64 bit double precision IEEE 754 binary floating point number 0 - 111 1110 0110 - 1001 1111 1100 1000 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:
0 - 111 1110 0110 - 1001 1111 1100 1000 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:
111 1110 0110


The last 52 bits contain the mantissa:
1001 1111 1100 1000 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.

111 1110 0110(2) =


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


1,024 + 512 + 256 + 128 + 64 + 32 + 0 + 0 + 4 + 2 + 0 =


1,024 + 512 + 256 + 128 + 64 + 32 + 4 + 2 =


2,022(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 = 2,022 - 1023 = 999


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 1100 1000 0000 0000 0000 0000 0000 0000 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 + 0 × 2-11 + 0 × 2-12 + 1 × 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 =


0.5 + 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 + 0 + 0.000 122 070 312 5 + 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 =


0.5 + 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 + 0.000 122 070 312 5 =


0.624 145 507 812 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 + 0.624 145 507 812 5) × 2999 =


1.624 145 507 812 5 × 2999 =


8 701 429 454 720 023 623 790 768 602 138 259 331 982 111 218 104 324 125 390 075 022 746 289 511 241 012 640 241 701 922 694 600 459 220 976 027 324 245 982 976 754 741 226 889 793 914 842 212 196 738 558 213 278 163 100 832 299 393 373 254 249 289 236 330 730 357 690 035 783 895 853 061 384 799 971 960 594 158 786 755 527 058 232 322 373 604 468 960 177 046 021 661 995 737 760 429 017 420 267 520

Conclusion:

0 - 111 1110 0110 - 1001 1111 1100 1000 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) =

8 701 429 454 720 023 623 790 768 602 138 259 331 982 111 218 104 324 125 390 075 022 746 289 511 241 012 640 241 701 922 694 600 459 220 976 027 324 245 982 976 754 741 226 889 793 914 842 212 196 738 558 213 278 163 100 832 299 393 373 254 249 289 236 330 730 357 690 035 783 895 853 061 384 799 971 960 594 158 786 755 527 058 232 322 373 604 468 960 177 046 021 661 995 737 760 429 017 420 267 520(10)

More operations of this kind:

0 - 111 1110 0110 - 1001 1111 1100 0111 1111 1111 1111 1111 1111 1111 1111 1111 1111 = ?

0 - 111 1110 0110 - 1001 1111 1100 1000 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)

0 - 111 1110 0110 - 1001 1111 1100 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 8 701 429 454 720 023 623 790 768 602 138 259 331 982 111 218 104 324 125 390 075 022 746 289 511 241 012 640 241 701 922 694 600 459 220 976 027 324 245 982 976 754 741 226 889 793 914 842 212 196 738 558 213 278 163 100 832 299 393 373 254 249 289 236 330 730 357 690 035 783 895 853 061 384 799 971 960 594 158 786 755 527 058 232 322 373 604 468 960 177 046 021 661 995 737 760 429 017 420 267 520 Nov 29 02:24 UTC (GMT)
0 - 100 0001 0101 - 0110 1111 0001 0101 0011 1111 0001 0000 1100 1010 1011 0010 0011 = 6 014 287.766 398 224 048 316 478 729 248 046 875 Nov 29 02:24 UTC (GMT)
1 - 110 0101 0000 - 1000 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = -49 640 200 897 257 972 404 079 097 978 727 269 628 811 002 290 009 621 118 112 111 610 068 844 667 781 371 868 551 182 958 595 724 101 592 211 045 887 143 229 965 176 367 845 502 194 845 326 383 513 570 366 373 534 645 597 332 511 391 744 Nov 29 02:21 UTC (GMT)
0 - 100 0001 0000 - 0001 1101 0001 0111 0111 0101 1100 1101 0101 0111 0011 0110 0100 = 145 966.920 328 999 985 940 754 413 604 736 328 125 Nov 29 02:20 UTC (GMT)
0 - 100 0000 0110 - 0010 0010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0101 = 145.199 999 999 999 960 209 606 797 434 389 591 217 041 015 625 Nov 29 02:16 UTC (GMT)
0 - 100 0000 0010 - 0101 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 10.5 Nov 29 02:15 UTC (GMT)
0 - 100 0001 1110 - 1010 1101 1100 1001 0110 1001 1010 1000 0111 1111 1111 1111 1100 = 3 605 312 724.249 998 092 651 367 187 5 Nov 29 02:13 UTC (GMT)
0 - 011 1111 0010 - 0110 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 = 0.000 167 846 679 687 500 081 315 162 936 412 832 550 558 960 065 245 628 356 933 593 75 Nov 29 02:12 UTC (GMT)
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 1100 0000 0011 1100 = 0 Nov 29 02:12 UTC (GMT)
0 - 100 0101 1010 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 = 2 475 880 078 570 760 549 798 248 448 Nov 29 02:12 UTC (GMT)
0 - 100 0001 1010 - 1111 0100 1101 1010 1101 0011 0110 1110 0010 1000 1000 1011 1100 = 262 592 155.442 449 450 492 858 886 718 75 Nov 29 02:12 UTC (GMT)
0 - 000 0000 0000 - 0000 0000 0000 0000 0000 0000 0000 0001 0000 0000 0000 0000 0000 = 0 Nov 29 02:10 UTC (GMT)
1 - 000 0100 1000 - 0010 1100 0011 1100 0010 1001 1111 1101 1111 1010 1000 1111 0100 = -0 Nov 29 02:09 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)