-0.016 738 891 601 562 496 761 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 761(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
-0.016 738 891 601 562 496 761(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. Start with the positive version of the number:

|-0.016 738 891 601 562 496 761| = 0.016 738 891 601 562 496 761


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.016 738 891 601 562 496 761.

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) 0.016 738 891 601 562 496 761 × 2 = 0 + 0.033 477 783 203 124 993 522;
  • 2) 0.033 477 783 203 124 993 522 × 2 = 0 + 0.066 955 566 406 249 987 044;
  • 3) 0.066 955 566 406 249 987 044 × 2 = 0 + 0.133 911 132 812 499 974 088;
  • 4) 0.133 911 132 812 499 974 088 × 2 = 0 + 0.267 822 265 624 999 948 176;
  • 5) 0.267 822 265 624 999 948 176 × 2 = 0 + 0.535 644 531 249 999 896 352;
  • 6) 0.535 644 531 249 999 896 352 × 2 = 1 + 0.071 289 062 499 999 792 704;
  • 7) 0.071 289 062 499 999 792 704 × 2 = 0 + 0.142 578 124 999 999 585 408;
  • 8) 0.142 578 124 999 999 585 408 × 2 = 0 + 0.285 156 249 999 999 170 816;
  • 9) 0.285 156 249 999 999 170 816 × 2 = 0 + 0.570 312 499 999 998 341 632;
  • 10) 0.570 312 499 999 998 341 632 × 2 = 1 + 0.140 624 999 999 996 683 264;
  • 11) 0.140 624 999 999 996 683 264 × 2 = 0 + 0.281 249 999 999 993 366 528;
  • 12) 0.281 249 999 999 993 366 528 × 2 = 0 + 0.562 499 999 999 986 733 056;
  • 13) 0.562 499 999 999 986 733 056 × 2 = 1 + 0.124 999 999 999 973 466 112;
  • 14) 0.124 999 999 999 973 466 112 × 2 = 0 + 0.249 999 999 999 946 932 224;
  • 15) 0.249 999 999 999 946 932 224 × 2 = 0 + 0.499 999 999 999 893 864 448;
  • 16) 0.499 999 999 999 893 864 448 × 2 = 0 + 0.999 999 999 999 787 728 896;
  • 17) 0.999 999 999 999 787 728 896 × 2 = 1 + 0.999 999 999 999 575 457 792;
  • 18) 0.999 999 999 999 575 457 792 × 2 = 1 + 0.999 999 999 999 150 915 584;
  • 19) 0.999 999 999 999 150 915 584 × 2 = 1 + 0.999 999 999 998 301 831 168;
  • 20) 0.999 999 999 998 301 831 168 × 2 = 1 + 0.999 999 999 996 603 662 336;
  • 21) 0.999 999 999 996 603 662 336 × 2 = 1 + 0.999 999 999 993 207 324 672;
  • 22) 0.999 999 999 993 207 324 672 × 2 = 1 + 0.999 999 999 986 414 649 344;
  • 23) 0.999 999 999 986 414 649 344 × 2 = 1 + 0.999 999 999 972 829 298 688;
  • 24) 0.999 999 999 972 829 298 688 × 2 = 1 + 0.999 999 999 945 658 597 376;
  • 25) 0.999 999 999 945 658 597 376 × 2 = 1 + 0.999 999 999 891 317 194 752;
  • 26) 0.999 999 999 891 317 194 752 × 2 = 1 + 0.999 999 999 782 634 389 504;
  • 27) 0.999 999 999 782 634 389 504 × 2 = 1 + 0.999 999 999 565 268 779 008;
  • 28) 0.999 999 999 565 268 779 008 × 2 = 1 + 0.999 999 999 130 537 558 016;
  • 29) 0.999 999 999 130 537 558 016 × 2 = 1 + 0.999 999 998 261 075 116 032;
  • 30) 0.999 999 998 261 075 116 032 × 2 = 1 + 0.999 999 996 522 150 232 064;
  • 31) 0.999 999 996 522 150 232 064 × 2 = 1 + 0.999 999 993 044 300 464 128;
  • 32) 0.999 999 993 044 300 464 128 × 2 = 1 + 0.999 999 986 088 600 928 256;
  • 33) 0.999 999 986 088 600 928 256 × 2 = 1 + 0.999 999 972 177 201 856 512;
  • 34) 0.999 999 972 177 201 856 512 × 2 = 1 + 0.999 999 944 354 403 713 024;
  • 35) 0.999 999 944 354 403 713 024 × 2 = 1 + 0.999 999 888 708 807 426 048;
  • 36) 0.999 999 888 708 807 426 048 × 2 = 1 + 0.999 999 777 417 614 852 096;
  • 37) 0.999 999 777 417 614 852 096 × 2 = 1 + 0.999 999 554 835 229 704 192;
  • 38) 0.999 999 554 835 229 704 192 × 2 = 1 + 0.999 999 109 670 459 408 384;
  • 39) 0.999 999 109 670 459 408 384 × 2 = 1 + 0.999 998 219 340 918 816 768;
  • 40) 0.999 998 219 340 918 816 768 × 2 = 1 + 0.999 996 438 681 837 633 536;
  • 41) 0.999 996 438 681 837 633 536 × 2 = 1 + 0.999 992 877 363 675 267 072;
  • 42) 0.999 992 877 363 675 267 072 × 2 = 1 + 0.999 985 754 727 350 534 144;
  • 43) 0.999 985 754 727 350 534 144 × 2 = 1 + 0.999 971 509 454 701 068 288;
  • 44) 0.999 971 509 454 701 068 288 × 2 = 1 + 0.999 943 018 909 402 136 576;
  • 45) 0.999 943 018 909 402 136 576 × 2 = 1 + 0.999 886 037 818 804 273 152;
  • 46) 0.999 886 037 818 804 273 152 × 2 = 1 + 0.999 772 075 637 608 546 304;
  • 47) 0.999 772 075 637 608 546 304 × 2 = 1 + 0.999 544 151 275 217 092 608;
  • 48) 0.999 544 151 275 217 092 608 × 2 = 1 + 0.999 088 302 550 434 185 216;
  • 49) 0.999 088 302 550 434 185 216 × 2 = 1 + 0.998 176 605 100 868 370 432;
  • 50) 0.998 176 605 100 868 370 432 × 2 = 1 + 0.996 353 210 201 736 740 864;
  • 51) 0.996 353 210 201 736 740 864 × 2 = 1 + 0.992 706 420 403 473 481 728;
  • 52) 0.992 706 420 403 473 481 728 × 2 = 1 + 0.985 412 840 806 946 963 456;
  • 53) 0.985 412 840 806 946 963 456 × 2 = 1 + 0.970 825 681 613 893 926 912;
  • 54) 0.970 825 681 613 893 926 912 × 2 = 1 + 0.941 651 363 227 787 853 824;
  • 55) 0.941 651 363 227 787 853 824 × 2 = 1 + 0.883 302 726 455 575 707 648;
  • 56) 0.883 302 726 455 575 707 648 × 2 = 1 + 0.766 605 452 911 151 415 296;
  • 57) 0.766 605 452 911 151 415 296 × 2 = 1 + 0.533 210 905 822 302 830 592;
  • 58) 0.533 210 905 822 302 830 592 × 2 = 1 + 0.066 421 811 644 605 661 184;

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).


5. 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:


0.016 738 891 601 562 496 761(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 761(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the right, so that only one non zero digit remains to the left of it:


0.016 738 891 601 562 496 761(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 11(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111(2) × 2-6


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): -6

Mantissa (not normalized):
1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-6 + 2(11-1) - 1 =

(-6 + 1 023)(10) =

1 017(10)


10. 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;
  • 1 017 ÷ 2 = 508 + 1;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

1017(10) =

011 1111 1001(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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

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

Exponent (11 bits) =
011 1111 1001

Mantissa (52 bits) =
0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number -0.016 738 891 601 562 496 761 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1001 - 0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

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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