-0.016 738 891 601 562 496 544 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 544(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
-0.016 738 891 601 562 496 544(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:

|-0.016 738 891 601 562 496 544| = 0.016 738 891 601 562 496 544


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: 0.016 738 891 601 562 496 544.

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) 0.016 738 891 601 562 496 544 × 2 = 0 + 0.033 477 783 203 124 993 088;
  • 2) 0.033 477 783 203 124 993 088 × 2 = 0 + 0.066 955 566 406 249 986 176;
  • 3) 0.066 955 566 406 249 986 176 × 2 = 0 + 0.133 911 132 812 499 972 352;
  • 4) 0.133 911 132 812 499 972 352 × 2 = 0 + 0.267 822 265 624 999 944 704;
  • 5) 0.267 822 265 624 999 944 704 × 2 = 0 + 0.535 644 531 249 999 889 408;
  • 6) 0.535 644 531 249 999 889 408 × 2 = 1 + 0.071 289 062 499 999 778 816;
  • 7) 0.071 289 062 499 999 778 816 × 2 = 0 + 0.142 578 124 999 999 557 632;
  • 8) 0.142 578 124 999 999 557 632 × 2 = 0 + 0.285 156 249 999 999 115 264;
  • 9) 0.285 156 249 999 999 115 264 × 2 = 0 + 0.570 312 499 999 998 230 528;
  • 10) 0.570 312 499 999 998 230 528 × 2 = 1 + 0.140 624 999 999 996 461 056;
  • 11) 0.140 624 999 999 996 461 056 × 2 = 0 + 0.281 249 999 999 992 922 112;
  • 12) 0.281 249 999 999 992 922 112 × 2 = 0 + 0.562 499 999 999 985 844 224;
  • 13) 0.562 499 999 999 985 844 224 × 2 = 1 + 0.124 999 999 999 971 688 448;
  • 14) 0.124 999 999 999 971 688 448 × 2 = 0 + 0.249 999 999 999 943 376 896;
  • 15) 0.249 999 999 999 943 376 896 × 2 = 0 + 0.499 999 999 999 886 753 792;
  • 16) 0.499 999 999 999 886 753 792 × 2 = 0 + 0.999 999 999 999 773 507 584;
  • 17) 0.999 999 999 999 773 507 584 × 2 = 1 + 0.999 999 999 999 547 015 168;
  • 18) 0.999 999 999 999 547 015 168 × 2 = 1 + 0.999 999 999 999 094 030 336;
  • 19) 0.999 999 999 999 094 030 336 × 2 = 1 + 0.999 999 999 998 188 060 672;
  • 20) 0.999 999 999 998 188 060 672 × 2 = 1 + 0.999 999 999 996 376 121 344;
  • 21) 0.999 999 999 996 376 121 344 × 2 = 1 + 0.999 999 999 992 752 242 688;
  • 22) 0.999 999 999 992 752 242 688 × 2 = 1 + 0.999 999 999 985 504 485 376;
  • 23) 0.999 999 999 985 504 485 376 × 2 = 1 + 0.999 999 999 971 008 970 752;
  • 24) 0.999 999 999 971 008 970 752 × 2 = 1 + 0.999 999 999 942 017 941 504;
  • 25) 0.999 999 999 942 017 941 504 × 2 = 1 + 0.999 999 999 884 035 883 008;
  • 26) 0.999 999 999 884 035 883 008 × 2 = 1 + 0.999 999 999 768 071 766 016;
  • 27) 0.999 999 999 768 071 766 016 × 2 = 1 + 0.999 999 999 536 143 532 032;
  • 28) 0.999 999 999 536 143 532 032 × 2 = 1 + 0.999 999 999 072 287 064 064;
  • 29) 0.999 999 999 072 287 064 064 × 2 = 1 + 0.999 999 998 144 574 128 128;
  • 30) 0.999 999 998 144 574 128 128 × 2 = 1 + 0.999 999 996 289 148 256 256;
  • 31) 0.999 999 996 289 148 256 256 × 2 = 1 + 0.999 999 992 578 296 512 512;
  • 32) 0.999 999 992 578 296 512 512 × 2 = 1 + 0.999 999 985 156 593 025 024;
  • 33) 0.999 999 985 156 593 025 024 × 2 = 1 + 0.999 999 970 313 186 050 048;
  • 34) 0.999 999 970 313 186 050 048 × 2 = 1 + 0.999 999 940 626 372 100 096;
  • 35) 0.999 999 940 626 372 100 096 × 2 = 1 + 0.999 999 881 252 744 200 192;
  • 36) 0.999 999 881 252 744 200 192 × 2 = 1 + 0.999 999 762 505 488 400 384;
  • 37) 0.999 999 762 505 488 400 384 × 2 = 1 + 0.999 999 525 010 976 800 768;
  • 38) 0.999 999 525 010 976 800 768 × 2 = 1 + 0.999 999 050 021 953 601 536;
  • 39) 0.999 999 050 021 953 601 536 × 2 = 1 + 0.999 998 100 043 907 203 072;
  • 40) 0.999 998 100 043 907 203 072 × 2 = 1 + 0.999 996 200 087 814 406 144;
  • 41) 0.999 996 200 087 814 406 144 × 2 = 1 + 0.999 992 400 175 628 812 288;
  • 42) 0.999 992 400 175 628 812 288 × 2 = 1 + 0.999 984 800 351 257 624 576;
  • 43) 0.999 984 800 351 257 624 576 × 2 = 1 + 0.999 969 600 702 515 249 152;
  • 44) 0.999 969 600 702 515 249 152 × 2 = 1 + 0.999 939 201 405 030 498 304;
  • 45) 0.999 939 201 405 030 498 304 × 2 = 1 + 0.999 878 402 810 060 996 608;
  • 46) 0.999 878 402 810 060 996 608 × 2 = 1 + 0.999 756 805 620 121 993 216;
  • 47) 0.999 756 805 620 121 993 216 × 2 = 1 + 0.999 513 611 240 243 986 432;
  • 48) 0.999 513 611 240 243 986 432 × 2 = 1 + 0.999 027 222 480 487 972 864;
  • 49) 0.999 027 222 480 487 972 864 × 2 = 1 + 0.998 054 444 960 975 945 728;
  • 50) 0.998 054 444 960 975 945 728 × 2 = 1 + 0.996 108 889 921 951 891 456;
  • 51) 0.996 108 889 921 951 891 456 × 2 = 1 + 0.992 217 779 843 903 782 912;
  • 52) 0.992 217 779 843 903 782 912 × 2 = 1 + 0.984 435 559 687 807 565 824;
  • 53) 0.984 435 559 687 807 565 824 × 2 = 1 + 0.968 871 119 375 615 131 648;
  • 54) 0.968 871 119 375 615 131 648 × 2 = 1 + 0.937 742 238 751 230 263 296;
  • 55) 0.937 742 238 751 230 263 296 × 2 = 1 + 0.875 484 477 502 460 526 592;
  • 56) 0.875 484 477 502 460 526 592 × 2 = 1 + 0.750 968 955 004 921 053 184;
  • 57) 0.750 968 955 004 921 053 184 × 2 = 1 + 0.501 937 910 009 842 106 368;
  • 58) 0.501 937 910 009 842 106 368 × 2 = 1 + 0.003 875 820 019 684 212 736;

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).


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:


0.016 738 891 601 562 496 544(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 544(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:


0.016 738 891 601 562 496 544(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 -0.016 738 891 601 562 496 544 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