50.569 600 888 243 378 733 477 649 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 50.569 600 888 243 378 733 477 649(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
50.569 600 888 243 378 733 477 649(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: 0.569 600 888 243 378 733 477 649.

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) 0.569 600 888 243 378 733 477 649 × 2 = 1 + 0.139 201 776 486 757 466 955 298;
  • 2) 0.139 201 776 486 757 466 955 298 × 2 = 0 + 0.278 403 552 973 514 933 910 596;
  • 3) 0.278 403 552 973 514 933 910 596 × 2 = 0 + 0.556 807 105 947 029 867 821 192;
  • 4) 0.556 807 105 947 029 867 821 192 × 2 = 1 + 0.113 614 211 894 059 735 642 384;
  • 5) 0.113 614 211 894 059 735 642 384 × 2 = 0 + 0.227 228 423 788 119 471 284 768;
  • 6) 0.227 228 423 788 119 471 284 768 × 2 = 0 + 0.454 456 847 576 238 942 569 536;
  • 7) 0.454 456 847 576 238 942 569 536 × 2 = 0 + 0.908 913 695 152 477 885 139 072;
  • 8) 0.908 913 695 152 477 885 139 072 × 2 = 1 + 0.817 827 390 304 955 770 278 144;
  • 9) 0.817 827 390 304 955 770 278 144 × 2 = 1 + 0.635 654 780 609 911 540 556 288;
  • 10) 0.635 654 780 609 911 540 556 288 × 2 = 1 + 0.271 309 561 219 823 081 112 576;
  • 11) 0.271 309 561 219 823 081 112 576 × 2 = 0 + 0.542 619 122 439 646 162 225 152;
  • 12) 0.542 619 122 439 646 162 225 152 × 2 = 1 + 0.085 238 244 879 292 324 450 304;
  • 13) 0.085 238 244 879 292 324 450 304 × 2 = 0 + 0.170 476 489 758 584 648 900 608;
  • 14) 0.170 476 489 758 584 648 900 608 × 2 = 0 + 0.340 952 979 517 169 297 801 216;
  • 15) 0.340 952 979 517 169 297 801 216 × 2 = 0 + 0.681 905 959 034 338 595 602 432;
  • 16) 0.681 905 959 034 338 595 602 432 × 2 = 1 + 0.363 811 918 068 677 191 204 864;
  • 17) 0.363 811 918 068 677 191 204 864 × 2 = 0 + 0.727 623 836 137 354 382 409 728;
  • 18) 0.727 623 836 137 354 382 409 728 × 2 = 1 + 0.455 247 672 274 708 764 819 456;
  • 19) 0.455 247 672 274 708 764 819 456 × 2 = 0 + 0.910 495 344 549 417 529 638 912;
  • 20) 0.910 495 344 549 417 529 638 912 × 2 = 1 + 0.820 990 689 098 835 059 277 824;
  • 21) 0.820 990 689 098 835 059 277 824 × 2 = 1 + 0.641 981 378 197 670 118 555 648;
  • 22) 0.641 981 378 197 670 118 555 648 × 2 = 1 + 0.283 962 756 395 340 237 111 296;
  • 23) 0.283 962 756 395 340 237 111 296 × 2 = 0 + 0.567 925 512 790 680 474 222 592;
  • 24) 0.567 925 512 790 680 474 222 592 × 2 = 1 + 0.135 851 025 581 360 948 445 184;
  • 25) 0.135 851 025 581 360 948 445 184 × 2 = 0 + 0.271 702 051 162 721 896 890 368;
  • 26) 0.271 702 051 162 721 896 890 368 × 2 = 0 + 0.543 404 102 325 443 793 780 736;
  • 27) 0.543 404 102 325 443 793 780 736 × 2 = 1 + 0.086 808 204 650 887 587 561 472;
  • 28) 0.086 808 204 650 887 587 561 472 × 2 = 0 + 0.173 616 409 301 775 175 122 944;
  • 29) 0.173 616 409 301 775 175 122 944 × 2 = 0 + 0.347 232 818 603 550 350 245 888;
  • 30) 0.347 232 818 603 550 350 245 888 × 2 = 0 + 0.694 465 637 207 100 700 491 776;
  • 31) 0.694 465 637 207 100 700 491 776 × 2 = 1 + 0.388 931 274 414 201 400 983 552;
  • 32) 0.388 931 274 414 201 400 983 552 × 2 = 0 + 0.777 862 548 828 402 801 967 104;
  • 33) 0.777 862 548 828 402 801 967 104 × 2 = 1 + 0.555 725 097 656 805 603 934 208;
  • 34) 0.555 725 097 656 805 603 934 208 × 2 = 1 + 0.111 450 195 313 611 207 868 416;
  • 35) 0.111 450 195 313 611 207 868 416 × 2 = 0 + 0.222 900 390 627 222 415 736 832;
  • 36) 0.222 900 390 627 222 415 736 832 × 2 = 0 + 0.445 800 781 254 444 831 473 664;
  • 37) 0.445 800 781 254 444 831 473 664 × 2 = 0 + 0.891 601 562 508 889 662 947 328;
  • 38) 0.891 601 562 508 889 662 947 328 × 2 = 1 + 0.783 203 125 017 779 325 894 656;
  • 39) 0.783 203 125 017 779 325 894 656 × 2 = 1 + 0.566 406 250 035 558 651 789 312;
  • 40) 0.566 406 250 035 558 651 789 312 × 2 = 1 + 0.132 812 500 071 117 303 578 624;
  • 41) 0.132 812 500 071 117 303 578 624 × 2 = 0 + 0.265 625 000 142 234 607 157 248;
  • 42) 0.265 625 000 142 234 607 157 248 × 2 = 0 + 0.531 250 000 284 469 214 314 496;
  • 43) 0.531 250 000 284 469 214 314 496 × 2 = 1 + 0.062 500 000 568 938 428 628 992;
  • 44) 0.062 500 000 568 938 428 628 992 × 2 = 0 + 0.125 000 001 137 876 857 257 984;
  • 45) 0.125 000 001 137 876 857 257 984 × 2 = 0 + 0.250 000 002 275 753 714 515 968;
  • 46) 0.250 000 002 275 753 714 515 968 × 2 = 0 + 0.500 000 004 551 507 429 031 936;
  • 47) 0.500 000 004 551 507 429 031 936 × 2 = 1 + 0.000 000 009 103 014 858 063 872;
  • 48) 0.000 000 009 103 014 858 063 872 × 2 = 0 + 0.000 000 018 206 029 716 127 744;
  • 49) 0.000 000 018 206 029 716 127 744 × 2 = 0 + 0.000 000 036 412 059 432 255 488;
  • 50) 0.000 000 036 412 059 432 255 488 × 2 = 0 + 0.000 000 072 824 118 864 510 976;
  • 51) 0.000 000 072 824 118 864 510 976 × 2 = 0 + 0.000 000 145 648 237 729 021 952;
  • 52) 0.000 000 145 648 237 729 021 952 × 2 = 0 + 0.000 000 291 296 475 458 043 904;
  • 53) 0.000 000 291 296 475 458 043 904 × 2 = 0 + 0.000 000 582 592 950 916 087 808;

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:


0.569 600 888 243 378 733 477 649(10) =

0.1001 0001 1101 0001 0101 1101 0010 0010 1100 0111 0010 0010 0000 0(2)

5. Positive number before normalization:

50.569 600 888 243 378 733 477 649(10) =

11 0010.1001 0001 1101 0001 0101 1101 0010 0010 1100 0111 0010 0010 0000 0(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:


50.569 600 888 243 378 733 477 649(10) =

11 0010.1001 0001 1101 0001 0101 1101 0010 0010 1100 0111 0010 0010 0000 0(2) =

11 0010.1001 0001 1101 0001 0101 1101 0010 0010 1100 0111 0010 0010 0000 0(2) × 20 =

1.1001 0100 1000 1110 1000 1010 1110 1001 0001 0110 0011 1001 0001 0000 00(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 0001 0000 00


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 0001 00 0000 =

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


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 0001


Decimal number 50.569 600 888 243 378 733 477 649 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 0001

» Convert decimal 50.569 600 888 243 378 733 477 746 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