-0.016 738 891 601 562 496 259 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 259(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
-0.016 738 891 601 562 496 259(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:

|-0.016 738 891 601 562 496 259| = 0.016 738 891 601 562 496 259


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: 0.016 738 891 601 562 496 259.

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) 0.016 738 891 601 562 496 259 × 2 = 0 + 0.033 477 783 203 124 992 518;
  • 2) 0.033 477 783 203 124 992 518 × 2 = 0 + 0.066 955 566 406 249 985 036;
  • 3) 0.066 955 566 406 249 985 036 × 2 = 0 + 0.133 911 132 812 499 970 072;
  • 4) 0.133 911 132 812 499 970 072 × 2 = 0 + 0.267 822 265 624 999 940 144;
  • 5) 0.267 822 265 624 999 940 144 × 2 = 0 + 0.535 644 531 249 999 880 288;
  • 6) 0.535 644 531 249 999 880 288 × 2 = 1 + 0.071 289 062 499 999 760 576;
  • 7) 0.071 289 062 499 999 760 576 × 2 = 0 + 0.142 578 124 999 999 521 152;
  • 8) 0.142 578 124 999 999 521 152 × 2 = 0 + 0.285 156 249 999 999 042 304;
  • 9) 0.285 156 249 999 999 042 304 × 2 = 0 + 0.570 312 499 999 998 084 608;
  • 10) 0.570 312 499 999 998 084 608 × 2 = 1 + 0.140 624 999 999 996 169 216;
  • 11) 0.140 624 999 999 996 169 216 × 2 = 0 + 0.281 249 999 999 992 338 432;
  • 12) 0.281 249 999 999 992 338 432 × 2 = 0 + 0.562 499 999 999 984 676 864;
  • 13) 0.562 499 999 999 984 676 864 × 2 = 1 + 0.124 999 999 999 969 353 728;
  • 14) 0.124 999 999 999 969 353 728 × 2 = 0 + 0.249 999 999 999 938 707 456;
  • 15) 0.249 999 999 999 938 707 456 × 2 = 0 + 0.499 999 999 999 877 414 912;
  • 16) 0.499 999 999 999 877 414 912 × 2 = 0 + 0.999 999 999 999 754 829 824;
  • 17) 0.999 999 999 999 754 829 824 × 2 = 1 + 0.999 999 999 999 509 659 648;
  • 18) 0.999 999 999 999 509 659 648 × 2 = 1 + 0.999 999 999 999 019 319 296;
  • 19) 0.999 999 999 999 019 319 296 × 2 = 1 + 0.999 999 999 998 038 638 592;
  • 20) 0.999 999 999 998 038 638 592 × 2 = 1 + 0.999 999 999 996 077 277 184;
  • 21) 0.999 999 999 996 077 277 184 × 2 = 1 + 0.999 999 999 992 154 554 368;
  • 22) 0.999 999 999 992 154 554 368 × 2 = 1 + 0.999 999 999 984 309 108 736;
  • 23) 0.999 999 999 984 309 108 736 × 2 = 1 + 0.999 999 999 968 618 217 472;
  • 24) 0.999 999 999 968 618 217 472 × 2 = 1 + 0.999 999 999 937 236 434 944;
  • 25) 0.999 999 999 937 236 434 944 × 2 = 1 + 0.999 999 999 874 472 869 888;
  • 26) 0.999 999 999 874 472 869 888 × 2 = 1 + 0.999 999 999 748 945 739 776;
  • 27) 0.999 999 999 748 945 739 776 × 2 = 1 + 0.999 999 999 497 891 479 552;
  • 28) 0.999 999 999 497 891 479 552 × 2 = 1 + 0.999 999 998 995 782 959 104;
  • 29) 0.999 999 998 995 782 959 104 × 2 = 1 + 0.999 999 997 991 565 918 208;
  • 30) 0.999 999 997 991 565 918 208 × 2 = 1 + 0.999 999 995 983 131 836 416;
  • 31) 0.999 999 995 983 131 836 416 × 2 = 1 + 0.999 999 991 966 263 672 832;
  • 32) 0.999 999 991 966 263 672 832 × 2 = 1 + 0.999 999 983 932 527 345 664;
  • 33) 0.999 999 983 932 527 345 664 × 2 = 1 + 0.999 999 967 865 054 691 328;
  • 34) 0.999 999 967 865 054 691 328 × 2 = 1 + 0.999 999 935 730 109 382 656;
  • 35) 0.999 999 935 730 109 382 656 × 2 = 1 + 0.999 999 871 460 218 765 312;
  • 36) 0.999 999 871 460 218 765 312 × 2 = 1 + 0.999 999 742 920 437 530 624;
  • 37) 0.999 999 742 920 437 530 624 × 2 = 1 + 0.999 999 485 840 875 061 248;
  • 38) 0.999 999 485 840 875 061 248 × 2 = 1 + 0.999 998 971 681 750 122 496;
  • 39) 0.999 998 971 681 750 122 496 × 2 = 1 + 0.999 997 943 363 500 244 992;
  • 40) 0.999 997 943 363 500 244 992 × 2 = 1 + 0.999 995 886 727 000 489 984;
  • 41) 0.999 995 886 727 000 489 984 × 2 = 1 + 0.999 991 773 454 000 979 968;
  • 42) 0.999 991 773 454 000 979 968 × 2 = 1 + 0.999 983 546 908 001 959 936;
  • 43) 0.999 983 546 908 001 959 936 × 2 = 1 + 0.999 967 093 816 003 919 872;
  • 44) 0.999 967 093 816 003 919 872 × 2 = 1 + 0.999 934 187 632 007 839 744;
  • 45) 0.999 934 187 632 007 839 744 × 2 = 1 + 0.999 868 375 264 015 679 488;
  • 46) 0.999 868 375 264 015 679 488 × 2 = 1 + 0.999 736 750 528 031 358 976;
  • 47) 0.999 736 750 528 031 358 976 × 2 = 1 + 0.999 473 501 056 062 717 952;
  • 48) 0.999 473 501 056 062 717 952 × 2 = 1 + 0.998 947 002 112 125 435 904;
  • 49) 0.998 947 002 112 125 435 904 × 2 = 1 + 0.997 894 004 224 250 871 808;
  • 50) 0.997 894 004 224 250 871 808 × 2 = 1 + 0.995 788 008 448 501 743 616;
  • 51) 0.995 788 008 448 501 743 616 × 2 = 1 + 0.991 576 016 897 003 487 232;
  • 52) 0.991 576 016 897 003 487 232 × 2 = 1 + 0.983 152 033 794 006 974 464;
  • 53) 0.983 152 033 794 006 974 464 × 2 = 1 + 0.966 304 067 588 013 948 928;
  • 54) 0.966 304 067 588 013 948 928 × 2 = 1 + 0.932 608 135 176 027 897 856;
  • 55) 0.932 608 135 176 027 897 856 × 2 = 1 + 0.865 216 270 352 055 795 712;
  • 56) 0.865 216 270 352 055 795 712 × 2 = 1 + 0.730 432 540 704 111 591 424;
  • 57) 0.730 432 540 704 111 591 424 × 2 = 1 + 0.460 865 081 408 223 182 848;
  • 58) 0.460 865 081 408 223 182 848 × 2 = 0 + 0.921 730 162 816 446 365 696;

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:


0.016 738 891 601 562 496 259(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 259(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:


0.016 738 891 601 562 496 259(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 -0.016 738 891 601 562 496 259 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