50.569 600 888 243 378 733 477 584 319 189 127 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 50.569 600 888 243 378 733 477 584 319 189 127(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
50.569 600 888 243 378 733 477 584 319 189 127(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: 0.569 600 888 243 378 733 477 584 319 189 127.

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) 0.569 600 888 243 378 733 477 584 319 189 127 × 2 = 1 + 0.139 201 776 486 757 466 955 168 638 378 254;
  • 2) 0.139 201 776 486 757 466 955 168 638 378 254 × 2 = 0 + 0.278 403 552 973 514 933 910 337 276 756 508;
  • 3) 0.278 403 552 973 514 933 910 337 276 756 508 × 2 = 0 + 0.556 807 105 947 029 867 820 674 553 513 016;
  • 4) 0.556 807 105 947 029 867 820 674 553 513 016 × 2 = 1 + 0.113 614 211 894 059 735 641 349 107 026 032;
  • 5) 0.113 614 211 894 059 735 641 349 107 026 032 × 2 = 0 + 0.227 228 423 788 119 471 282 698 214 052 064;
  • 6) 0.227 228 423 788 119 471 282 698 214 052 064 × 2 = 0 + 0.454 456 847 576 238 942 565 396 428 104 128;
  • 7) 0.454 456 847 576 238 942 565 396 428 104 128 × 2 = 0 + 0.908 913 695 152 477 885 130 792 856 208 256;
  • 8) 0.908 913 695 152 477 885 130 792 856 208 256 × 2 = 1 + 0.817 827 390 304 955 770 261 585 712 416 512;
  • 9) 0.817 827 390 304 955 770 261 585 712 416 512 × 2 = 1 + 0.635 654 780 609 911 540 523 171 424 833 024;
  • 10) 0.635 654 780 609 911 540 523 171 424 833 024 × 2 = 1 + 0.271 309 561 219 823 081 046 342 849 666 048;
  • 11) 0.271 309 561 219 823 081 046 342 849 666 048 × 2 = 0 + 0.542 619 122 439 646 162 092 685 699 332 096;
  • 12) 0.542 619 122 439 646 162 092 685 699 332 096 × 2 = 1 + 0.085 238 244 879 292 324 185 371 398 664 192;
  • 13) 0.085 238 244 879 292 324 185 371 398 664 192 × 2 = 0 + 0.170 476 489 758 584 648 370 742 797 328 384;
  • 14) 0.170 476 489 758 584 648 370 742 797 328 384 × 2 = 0 + 0.340 952 979 517 169 296 741 485 594 656 768;
  • 15) 0.340 952 979 517 169 296 741 485 594 656 768 × 2 = 0 + 0.681 905 959 034 338 593 482 971 189 313 536;
  • 16) 0.681 905 959 034 338 593 482 971 189 313 536 × 2 = 1 + 0.363 811 918 068 677 186 965 942 378 627 072;
  • 17) 0.363 811 918 068 677 186 965 942 378 627 072 × 2 = 0 + 0.727 623 836 137 354 373 931 884 757 254 144;
  • 18) 0.727 623 836 137 354 373 931 884 757 254 144 × 2 = 1 + 0.455 247 672 274 708 747 863 769 514 508 288;
  • 19) 0.455 247 672 274 708 747 863 769 514 508 288 × 2 = 0 + 0.910 495 344 549 417 495 727 539 029 016 576;
  • 20) 0.910 495 344 549 417 495 727 539 029 016 576 × 2 = 1 + 0.820 990 689 098 834 991 455 078 058 033 152;
  • 21) 0.820 990 689 098 834 991 455 078 058 033 152 × 2 = 1 + 0.641 981 378 197 669 982 910 156 116 066 304;
  • 22) 0.641 981 378 197 669 982 910 156 116 066 304 × 2 = 1 + 0.283 962 756 395 339 965 820 312 232 132 608;
  • 23) 0.283 962 756 395 339 965 820 312 232 132 608 × 2 = 0 + 0.567 925 512 790 679 931 640 624 464 265 216;
  • 24) 0.567 925 512 790 679 931 640 624 464 265 216 × 2 = 1 + 0.135 851 025 581 359 863 281 248 928 530 432;
  • 25) 0.135 851 025 581 359 863 281 248 928 530 432 × 2 = 0 + 0.271 702 051 162 719 726 562 497 857 060 864;
  • 26) 0.271 702 051 162 719 726 562 497 857 060 864 × 2 = 0 + 0.543 404 102 325 439 453 124 995 714 121 728;
  • 27) 0.543 404 102 325 439 453 124 995 714 121 728 × 2 = 1 + 0.086 808 204 650 878 906 249 991 428 243 456;
  • 28) 0.086 808 204 650 878 906 249 991 428 243 456 × 2 = 0 + 0.173 616 409 301 757 812 499 982 856 486 912;
  • 29) 0.173 616 409 301 757 812 499 982 856 486 912 × 2 = 0 + 0.347 232 818 603 515 624 999 965 712 973 824;
  • 30) 0.347 232 818 603 515 624 999 965 712 973 824 × 2 = 0 + 0.694 465 637 207 031 249 999 931 425 947 648;
  • 31) 0.694 465 637 207 031 249 999 931 425 947 648 × 2 = 1 + 0.388 931 274 414 062 499 999 862 851 895 296;
  • 32) 0.388 931 274 414 062 499 999 862 851 895 296 × 2 = 0 + 0.777 862 548 828 124 999 999 725 703 790 592;
  • 33) 0.777 862 548 828 124 999 999 725 703 790 592 × 2 = 1 + 0.555 725 097 656 249 999 999 451 407 581 184;
  • 34) 0.555 725 097 656 249 999 999 451 407 581 184 × 2 = 1 + 0.111 450 195 312 499 999 998 902 815 162 368;
  • 35) 0.111 450 195 312 499 999 998 902 815 162 368 × 2 = 0 + 0.222 900 390 624 999 999 997 805 630 324 736;
  • 36) 0.222 900 390 624 999 999 997 805 630 324 736 × 2 = 0 + 0.445 800 781 249 999 999 995 611 260 649 472;
  • 37) 0.445 800 781 249 999 999 995 611 260 649 472 × 2 = 0 + 0.891 601 562 499 999 999 991 222 521 298 944;
  • 38) 0.891 601 562 499 999 999 991 222 521 298 944 × 2 = 1 + 0.783 203 124 999 999 999 982 445 042 597 888;
  • 39) 0.783 203 124 999 999 999 982 445 042 597 888 × 2 = 1 + 0.566 406 249 999 999 999 964 890 085 195 776;
  • 40) 0.566 406 249 999 999 999 964 890 085 195 776 × 2 = 1 + 0.132 812 499 999 999 999 929 780 170 391 552;
  • 41) 0.132 812 499 999 999 999 929 780 170 391 552 × 2 = 0 + 0.265 624 999 999 999 999 859 560 340 783 104;
  • 42) 0.265 624 999 999 999 999 859 560 340 783 104 × 2 = 0 + 0.531 249 999 999 999 999 719 120 681 566 208;
  • 43) 0.531 249 999 999 999 999 719 120 681 566 208 × 2 = 1 + 0.062 499 999 999 999 999 438 241 363 132 416;
  • 44) 0.062 499 999 999 999 999 438 241 363 132 416 × 2 = 0 + 0.124 999 999 999 999 998 876 482 726 264 832;
  • 45) 0.124 999 999 999 999 998 876 482 726 264 832 × 2 = 0 + 0.249 999 999 999 999 997 752 965 452 529 664;
  • 46) 0.249 999 999 999 999 997 752 965 452 529 664 × 2 = 0 + 0.499 999 999 999 999 995 505 930 905 059 328;
  • 47) 0.499 999 999 999 999 995 505 930 905 059 328 × 2 = 0 + 0.999 999 999 999 999 991 011 861 810 118 656;
  • 48) 0.999 999 999 999 999 991 011 861 810 118 656 × 2 = 1 + 0.999 999 999 999 999 982 023 723 620 237 312;
  • 49) 0.999 999 999 999 999 982 023 723 620 237 312 × 2 = 1 + 0.999 999 999 999 999 964 047 447 240 474 624;
  • 50) 0.999 999 999 999 999 964 047 447 240 474 624 × 2 = 1 + 0.999 999 999 999 999 928 094 894 480 949 248;
  • 51) 0.999 999 999 999 999 928 094 894 480 949 248 × 2 = 1 + 0.999 999 999 999 999 856 189 788 961 898 496;
  • 52) 0.999 999 999 999 999 856 189 788 961 898 496 × 2 = 1 + 0.999 999 999 999 999 712 379 577 923 796 992;
  • 53) 0.999 999 999 999 999 712 379 577 923 796 992 × 2 = 1 + 0.999 999 999 999 999 424 759 155 847 593 984;

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:


0.569 600 888 243 378 733 477 584 319 189 127(10) =

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

5. Positive number before normalization:

50.569 600 888 243 378 733 477 584 319 189 127(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:


50.569 600 888 243 378 733 477 584 319 189 127(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 50.569 600 888 243 378 733 477 584 319 189 127 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 50.569 600 888 243 378 733 477 584 319 189 076 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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