-0.016 738 891 601 562 496 530 338 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 530 338(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
-0.016 738 891 601 562 496 530 338(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:

|-0.016 738 891 601 562 496 530 338| = 0.016 738 891 601 562 496 530 338


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: 0.016 738 891 601 562 496 530 338.

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) 0.016 738 891 601 562 496 530 338 × 2 = 0 + 0.033 477 783 203 124 993 060 676;
  • 2) 0.033 477 783 203 124 993 060 676 × 2 = 0 + 0.066 955 566 406 249 986 121 352;
  • 3) 0.066 955 566 406 249 986 121 352 × 2 = 0 + 0.133 911 132 812 499 972 242 704;
  • 4) 0.133 911 132 812 499 972 242 704 × 2 = 0 + 0.267 822 265 624 999 944 485 408;
  • 5) 0.267 822 265 624 999 944 485 408 × 2 = 0 + 0.535 644 531 249 999 888 970 816;
  • 6) 0.535 644 531 249 999 888 970 816 × 2 = 1 + 0.071 289 062 499 999 777 941 632;
  • 7) 0.071 289 062 499 999 777 941 632 × 2 = 0 + 0.142 578 124 999 999 555 883 264;
  • 8) 0.142 578 124 999 999 555 883 264 × 2 = 0 + 0.285 156 249 999 999 111 766 528;
  • 9) 0.285 156 249 999 999 111 766 528 × 2 = 0 + 0.570 312 499 999 998 223 533 056;
  • 10) 0.570 312 499 999 998 223 533 056 × 2 = 1 + 0.140 624 999 999 996 447 066 112;
  • 11) 0.140 624 999 999 996 447 066 112 × 2 = 0 + 0.281 249 999 999 992 894 132 224;
  • 12) 0.281 249 999 999 992 894 132 224 × 2 = 0 + 0.562 499 999 999 985 788 264 448;
  • 13) 0.562 499 999 999 985 788 264 448 × 2 = 1 + 0.124 999 999 999 971 576 528 896;
  • 14) 0.124 999 999 999 971 576 528 896 × 2 = 0 + 0.249 999 999 999 943 153 057 792;
  • 15) 0.249 999 999 999 943 153 057 792 × 2 = 0 + 0.499 999 999 999 886 306 115 584;
  • 16) 0.499 999 999 999 886 306 115 584 × 2 = 0 + 0.999 999 999 999 772 612 231 168;
  • 17) 0.999 999 999 999 772 612 231 168 × 2 = 1 + 0.999 999 999 999 545 224 462 336;
  • 18) 0.999 999 999 999 545 224 462 336 × 2 = 1 + 0.999 999 999 999 090 448 924 672;
  • 19) 0.999 999 999 999 090 448 924 672 × 2 = 1 + 0.999 999 999 998 180 897 849 344;
  • 20) 0.999 999 999 998 180 897 849 344 × 2 = 1 + 0.999 999 999 996 361 795 698 688;
  • 21) 0.999 999 999 996 361 795 698 688 × 2 = 1 + 0.999 999 999 992 723 591 397 376;
  • 22) 0.999 999 999 992 723 591 397 376 × 2 = 1 + 0.999 999 999 985 447 182 794 752;
  • 23) 0.999 999 999 985 447 182 794 752 × 2 = 1 + 0.999 999 999 970 894 365 589 504;
  • 24) 0.999 999 999 970 894 365 589 504 × 2 = 1 + 0.999 999 999 941 788 731 179 008;
  • 25) 0.999 999 999 941 788 731 179 008 × 2 = 1 + 0.999 999 999 883 577 462 358 016;
  • 26) 0.999 999 999 883 577 462 358 016 × 2 = 1 + 0.999 999 999 767 154 924 716 032;
  • 27) 0.999 999 999 767 154 924 716 032 × 2 = 1 + 0.999 999 999 534 309 849 432 064;
  • 28) 0.999 999 999 534 309 849 432 064 × 2 = 1 + 0.999 999 999 068 619 698 864 128;
  • 29) 0.999 999 999 068 619 698 864 128 × 2 = 1 + 0.999 999 998 137 239 397 728 256;
  • 30) 0.999 999 998 137 239 397 728 256 × 2 = 1 + 0.999 999 996 274 478 795 456 512;
  • 31) 0.999 999 996 274 478 795 456 512 × 2 = 1 + 0.999 999 992 548 957 590 913 024;
  • 32) 0.999 999 992 548 957 590 913 024 × 2 = 1 + 0.999 999 985 097 915 181 826 048;
  • 33) 0.999 999 985 097 915 181 826 048 × 2 = 1 + 0.999 999 970 195 830 363 652 096;
  • 34) 0.999 999 970 195 830 363 652 096 × 2 = 1 + 0.999 999 940 391 660 727 304 192;
  • 35) 0.999 999 940 391 660 727 304 192 × 2 = 1 + 0.999 999 880 783 321 454 608 384;
  • 36) 0.999 999 880 783 321 454 608 384 × 2 = 1 + 0.999 999 761 566 642 909 216 768;
  • 37) 0.999 999 761 566 642 909 216 768 × 2 = 1 + 0.999 999 523 133 285 818 433 536;
  • 38) 0.999 999 523 133 285 818 433 536 × 2 = 1 + 0.999 999 046 266 571 636 867 072;
  • 39) 0.999 999 046 266 571 636 867 072 × 2 = 1 + 0.999 998 092 533 143 273 734 144;
  • 40) 0.999 998 092 533 143 273 734 144 × 2 = 1 + 0.999 996 185 066 286 547 468 288;
  • 41) 0.999 996 185 066 286 547 468 288 × 2 = 1 + 0.999 992 370 132 573 094 936 576;
  • 42) 0.999 992 370 132 573 094 936 576 × 2 = 1 + 0.999 984 740 265 146 189 873 152;
  • 43) 0.999 984 740 265 146 189 873 152 × 2 = 1 + 0.999 969 480 530 292 379 746 304;
  • 44) 0.999 969 480 530 292 379 746 304 × 2 = 1 + 0.999 938 961 060 584 759 492 608;
  • 45) 0.999 938 961 060 584 759 492 608 × 2 = 1 + 0.999 877 922 121 169 518 985 216;
  • 46) 0.999 877 922 121 169 518 985 216 × 2 = 1 + 0.999 755 844 242 339 037 970 432;
  • 47) 0.999 755 844 242 339 037 970 432 × 2 = 1 + 0.999 511 688 484 678 075 940 864;
  • 48) 0.999 511 688 484 678 075 940 864 × 2 = 1 + 0.999 023 376 969 356 151 881 728;
  • 49) 0.999 023 376 969 356 151 881 728 × 2 = 1 + 0.998 046 753 938 712 303 763 456;
  • 50) 0.998 046 753 938 712 303 763 456 × 2 = 1 + 0.996 093 507 877 424 607 526 912;
  • 51) 0.996 093 507 877 424 607 526 912 × 2 = 1 + 0.992 187 015 754 849 215 053 824;
  • 52) 0.992 187 015 754 849 215 053 824 × 2 = 1 + 0.984 374 031 509 698 430 107 648;
  • 53) 0.984 374 031 509 698 430 107 648 × 2 = 1 + 0.968 748 063 019 396 860 215 296;
  • 54) 0.968 748 063 019 396 860 215 296 × 2 = 1 + 0.937 496 126 038 793 720 430 592;
  • 55) 0.937 496 126 038 793 720 430 592 × 2 = 1 + 0.874 992 252 077 587 440 861 184;
  • 56) 0.874 992 252 077 587 440 861 184 × 2 = 1 + 0.749 984 504 155 174 881 722 368;
  • 57) 0.749 984 504 155 174 881 722 368 × 2 = 1 + 0.499 969 008 310 349 763 444 736;
  • 58) 0.499 969 008 310 349 763 444 736 × 2 = 0 + 0.999 938 016 620 699 526 889 472;

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:


0.016 738 891 601 562 496 530 338(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2)

6. Positive number before normalization:

0.016 738 891 601 562 496 530 338(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(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:


0.016 738 891 601 562 496 530 338(10) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) =

0.0000 0100 0100 1000 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 10(2) × 20 =

1.0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110(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 1110


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 1110 =

0001 0010 0011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110


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 1110


Decimal number -0.016 738 891 601 562 496 530 338 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 1110

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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