50.569 600 888 243 378 733 477 584 319 189 158 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 50.569 600 888 243 378 733 477 584 319 189 158(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
50.569 600 888 243 378 733 477 584 319 189 158(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: 0.569 600 888 243 378 733 477 584 319 189 158.

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) 0.569 600 888 243 378 733 477 584 319 189 158 × 2 = 1 + 0.139 201 776 486 757 466 955 168 638 378 316;
  • 2) 0.139 201 776 486 757 466 955 168 638 378 316 × 2 = 0 + 0.278 403 552 973 514 933 910 337 276 756 632;
  • 3) 0.278 403 552 973 514 933 910 337 276 756 632 × 2 = 0 + 0.556 807 105 947 029 867 820 674 553 513 264;
  • 4) 0.556 807 105 947 029 867 820 674 553 513 264 × 2 = 1 + 0.113 614 211 894 059 735 641 349 107 026 528;
  • 5) 0.113 614 211 894 059 735 641 349 107 026 528 × 2 = 0 + 0.227 228 423 788 119 471 282 698 214 053 056;
  • 6) 0.227 228 423 788 119 471 282 698 214 053 056 × 2 = 0 + 0.454 456 847 576 238 942 565 396 428 106 112;
  • 7) 0.454 456 847 576 238 942 565 396 428 106 112 × 2 = 0 + 0.908 913 695 152 477 885 130 792 856 212 224;
  • 8) 0.908 913 695 152 477 885 130 792 856 212 224 × 2 = 1 + 0.817 827 390 304 955 770 261 585 712 424 448;
  • 9) 0.817 827 390 304 955 770 261 585 712 424 448 × 2 = 1 + 0.635 654 780 609 911 540 523 171 424 848 896;
  • 10) 0.635 654 780 609 911 540 523 171 424 848 896 × 2 = 1 + 0.271 309 561 219 823 081 046 342 849 697 792;
  • 11) 0.271 309 561 219 823 081 046 342 849 697 792 × 2 = 0 + 0.542 619 122 439 646 162 092 685 699 395 584;
  • 12) 0.542 619 122 439 646 162 092 685 699 395 584 × 2 = 1 + 0.085 238 244 879 292 324 185 371 398 791 168;
  • 13) 0.085 238 244 879 292 324 185 371 398 791 168 × 2 = 0 + 0.170 476 489 758 584 648 370 742 797 582 336;
  • 14) 0.170 476 489 758 584 648 370 742 797 582 336 × 2 = 0 + 0.340 952 979 517 169 296 741 485 595 164 672;
  • 15) 0.340 952 979 517 169 296 741 485 595 164 672 × 2 = 0 + 0.681 905 959 034 338 593 482 971 190 329 344;
  • 16) 0.681 905 959 034 338 593 482 971 190 329 344 × 2 = 1 + 0.363 811 918 068 677 186 965 942 380 658 688;
  • 17) 0.363 811 918 068 677 186 965 942 380 658 688 × 2 = 0 + 0.727 623 836 137 354 373 931 884 761 317 376;
  • 18) 0.727 623 836 137 354 373 931 884 761 317 376 × 2 = 1 + 0.455 247 672 274 708 747 863 769 522 634 752;
  • 19) 0.455 247 672 274 708 747 863 769 522 634 752 × 2 = 0 + 0.910 495 344 549 417 495 727 539 045 269 504;
  • 20) 0.910 495 344 549 417 495 727 539 045 269 504 × 2 = 1 + 0.820 990 689 098 834 991 455 078 090 539 008;
  • 21) 0.820 990 689 098 834 991 455 078 090 539 008 × 2 = 1 + 0.641 981 378 197 669 982 910 156 181 078 016;
  • 22) 0.641 981 378 197 669 982 910 156 181 078 016 × 2 = 1 + 0.283 962 756 395 339 965 820 312 362 156 032;
  • 23) 0.283 962 756 395 339 965 820 312 362 156 032 × 2 = 0 + 0.567 925 512 790 679 931 640 624 724 312 064;
  • 24) 0.567 925 512 790 679 931 640 624 724 312 064 × 2 = 1 + 0.135 851 025 581 359 863 281 249 448 624 128;
  • 25) 0.135 851 025 581 359 863 281 249 448 624 128 × 2 = 0 + 0.271 702 051 162 719 726 562 498 897 248 256;
  • 26) 0.271 702 051 162 719 726 562 498 897 248 256 × 2 = 0 + 0.543 404 102 325 439 453 124 997 794 496 512;
  • 27) 0.543 404 102 325 439 453 124 997 794 496 512 × 2 = 1 + 0.086 808 204 650 878 906 249 995 588 993 024;
  • 28) 0.086 808 204 650 878 906 249 995 588 993 024 × 2 = 0 + 0.173 616 409 301 757 812 499 991 177 986 048;
  • 29) 0.173 616 409 301 757 812 499 991 177 986 048 × 2 = 0 + 0.347 232 818 603 515 624 999 982 355 972 096;
  • 30) 0.347 232 818 603 515 624 999 982 355 972 096 × 2 = 0 + 0.694 465 637 207 031 249 999 964 711 944 192;
  • 31) 0.694 465 637 207 031 249 999 964 711 944 192 × 2 = 1 + 0.388 931 274 414 062 499 999 929 423 888 384;
  • 32) 0.388 931 274 414 062 499 999 929 423 888 384 × 2 = 0 + 0.777 862 548 828 124 999 999 858 847 776 768;
  • 33) 0.777 862 548 828 124 999 999 858 847 776 768 × 2 = 1 + 0.555 725 097 656 249 999 999 717 695 553 536;
  • 34) 0.555 725 097 656 249 999 999 717 695 553 536 × 2 = 1 + 0.111 450 195 312 499 999 999 435 391 107 072;
  • 35) 0.111 450 195 312 499 999 999 435 391 107 072 × 2 = 0 + 0.222 900 390 624 999 999 998 870 782 214 144;
  • 36) 0.222 900 390 624 999 999 998 870 782 214 144 × 2 = 0 + 0.445 800 781 249 999 999 997 741 564 428 288;
  • 37) 0.445 800 781 249 999 999 997 741 564 428 288 × 2 = 0 + 0.891 601 562 499 999 999 995 483 128 856 576;
  • 38) 0.891 601 562 499 999 999 995 483 128 856 576 × 2 = 1 + 0.783 203 124 999 999 999 990 966 257 713 152;
  • 39) 0.783 203 124 999 999 999 990 966 257 713 152 × 2 = 1 + 0.566 406 249 999 999 999 981 932 515 426 304;
  • 40) 0.566 406 249 999 999 999 981 932 515 426 304 × 2 = 1 + 0.132 812 499 999 999 999 963 865 030 852 608;
  • 41) 0.132 812 499 999 999 999 963 865 030 852 608 × 2 = 0 + 0.265 624 999 999 999 999 927 730 061 705 216;
  • 42) 0.265 624 999 999 999 999 927 730 061 705 216 × 2 = 0 + 0.531 249 999 999 999 999 855 460 123 410 432;
  • 43) 0.531 249 999 999 999 999 855 460 123 410 432 × 2 = 1 + 0.062 499 999 999 999 999 710 920 246 820 864;
  • 44) 0.062 499 999 999 999 999 710 920 246 820 864 × 2 = 0 + 0.124 999 999 999 999 999 421 840 493 641 728;
  • 45) 0.124 999 999 999 999 999 421 840 493 641 728 × 2 = 0 + 0.249 999 999 999 999 998 843 680 987 283 456;
  • 46) 0.249 999 999 999 999 998 843 680 987 283 456 × 2 = 0 + 0.499 999 999 999 999 997 687 361 974 566 912;
  • 47) 0.499 999 999 999 999 997 687 361 974 566 912 × 2 = 0 + 0.999 999 999 999 999 995 374 723 949 133 824;
  • 48) 0.999 999 999 999 999 995 374 723 949 133 824 × 2 = 1 + 0.999 999 999 999 999 990 749 447 898 267 648;
  • 49) 0.999 999 999 999 999 990 749 447 898 267 648 × 2 = 1 + 0.999 999 999 999 999 981 498 895 796 535 296;
  • 50) 0.999 999 999 999 999 981 498 895 796 535 296 × 2 = 1 + 0.999 999 999 999 999 962 997 791 593 070 592;
  • 51) 0.999 999 999 999 999 962 997 791 593 070 592 × 2 = 1 + 0.999 999 999 999 999 925 995 583 186 141 184;
  • 52) 0.999 999 999 999 999 925 995 583 186 141 184 × 2 = 1 + 0.999 999 999 999 999 851 991 166 372 282 368;
  • 53) 0.999 999 999 999 999 851 991 166 372 282 368 × 2 = 1 + 0.999 999 999 999 999 703 982 332 744 564 736;

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:


0.569 600 888 243 378 733 477 584 319 189 158(10) =

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

5. Positive number before normalization:

50.569 600 888 243 378 733 477 584 319 189 158(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:


50.569 600 888 243 378 733 477 584 319 189 158(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 50.569 600 888 243 378 733 477 584 319 189 158 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 11 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