-0.016 738 891 601 562 496 187 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 187(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
-0.016 738 891 601 562 496 187(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:

|-0.016 738 891 601 562 496 187| = 0.016 738 891 601 562 496 187


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: 0.016 738 891 601 562 496 187.

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) 0.016 738 891 601 562 496 187 × 2 = 0 + 0.033 477 783 203 124 992 374;
  • 2) 0.033 477 783 203 124 992 374 × 2 = 0 + 0.066 955 566 406 249 984 748;
  • 3) 0.066 955 566 406 249 984 748 × 2 = 0 + 0.133 911 132 812 499 969 496;
  • 4) 0.133 911 132 812 499 969 496 × 2 = 0 + 0.267 822 265 624 999 938 992;
  • 5) 0.267 822 265 624 999 938 992 × 2 = 0 + 0.535 644 531 249 999 877 984;
  • 6) 0.535 644 531 249 999 877 984 × 2 = 1 + 0.071 289 062 499 999 755 968;
  • 7) 0.071 289 062 499 999 755 968 × 2 = 0 + 0.142 578 124 999 999 511 936;
  • 8) 0.142 578 124 999 999 511 936 × 2 = 0 + 0.285 156 249 999 999 023 872;
  • 9) 0.285 156 249 999 999 023 872 × 2 = 0 + 0.570 312 499 999 998 047 744;
  • 10) 0.570 312 499 999 998 047 744 × 2 = 1 + 0.140 624 999 999 996 095 488;
  • 11) 0.140 624 999 999 996 095 488 × 2 = 0 + 0.281 249 999 999 992 190 976;
  • 12) 0.281 249 999 999 992 190 976 × 2 = 0 + 0.562 499 999 999 984 381 952;
  • 13) 0.562 499 999 999 984 381 952 × 2 = 1 + 0.124 999 999 999 968 763 904;
  • 14) 0.124 999 999 999 968 763 904 × 2 = 0 + 0.249 999 999 999 937 527 808;
  • 15) 0.249 999 999 999 937 527 808 × 2 = 0 + 0.499 999 999 999 875 055 616;
  • 16) 0.499 999 999 999 875 055 616 × 2 = 0 + 0.999 999 999 999 750 111 232;
  • 17) 0.999 999 999 999 750 111 232 × 2 = 1 + 0.999 999 999 999 500 222 464;
  • 18) 0.999 999 999 999 500 222 464 × 2 = 1 + 0.999 999 999 999 000 444 928;
  • 19) 0.999 999 999 999 000 444 928 × 2 = 1 + 0.999 999 999 998 000 889 856;
  • 20) 0.999 999 999 998 000 889 856 × 2 = 1 + 0.999 999 999 996 001 779 712;
  • 21) 0.999 999 999 996 001 779 712 × 2 = 1 + 0.999 999 999 992 003 559 424;
  • 22) 0.999 999 999 992 003 559 424 × 2 = 1 + 0.999 999 999 984 007 118 848;
  • 23) 0.999 999 999 984 007 118 848 × 2 = 1 + 0.999 999 999 968 014 237 696;
  • 24) 0.999 999 999 968 014 237 696 × 2 = 1 + 0.999 999 999 936 028 475 392;
  • 25) 0.999 999 999 936 028 475 392 × 2 = 1 + 0.999 999 999 872 056 950 784;
  • 26) 0.999 999 999 872 056 950 784 × 2 = 1 + 0.999 999 999 744 113 901 568;
  • 27) 0.999 999 999 744 113 901 568 × 2 = 1 + 0.999 999 999 488 227 803 136;
  • 28) 0.999 999 999 488 227 803 136 × 2 = 1 + 0.999 999 998 976 455 606 272;
  • 29) 0.999 999 998 976 455 606 272 × 2 = 1 + 0.999 999 997 952 911 212 544;
  • 30) 0.999 999 997 952 911 212 544 × 2 = 1 + 0.999 999 995 905 822 425 088;
  • 31) 0.999 999 995 905 822 425 088 × 2 = 1 + 0.999 999 991 811 644 850 176;
  • 32) 0.999 999 991 811 644 850 176 × 2 = 1 + 0.999 999 983 623 289 700 352;
  • 33) 0.999 999 983 623 289 700 352 × 2 = 1 + 0.999 999 967 246 579 400 704;
  • 34) 0.999 999 967 246 579 400 704 × 2 = 1 + 0.999 999 934 493 158 801 408;
  • 35) 0.999 999 934 493 158 801 408 × 2 = 1 + 0.999 999 868 986 317 602 816;
  • 36) 0.999 999 868 986 317 602 816 × 2 = 1 + 0.999 999 737 972 635 205 632;
  • 37) 0.999 999 737 972 635 205 632 × 2 = 1 + 0.999 999 475 945 270 411 264;
  • 38) 0.999 999 475 945 270 411 264 × 2 = 1 + 0.999 998 951 890 540 822 528;
  • 39) 0.999 998 951 890 540 822 528 × 2 = 1 + 0.999 997 903 781 081 645 056;
  • 40) 0.999 997 903 781 081 645 056 × 2 = 1 + 0.999 995 807 562 163 290 112;
  • 41) 0.999 995 807 562 163 290 112 × 2 = 1 + 0.999 991 615 124 326 580 224;
  • 42) 0.999 991 615 124 326 580 224 × 2 = 1 + 0.999 983 230 248 653 160 448;
  • 43) 0.999 983 230 248 653 160 448 × 2 = 1 + 0.999 966 460 497 306 320 896;
  • 44) 0.999 966 460 497 306 320 896 × 2 = 1 + 0.999 932 920 994 612 641 792;
  • 45) 0.999 932 920 994 612 641 792 × 2 = 1 + 0.999 865 841 989 225 283 584;
  • 46) 0.999 865 841 989 225 283 584 × 2 = 1 + 0.999 731 683 978 450 567 168;
  • 47) 0.999 731 683 978 450 567 168 × 2 = 1 + 0.999 463 367 956 901 134 336;
  • 48) 0.999 463 367 956 901 134 336 × 2 = 1 + 0.998 926 735 913 802 268 672;
  • 49) 0.998 926 735 913 802 268 672 × 2 = 1 + 0.997 853 471 827 604 537 344;
  • 50) 0.997 853 471 827 604 537 344 × 2 = 1 + 0.995 706 943 655 209 074 688;
  • 51) 0.995 706 943 655 209 074 688 × 2 = 1 + 0.991 413 887 310 418 149 376;
  • 52) 0.991 413 887 310 418 149 376 × 2 = 1 + 0.982 827 774 620 836 298 752;
  • 53) 0.982 827 774 620 836 298 752 × 2 = 1 + 0.965 655 549 241 672 597 504;
  • 54) 0.965 655 549 241 672 597 504 × 2 = 1 + 0.931 311 098 483 345 195 008;
  • 55) 0.931 311 098 483 345 195 008 × 2 = 1 + 0.862 622 196 966 690 390 016;
  • 56) 0.862 622 196 966 690 390 016 × 2 = 1 + 0.725 244 393 933 380 780 032;
  • 57) 0.725 244 393 933 380 780 032 × 2 = 1 + 0.450 488 787 866 761 560 064;
  • 58) 0.450 488 787 866 761 560 064 × 2 = 0 + 0.900 977 575 733 523 120 128;

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:


0.016 738 891 601 562 496 187(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 187(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:


0.016 738 891 601 562 496 187(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 -0.016 738 891 601 562 496 187 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