 + 0 × 2-27 + 0 × 2-28 + 1 × 2-29 + 1 × 2-30 + 1 × 2-31 + 0 × 2-32 + 0 × 2-33 + 0 × 2-34 + 0 × 2-35 + 0 × 2-36 + 0 × 2-37 + 1 × 2-38 + 0 × 2-39 + 0 × 2-40 + 0 × 2-41 + 0 × 2-42 + 0 × 2-43 + 0 × 2-44 + 1 × 2-45 + 0 × 2-46 + 1 × 2-47 + 0 × 2-48 + 1 × 2-49 + 0 × 2-50 + 0 × 2-51 + 0 × 2-52 =
    0.5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 488 281 25 + 0 + 0 + 0 + 0 + 0.000 015 258 789 062 5 + 0 + 0.000 003 814 697 265 625 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 014 901 161 193 847 656 25 + 0 + 0 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 0 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 + 0 + 0 + 0 =
    0.5 + 0.000 488 281 25 + 0.000 015 258 789 062 5 + 0.000 003 814 697 265 625 + 0.000 000 014 901 161 193 847 656 25 + 0.000 000 001 862 645 149 230 957 031 25 + 0.000 000 000 931 322 574 615 478 515 625 + 0.000 000 000 465 661 287 307 739 257 812 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 000 028 421 709 430 404 007 434 844 970 703 125 + 0.000 000 000 000 007 105 427 357 601 001 858 711 242 675 781 25 + 0.000 000 000 000 001 776 356 839 400 250 464 677 810 668 945 312 5 =
    0.500 507 372 900 793 612 302 550 172 898 918 390 274 047 851 562 5(10)
  • 5. Put all the numbers into expression to calculate the double precision floating point decimal value:
    (-1)Sign × (1 + Mantissa) × 2(Exponent adjusted) =
    (-1)1 × (1 + 0.500 507 372 900 793 612 302 550 172 898 918 390 274 047 851 562 5) × 262 =
    -1.500 507 372 900 793 612 302 550 172 898 918 390 274 047 851 562 5 × 262 =
    -6 919 868 872 153 800 704(10)
  • 1 - 100 0011 1101 - 1000 0000 0010 0001 0100 0000 0100 1110 0000 0100 0000 1010 1000 converted from 64 bit double precision IEEE 754 binary floating point representation to a decimal number (float) in decimal system (in base 10) = -6 919 868 872 153 800 704(10)