64 bit double precision IEEE 754 binary floating point number 0 - 100 0000 1100 - 1001 1111 1111 1101 0101 0111 1000 0000 0100 1000 0000 0000 0000 converted to decimal base ten (double)

64 bit double precision IEEE 754 binary floating point 0 - 100 0000 1100 - 1001 1111 1111 1101 0101 0111 1000 0000 0100 1000 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 1100


The last 52 bits contain the mantissa:
1001 1111 1111 1101 0101 0111 1000 0000 0100 1000 0000 0000 0000

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

The exponent is allways a positive integer.

100 0000 1100(2) =


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


1,024 + 0 + 0 + 0 + 0 + 0 + 0 + 8 + 4 + 0 + 0 =


1,024 + 8 + 4 =


1,036(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 = 1,036 - 1023 = 13


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)

1001 1111 1111 1101 0101 0111 1000 0000 0100 1000 0000 0000 0000(2) =

1 × 2-1 + 0 × 2-2 + 0 × 2-3 + 1 × 2-4 + 1 × 2-5 + 1 × 2-6 + 1 × 2-7 + 1 × 2-8 + 1 × 2-9 + 1 × 2-10 + 1 × 2-11 + 1 × 2-12 + 1 × 2-13 + 1 × 2-14 + 0 × 2-15 + 1 × 2-16 + 0 × 2-17 + 1 × 2-18 + 0 × 2-19 + 1 × 2-20 + 0 × 2-21 + 1 × 2-22 + 1 × 2-23 + 1 × 2-24 + 1 × 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 + 1 × 2-34 + 0 × 2-35 + 0 × 2-36 + 1 × 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 =


0.5 + 0 + 0 + 0.062 5 + 0.031 25 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0 + 0.000 015 258 789 062 5 + 0 + 0.000 003 814 697 265 625 + 0 + 0.000 000 953 674 316 406 25 + 0 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 + 0.000 000 059 604 644 775 390 625 + 0.000 000 029 802 322 387 695 312 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0.000 000 000 058 207 660 913 467 407 226 562 5 + 0 + 0 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 =


0.5 + 0.062 5 + 0.031 25 + 0.015 625 + 0.007 812 5 + 0.003 906 25 + 0.001 953 125 + 0.000 976 562 5 + 0.000 488 281 25 + 0.000 244 140 625 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0.000 015 258 789 062 5 + 0.000 003 814 697 265 625 + 0.000 000 953 674 316 406 25 + 0.000 000 238 418 579 101 562 5 + 0.000 000 119 209 289 550 781 25 + 0.000 000 059 604 644 775 390 625 + 0.000 000 029 802 322 387 695 312 5 + 0.000 000 000 058 207 660 913 467 407 226 562 5 + 0.000 000 000 007 275 957 614 183 425 903 320 312 5 =


0.624 959 439 104 713 965 207 338 333 129 882 812 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)0 × (1 + 0.624 959 439 104 713 965 207 338 333 129 882 812 5) × 213 =


1.624 959 439 104 713 965 207 338 333 129 882 812 5 × 213 =


13 311.667 725 145 816 802 978 515 625

0 - 100 0000 1100 - 1001 1111 1111 1101 0101 0111 1000 0000 0100 1000 0000 0000 0000 converted from 64 bit double precision IEEE 754 binary floating point to base ten decimal system (double) =
13 311.667 725 145 816 802 978 515 625(10)

More operations of this kind:

0 - 100 0000 1100 - 1001 1111 1111 1101 0101 0111 1000 0000 0100 0111 1111 1111 1111 = ?

0 - 100 0000 1100 - 1001 1111 1111 1101 0101 0111 1000 0000 0100 1000 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)

0 - 100 0000 1100 - 1001 1111 1111 1101 0101 0111 1000 0000 0100 1000 0000 0000 0000 = ? Jun 13 23:42 UTC (GMT)
1 - 011 1111 1000 - 0100 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = ? Jun 13 23:42 UTC (GMT)
0 - 000 0000 0000 - 1100 0001 1101 1111 1111 1111 1111 1111 1111 1111 1111 1111 1001 = ? Jun 13 23:41 UTC (GMT)
1 - 011 1111 1001 - 1011 1111 1001 0110 0000 0010 0110 0100 0101 1110 0100 1110 0101 = ? Jun 13 23:39 UTC (GMT)
0 - 100 0000 0111 - 0010 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1100 = ? Jun 13 23:38 UTC (GMT)
1 - 000 0100 0100 - 1100 1011 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0111 = ? Jun 13 23:38 UTC (GMT)
0 - 000 1111 1101 - 0110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 = ? Jun 13 23:37 UTC (GMT)
0 - 000 1100 0000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = ? Jun 13 23:37 UTC (GMT)
1 - 100 0000 1000 - 0100 0010 0111 0111 0110 0010 1111 0101 0111 0000 0101 1000 0000 = ? Jun 13 23:37 UTC (GMT)
0 - 100 0000 0100 - 1111 1100 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 = ? Jun 13 23:36 UTC (GMT)
0 - 011 1111 1111 - 0101 0011 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 = ? Jun 13 23:35 UTC (GMT)
0 - 100 0001 1000 - 1001 0011 1100 1011 1111 1110 1000 0101 1101 0010 0100 0101 0010 = ? Jun 13 23:33 UTC (GMT)
1 - 100 0000 0011 - 0010 0001 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = ? Jun 13 23:33 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)