-0.016 738 891 601 562 496 463 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.016 738 891 601 562 496 463(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
-0.016 738 891 601 562 496 463(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:

|-0.016 738 891 601 562 496 463| = 0.016 738 891 601 562 496 463


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: 0.016 738 891 601 562 496 463.

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) 0.016 738 891 601 562 496 463 × 2 = 0 + 0.033 477 783 203 124 992 926;
  • 2) 0.033 477 783 203 124 992 926 × 2 = 0 + 0.066 955 566 406 249 985 852;
  • 3) 0.066 955 566 406 249 985 852 × 2 = 0 + 0.133 911 132 812 499 971 704;
  • 4) 0.133 911 132 812 499 971 704 × 2 = 0 + 0.267 822 265 624 999 943 408;
  • 5) 0.267 822 265 624 999 943 408 × 2 = 0 + 0.535 644 531 249 999 886 816;
  • 6) 0.535 644 531 249 999 886 816 × 2 = 1 + 0.071 289 062 499 999 773 632;
  • 7) 0.071 289 062 499 999 773 632 × 2 = 0 + 0.142 578 124 999 999 547 264;
  • 8) 0.142 578 124 999 999 547 264 × 2 = 0 + 0.285 156 249 999 999 094 528;
  • 9) 0.285 156 249 999 999 094 528 × 2 = 0 + 0.570 312 499 999 998 189 056;
  • 10) 0.570 312 499 999 998 189 056 × 2 = 1 + 0.140 624 999 999 996 378 112;
  • 11) 0.140 624 999 999 996 378 112 × 2 = 0 + 0.281 249 999 999 992 756 224;
  • 12) 0.281 249 999 999 992 756 224 × 2 = 0 + 0.562 499 999 999 985 512 448;
  • 13) 0.562 499 999 999 985 512 448 × 2 = 1 + 0.124 999 999 999 971 024 896;
  • 14) 0.124 999 999 999 971 024 896 × 2 = 0 + 0.249 999 999 999 942 049 792;
  • 15) 0.249 999 999 999 942 049 792 × 2 = 0 + 0.499 999 999 999 884 099 584;
  • 16) 0.499 999 999 999 884 099 584 × 2 = 0 + 0.999 999 999 999 768 199 168;
  • 17) 0.999 999 999 999 768 199 168 × 2 = 1 + 0.999 999 999 999 536 398 336;
  • 18) 0.999 999 999 999 536 398 336 × 2 = 1 + 0.999 999 999 999 072 796 672;
  • 19) 0.999 999 999 999 072 796 672 × 2 = 1 + 0.999 999 999 998 145 593 344;
  • 20) 0.999 999 999 998 145 593 344 × 2 = 1 + 0.999 999 999 996 291 186 688;
  • 21) 0.999 999 999 996 291 186 688 × 2 = 1 + 0.999 999 999 992 582 373 376;
  • 22) 0.999 999 999 992 582 373 376 × 2 = 1 + 0.999 999 999 985 164 746 752;
  • 23) 0.999 999 999 985 164 746 752 × 2 = 1 + 0.999 999 999 970 329 493 504;
  • 24) 0.999 999 999 970 329 493 504 × 2 = 1 + 0.999 999 999 940 658 987 008;
  • 25) 0.999 999 999 940 658 987 008 × 2 = 1 + 0.999 999 999 881 317 974 016;
  • 26) 0.999 999 999 881 317 974 016 × 2 = 1 + 0.999 999 999 762 635 948 032;
  • 27) 0.999 999 999 762 635 948 032 × 2 = 1 + 0.999 999 999 525 271 896 064;
  • 28) 0.999 999 999 525 271 896 064 × 2 = 1 + 0.999 999 999 050 543 792 128;
  • 29) 0.999 999 999 050 543 792 128 × 2 = 1 + 0.999 999 998 101 087 584 256;
  • 30) 0.999 999 998 101 087 584 256 × 2 = 1 + 0.999 999 996 202 175 168 512;
  • 31) 0.999 999 996 202 175 168 512 × 2 = 1 + 0.999 999 992 404 350 337 024;
  • 32) 0.999 999 992 404 350 337 024 × 2 = 1 + 0.999 999 984 808 700 674 048;
  • 33) 0.999 999 984 808 700 674 048 × 2 = 1 + 0.999 999 969 617 401 348 096;
  • 34) 0.999 999 969 617 401 348 096 × 2 = 1 + 0.999 999 939 234 802 696 192;
  • 35) 0.999 999 939 234 802 696 192 × 2 = 1 + 0.999 999 878 469 605 392 384;
  • 36) 0.999 999 878 469 605 392 384 × 2 = 1 + 0.999 999 756 939 210 784 768;
  • 37) 0.999 999 756 939 210 784 768 × 2 = 1 + 0.999 999 513 878 421 569 536;
  • 38) 0.999 999 513 878 421 569 536 × 2 = 1 + 0.999 999 027 756 843 139 072;
  • 39) 0.999 999 027 756 843 139 072 × 2 = 1 + 0.999 998 055 513 686 278 144;
  • 40) 0.999 998 055 513 686 278 144 × 2 = 1 + 0.999 996 111 027 372 556 288;
  • 41) 0.999 996 111 027 372 556 288 × 2 = 1 + 0.999 992 222 054 745 112 576;
  • 42) 0.999 992 222 054 745 112 576 × 2 = 1 + 0.999 984 444 109 490 225 152;
  • 43) 0.999 984 444 109 490 225 152 × 2 = 1 + 0.999 968 888 218 980 450 304;
  • 44) 0.999 968 888 218 980 450 304 × 2 = 1 + 0.999 937 776 437 960 900 608;
  • 45) 0.999 937 776 437 960 900 608 × 2 = 1 + 0.999 875 552 875 921 801 216;
  • 46) 0.999 875 552 875 921 801 216 × 2 = 1 + 0.999 751 105 751 843 602 432;
  • 47) 0.999 751 105 751 843 602 432 × 2 = 1 + 0.999 502 211 503 687 204 864;
  • 48) 0.999 502 211 503 687 204 864 × 2 = 1 + 0.999 004 423 007 374 409 728;
  • 49) 0.999 004 423 007 374 409 728 × 2 = 1 + 0.998 008 846 014 748 819 456;
  • 50) 0.998 008 846 014 748 819 456 × 2 = 1 + 0.996 017 692 029 497 638 912;
  • 51) 0.996 017 692 029 497 638 912 × 2 = 1 + 0.992 035 384 058 995 277 824;
  • 52) 0.992 035 384 058 995 277 824 × 2 = 1 + 0.984 070 768 117 990 555 648;
  • 53) 0.984 070 768 117 990 555 648 × 2 = 1 + 0.968 141 536 235 981 111 296;
  • 54) 0.968 141 536 235 981 111 296 × 2 = 1 + 0.936 283 072 471 962 222 592;
  • 55) 0.936 283 072 471 962 222 592 × 2 = 1 + 0.872 566 144 943 924 445 184;
  • 56) 0.872 566 144 943 924 445 184 × 2 = 1 + 0.745 132 289 887 848 890 368;
  • 57) 0.745 132 289 887 848 890 368 × 2 = 1 + 0.490 264 579 775 697 780 736;
  • 58) 0.490 264 579 775 697 780 736 × 2 = 0 + 0.980 529 159 551 395 561 472;

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:


0.016 738 891 601 562 496 463(10) =

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

6. Positive number before normalization:

0.016 738 891 601 562 496 463(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:


0.016 738 891 601 562 496 463(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 -0.016 738 891 601 562 496 463 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