64 bit double precision IEEE 754 binary floating point number 0 - 011 0110 1010 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 0110 0110 0110 1000 converted to decimal base ten (double)

64 bit double precision IEEE 754 binary floating point 0 - 011 0110 1010 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 0110 0110 0110 1000 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:
011 0110 1010


The last 52 bits contain the mantissa:
1100 1100 1100 1100 1100 1100 1100 1100 1100 0110 0110 0110 1000

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

The exponent is allways a positive integer.

011 0110 1010(2) =


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


0 + 512 + 256 + 0 + 64 + 32 + 0 + 8 + 0 + 2 + 0 =


512 + 256 + 64 + 32 + 8 + 2 =


874(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 = 874 - 1023 = -149


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)

1100 1100 1100 1100 1100 1100 1100 1100 1100 0110 0110 0110 1000(2) =

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


0.5 + 0.25 + 0 + 0 + 0.031 25 + 0.015 625 + 0 + 0 + 0.001 953 125 + 0.000 976 562 5 + 0 + 0 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0 + 0 + 0.000 007 629 394 531 25 + 0.000 003 814 697 265 625 + 0 + 0 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0 + 0 + 0.000 000 029 802 322 387 695 312 5 + 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 + 0 + 0.000 000 000 116 415 321 826 934 814 453 125 + 0.000 000 000 058 207 660 913 467 407 226 562 5 + 0 + 0 + 0 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0 + 0 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0 + 0 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 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 + 0.25 + 0.031 25 + 0.015 625 + 0.001 953 125 + 0.000 976 562 5 + 0.000 122 070 312 5 + 0.000 061 035 156 25 + 0.000 007 629 394 531 25 + 0.000 003 814 697 265 625 + 0.000 000 476 837 158 203 125 + 0.000 000 238 418 579 101 562 5 + 0.000 000 029 802 322 387 695 312 5 + 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 + 0.000 000 000 116 415 321 826 934 814 453 125 + 0.000 000 000 058 207 660 913 467 407 226 562 5 + 0.000 000 000 003 637 978 807 091 712 951 660 156 25 + 0.000 000 000 001 818 989 403 545 856 475 830 078 125 + 0.000 000 000 000 227 373 675 443 232 059 478 759 765 625 + 0.000 000 000 000 113 686 837 721 616 029 739 379 882 812 5 + 0.000 000 000 000 014 210 854 715 202 003 717 422 485 351 562 5 + 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 =


0.799 999 999 994 179 589 180 021 139 327 436 685 562 133 789 062 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 + 0.799 999 999 994 179 589 180 021 139 327 436 685 562 133 789 062 5) × 2-149 =


1.799 999 999 994 179 589 180 021 139 327 436 685 562 133 789 062 5 × 2-149 =


0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 002 522 337 235 776 514 594 918 937 323 154 797 740 193 025 113 343 041 181 060 310 200 510 296 301 354 541 111 796 108 604 098 465 522 213 268 514 705 071 521 309 065 474 270 028 062 164 783 477 783 203 125

0 - 011 0110 1010 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 0110 0110 0110 1000 converted from 64 bit double precision IEEE 754 binary floating point to base ten decimal system (double) =
0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 002 522 337 235 776 514 594 918 937 323 154 797 740 193 025 113 343 041 181 060 310 200 510 296 301 354 541 111 796 108 604 098 465 522 213 268 514 705 071 521 309 065 474 270 028 062 164 783 477 783 203 125(10)

More operations of this kind:

0 - 011 0110 1010 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 0110 0110 0110 0111 = ?

0 - 011 0110 1010 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 0110 0110 0110 1001 = ?


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 - 011 0110 1010 - 1100 1100 1100 1100 1100 1100 1100 1100 1100 0110 0110 0110 1000 = ? Apr 14 11:25 UTC (GMT)
0 - 011 1111 1111 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 0000 0000 0001 = ? Apr 14 11:25 UTC (GMT)
0 - 100 0111 1111 - 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0110 = ? Apr 14 11:25 UTC (GMT)
0 - 011 1010 0110 - 0110 0001 1010 1111 0001 1101 0110 1010 1000 0011 0000 0111 1000 = ? Apr 14 11:24 UTC (GMT)
0 - 101 1100 0110 - 0110 1110 0110 0011 0000 1100 0111 0011 0101 1010 1010 0101 0110 = ? Apr 14 11:24 UTC (GMT)
0 - 100 0000 0000 - 0000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1010 = ? Apr 14 11:23 UTC (GMT)
0 - 100 0000 1000 - 1011 1110 0101 1001 0100 0110 0000 0000 0000 0000 0000 0000 0101 = ? Apr 14 11:22 UTC (GMT)
0 - 001 0000 0111 - 1101 1111 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = ? Apr 14 11:22 UTC (GMT)
0 - 000 0111 1110 - 1111 1111 1111 1111 1111 1110 0000 0000 0000 0000 0000 0000 0000 = ? Apr 14 11:21 UTC (GMT)
0 - 101 0000 0100 - 0000 0001 0100 0111 0111 0000 0000 0000 0000 0000 0000 0000 0001 = ? Apr 14 11:21 UTC (GMT)
1 - 000 0000 1101 - 1101 1010 1010 0011 1101 0111 0000 1010 0011 1101 0110 0000 0000 = ? Apr 14 11:21 UTC (GMT)
0 - 100 0001 1101 - 0111 0000 0000 0000 1100 1101 1101 1110 0001 0010 1111 1111 1110 = ? Apr 14 11:21 UTC (GMT)
0 - 101 0101 0100 - 1010 1001 0101 0101 0101 0010 1010 1010 1010 1010 1001 0101 0110 = ? Apr 14 11:20 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)