-0.016 738 891 601 562 496 530 554 33 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 554 33(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
-0.016 738 891 601 562 496 530 554 33(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:

|-0.016 738 891 601 562 496 530 554 33| = 0.016 738 891 601 562 496 530 554 33


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: 0.016 738 891 601 562 496 530 554 33.

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) 0.016 738 891 601 562 496 530 554 33 × 2 = 0 + 0.033 477 783 203 124 993 061 108 66;
  • 2) 0.033 477 783 203 124 993 061 108 66 × 2 = 0 + 0.066 955 566 406 249 986 122 217 32;
  • 3) 0.066 955 566 406 249 986 122 217 32 × 2 = 0 + 0.133 911 132 812 499 972 244 434 64;
  • 4) 0.133 911 132 812 499 972 244 434 64 × 2 = 0 + 0.267 822 265 624 999 944 488 869 28;
  • 5) 0.267 822 265 624 999 944 488 869 28 × 2 = 0 + 0.535 644 531 249 999 888 977 738 56;
  • 6) 0.535 644 531 249 999 888 977 738 56 × 2 = 1 + 0.071 289 062 499 999 777 955 477 12;
  • 7) 0.071 289 062 499 999 777 955 477 12 × 2 = 0 + 0.142 578 124 999 999 555 910 954 24;
  • 8) 0.142 578 124 999 999 555 910 954 24 × 2 = 0 + 0.285 156 249 999 999 111 821 908 48;
  • 9) 0.285 156 249 999 999 111 821 908 48 × 2 = 0 + 0.570 312 499 999 998 223 643 816 96;
  • 10) 0.570 312 499 999 998 223 643 816 96 × 2 = 1 + 0.140 624 999 999 996 447 287 633 92;
  • 11) 0.140 624 999 999 996 447 287 633 92 × 2 = 0 + 0.281 249 999 999 992 894 575 267 84;
  • 12) 0.281 249 999 999 992 894 575 267 84 × 2 = 0 + 0.562 499 999 999 985 789 150 535 68;
  • 13) 0.562 499 999 999 985 789 150 535 68 × 2 = 1 + 0.124 999 999 999 971 578 301 071 36;
  • 14) 0.124 999 999 999 971 578 301 071 36 × 2 = 0 + 0.249 999 999 999 943 156 602 142 72;
  • 15) 0.249 999 999 999 943 156 602 142 72 × 2 = 0 + 0.499 999 999 999 886 313 204 285 44;
  • 16) 0.499 999 999 999 886 313 204 285 44 × 2 = 0 + 0.999 999 999 999 772 626 408 570 88;
  • 17) 0.999 999 999 999 772 626 408 570 88 × 2 = 1 + 0.999 999 999 999 545 252 817 141 76;
  • 18) 0.999 999 999 999 545 252 817 141 76 × 2 = 1 + 0.999 999 999 999 090 505 634 283 52;
  • 19) 0.999 999 999 999 090 505 634 283 52 × 2 = 1 + 0.999 999 999 998 181 011 268 567 04;
  • 20) 0.999 999 999 998 181 011 268 567 04 × 2 = 1 + 0.999 999 999 996 362 022 537 134 08;
  • 21) 0.999 999 999 996 362 022 537 134 08 × 2 = 1 + 0.999 999 999 992 724 045 074 268 16;
  • 22) 0.999 999 999 992 724 045 074 268 16 × 2 = 1 + 0.999 999 999 985 448 090 148 536 32;
  • 23) 0.999 999 999 985 448 090 148 536 32 × 2 = 1 + 0.999 999 999 970 896 180 297 072 64;
  • 24) 0.999 999 999 970 896 180 297 072 64 × 2 = 1 + 0.999 999 999 941 792 360 594 145 28;
  • 25) 0.999 999 999 941 792 360 594 145 28 × 2 = 1 + 0.999 999 999 883 584 721 188 290 56;
  • 26) 0.999 999 999 883 584 721 188 290 56 × 2 = 1 + 0.999 999 999 767 169 442 376 581 12;
  • 27) 0.999 999 999 767 169 442 376 581 12 × 2 = 1 + 0.999 999 999 534 338 884 753 162 24;
  • 28) 0.999 999 999 534 338 884 753 162 24 × 2 = 1 + 0.999 999 999 068 677 769 506 324 48;
  • 29) 0.999 999 999 068 677 769 506 324 48 × 2 = 1 + 0.999 999 998 137 355 539 012 648 96;
  • 30) 0.999 999 998 137 355 539 012 648 96 × 2 = 1 + 0.999 999 996 274 711 078 025 297 92;
  • 31) 0.999 999 996 274 711 078 025 297 92 × 2 = 1 + 0.999 999 992 549 422 156 050 595 84;
  • 32) 0.999 999 992 549 422 156 050 595 84 × 2 = 1 + 0.999 999 985 098 844 312 101 191 68;
  • 33) 0.999 999 985 098 844 312 101 191 68 × 2 = 1 + 0.999 999 970 197 688 624 202 383 36;
  • 34) 0.999 999 970 197 688 624 202 383 36 × 2 = 1 + 0.999 999 940 395 377 248 404 766 72;
  • 35) 0.999 999 940 395 377 248 404 766 72 × 2 = 1 + 0.999 999 880 790 754 496 809 533 44;
  • 36) 0.999 999 880 790 754 496 809 533 44 × 2 = 1 + 0.999 999 761 581 508 993 619 066 88;
  • 37) 0.999 999 761 581 508 993 619 066 88 × 2 = 1 + 0.999 999 523 163 017 987 238 133 76;
  • 38) 0.999 999 523 163 017 987 238 133 76 × 2 = 1 + 0.999 999 046 326 035 974 476 267 52;
  • 39) 0.999 999 046 326 035 974 476 267 52 × 2 = 1 + 0.999 998 092 652 071 948 952 535 04;
  • 40) 0.999 998 092 652 071 948 952 535 04 × 2 = 1 + 0.999 996 185 304 143 897 905 070 08;
  • 41) 0.999 996 185 304 143 897 905 070 08 × 2 = 1 + 0.999 992 370 608 287 795 810 140 16;
  • 42) 0.999 992 370 608 287 795 810 140 16 × 2 = 1 + 0.999 984 741 216 575 591 620 280 32;
  • 43) 0.999 984 741 216 575 591 620 280 32 × 2 = 1 + 0.999 969 482 433 151 183 240 560 64;
  • 44) 0.999 969 482 433 151 183 240 560 64 × 2 = 1 + 0.999 938 964 866 302 366 481 121 28;
  • 45) 0.999 938 964 866 302 366 481 121 28 × 2 = 1 + 0.999 877 929 732 604 732 962 242 56;
  • 46) 0.999 877 929 732 604 732 962 242 56 × 2 = 1 + 0.999 755 859 465 209 465 924 485 12;
  • 47) 0.999 755 859 465 209 465 924 485 12 × 2 = 1 + 0.999 511 718 930 418 931 848 970 24;
  • 48) 0.999 511 718 930 418 931 848 970 24 × 2 = 1 + 0.999 023 437 860 837 863 697 940 48;
  • 49) 0.999 023 437 860 837 863 697 940 48 × 2 = 1 + 0.998 046 875 721 675 727 395 880 96;
  • 50) 0.998 046 875 721 675 727 395 880 96 × 2 = 1 + 0.996 093 751 443 351 454 791 761 92;
  • 51) 0.996 093 751 443 351 454 791 761 92 × 2 = 1 + 0.992 187 502 886 702 909 583 523 84;
  • 52) 0.992 187 502 886 702 909 583 523 84 × 2 = 1 + 0.984 375 005 773 405 819 167 047 68;
  • 53) 0.984 375 005 773 405 819 167 047 68 × 2 = 1 + 0.968 750 011 546 811 638 334 095 36;
  • 54) 0.968 750 011 546 811 638 334 095 36 × 2 = 1 + 0.937 500 023 093 623 276 668 190 72;
  • 55) 0.937 500 023 093 623 276 668 190 72 × 2 = 1 + 0.875 000 046 187 246 553 336 381 44;
  • 56) 0.875 000 046 187 246 553 336 381 44 × 2 = 1 + 0.750 000 092 374 493 106 672 762 88;
  • 57) 0.750 000 092 374 493 106 672 762 88 × 2 = 1 + 0.500 000 184 748 986 213 345 525 76;
  • 58) 0.500 000 184 748 986 213 345 525 76 × 2 = 1 + 0.000 000 369 497 972 426 691 051 52;

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:


0.016 738 891 601 562 496 530 554 33(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 530 554 33(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:


0.016 738 891 601 562 496 530 554 33(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 -0.016 738 891 601 562 496 530 554 33 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