50.569 600 888 243 378 733 477 577 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 50.569 600 888 243 378 733 477 577(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
50.569 600 888 243 378 733 477 577(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 50.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

50(10) =

11 0010(2)


3. Convert to binary (base 2) the fractional part: 0.569 600 888 243 378 733 477 577.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.569 600 888 243 378 733 477 577 × 2 = 1 + 0.139 201 776 486 757 466 955 154;
  • 2) 0.139 201 776 486 757 466 955 154 × 2 = 0 + 0.278 403 552 973 514 933 910 308;
  • 3) 0.278 403 552 973 514 933 910 308 × 2 = 0 + 0.556 807 105 947 029 867 820 616;
  • 4) 0.556 807 105 947 029 867 820 616 × 2 = 1 + 0.113 614 211 894 059 735 641 232;
  • 5) 0.113 614 211 894 059 735 641 232 × 2 = 0 + 0.227 228 423 788 119 471 282 464;
  • 6) 0.227 228 423 788 119 471 282 464 × 2 = 0 + 0.454 456 847 576 238 942 564 928;
  • 7) 0.454 456 847 576 238 942 564 928 × 2 = 0 + 0.908 913 695 152 477 885 129 856;
  • 8) 0.908 913 695 152 477 885 129 856 × 2 = 1 + 0.817 827 390 304 955 770 259 712;
  • 9) 0.817 827 390 304 955 770 259 712 × 2 = 1 + 0.635 654 780 609 911 540 519 424;
  • 10) 0.635 654 780 609 911 540 519 424 × 2 = 1 + 0.271 309 561 219 823 081 038 848;
  • 11) 0.271 309 561 219 823 081 038 848 × 2 = 0 + 0.542 619 122 439 646 162 077 696;
  • 12) 0.542 619 122 439 646 162 077 696 × 2 = 1 + 0.085 238 244 879 292 324 155 392;
  • 13) 0.085 238 244 879 292 324 155 392 × 2 = 0 + 0.170 476 489 758 584 648 310 784;
  • 14) 0.170 476 489 758 584 648 310 784 × 2 = 0 + 0.340 952 979 517 169 296 621 568;
  • 15) 0.340 952 979 517 169 296 621 568 × 2 = 0 + 0.681 905 959 034 338 593 243 136;
  • 16) 0.681 905 959 034 338 593 243 136 × 2 = 1 + 0.363 811 918 068 677 186 486 272;
  • 17) 0.363 811 918 068 677 186 486 272 × 2 = 0 + 0.727 623 836 137 354 372 972 544;
  • 18) 0.727 623 836 137 354 372 972 544 × 2 = 1 + 0.455 247 672 274 708 745 945 088;
  • 19) 0.455 247 672 274 708 745 945 088 × 2 = 0 + 0.910 495 344 549 417 491 890 176;
  • 20) 0.910 495 344 549 417 491 890 176 × 2 = 1 + 0.820 990 689 098 834 983 780 352;
  • 21) 0.820 990 689 098 834 983 780 352 × 2 = 1 + 0.641 981 378 197 669 967 560 704;
  • 22) 0.641 981 378 197 669 967 560 704 × 2 = 1 + 0.283 962 756 395 339 935 121 408;
  • 23) 0.283 962 756 395 339 935 121 408 × 2 = 0 + 0.567 925 512 790 679 870 242 816;
  • 24) 0.567 925 512 790 679 870 242 816 × 2 = 1 + 0.135 851 025 581 359 740 485 632;
  • 25) 0.135 851 025 581 359 740 485 632 × 2 = 0 + 0.271 702 051 162 719 480 971 264;
  • 26) 0.271 702 051 162 719 480 971 264 × 2 = 0 + 0.543 404 102 325 438 961 942 528;
  • 27) 0.543 404 102 325 438 961 942 528 × 2 = 1 + 0.086 808 204 650 877 923 885 056;
  • 28) 0.086 808 204 650 877 923 885 056 × 2 = 0 + 0.173 616 409 301 755 847 770 112;
  • 29) 0.173 616 409 301 755 847 770 112 × 2 = 0 + 0.347 232 818 603 511 695 540 224;
  • 30) 0.347 232 818 603 511 695 540 224 × 2 = 0 + 0.694 465 637 207 023 391 080 448;
  • 31) 0.694 465 637 207 023 391 080 448 × 2 = 1 + 0.388 931 274 414 046 782 160 896;
  • 32) 0.388 931 274 414 046 782 160 896 × 2 = 0 + 0.777 862 548 828 093 564 321 792;
  • 33) 0.777 862 548 828 093 564 321 792 × 2 = 1 + 0.555 725 097 656 187 128 643 584;
  • 34) 0.555 725 097 656 187 128 643 584 × 2 = 1 + 0.111 450 195 312 374 257 287 168;
  • 35) 0.111 450 195 312 374 257 287 168 × 2 = 0 + 0.222 900 390 624 748 514 574 336;
  • 36) 0.222 900 390 624 748 514 574 336 × 2 = 0 + 0.445 800 781 249 497 029 148 672;
  • 37) 0.445 800 781 249 497 029 148 672 × 2 = 0 + 0.891 601 562 498 994 058 297 344;
  • 38) 0.891 601 562 498 994 058 297 344 × 2 = 1 + 0.783 203 124 997 988 116 594 688;
  • 39) 0.783 203 124 997 988 116 594 688 × 2 = 1 + 0.566 406 249 995 976 233 189 376;
  • 40) 0.566 406 249 995 976 233 189 376 × 2 = 1 + 0.132 812 499 991 952 466 378 752;
  • 41) 0.132 812 499 991 952 466 378 752 × 2 = 0 + 0.265 624 999 983 904 932 757 504;
  • 42) 0.265 624 999 983 904 932 757 504 × 2 = 0 + 0.531 249 999 967 809 865 515 008;
  • 43) 0.531 249 999 967 809 865 515 008 × 2 = 1 + 0.062 499 999 935 619 731 030 016;
  • 44) 0.062 499 999 935 619 731 030 016 × 2 = 0 + 0.124 999 999 871 239 462 060 032;
  • 45) 0.124 999 999 871 239 462 060 032 × 2 = 0 + 0.249 999 999 742 478 924 120 064;
  • 46) 0.249 999 999 742 478 924 120 064 × 2 = 0 + 0.499 999 999 484 957 848 240 128;
  • 47) 0.499 999 999 484 957 848 240 128 × 2 = 0 + 0.999 999 998 969 915 696 480 256;
  • 48) 0.999 999 998 969 915 696 480 256 × 2 = 1 + 0.999 999 997 939 831 392 960 512;
  • 49) 0.999 999 997 939 831 392 960 512 × 2 = 1 + 0.999 999 995 879 662 785 921 024;
  • 50) 0.999 999 995 879 662 785 921 024 × 2 = 1 + 0.999 999 991 759 325 571 842 048;
  • 51) 0.999 999 991 759 325 571 842 048 × 2 = 1 + 0.999 999 983 518 651 143 684 096;
  • 52) 0.999 999 983 518 651 143 684 096 × 2 = 1 + 0.999 999 967 037 302 287 368 192;
  • 53) 0.999 999 967 037 302 287 368 192 × 2 = 1 + 0.999 999 934 074 604 574 736 384;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.569 600 888 243 378 733 477 577(10) =

0.1001 0001 1101 0001 0101 1101 0010 0010 1100 0111 0010 0001 1111 1(2)

5. Positive number before normalization:

50.569 600 888 243 378 733 477 577(10) =

11 0010.1001 0001 1101 0001 0101 1101 0010 0010 1100 0111 0010 0001 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left, so that only one non zero digit remains to the left of it:


50.569 600 888 243 378 733 477 577(10) =

11 0010.1001 0001 1101 0001 0101 1101 0010 0010 1100 0111 0010 0001 1111 1(2) =

11 0010.1001 0001 1101 0001 0101 1101 0010 0010 1100 0111 0010 0001 1111 1(2) × 20 =

1.1001 0100 1000 1110 1000 1010 1110 1001 0001 0110 0011 1001 0000 1111 11(2) × 25


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): 5

Mantissa (not normalized):
1.1001 0100 1000 1110 1000 1010 1110 1001 0001 0110 0011 1001 0000 1111 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

5 + 2(11-1) - 1 =

(5 + 1 023)(10) =

1 028(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 028 ÷ 2 = 514 + 0;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1028(10) =

100 0000 0100(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1001 0100 1000 1110 1000 1010 1110 1001 0001 0110 0011 1001 0000 11 1111 =

1001 0100 1000 1110 1000 1010 1110 1001 0001 0110 0011 1001 0000


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0100

Mantissa (52 bits) =
1001 0100 1000 1110 1000 1010 1110 1001 0001 0110 0011 1001 0000


Decimal number 50.569 600 888 243 378 733 477 577 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0100 - 1001 0100 1000 1110 1000 1010 1110 1001 0001 0110 0011 1001 0000

» Convert decimal 50.569 600 888 243 378 733 477 649 to 64 bit double precision IEEE 754 binary floating point representation standard

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

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

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100