-0.000 084 993 381 976 750 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 084 993 381 976 750 4(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
-0.000 084 993 381 976 750 4(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:

|-0.000 084 993 381 976 750 4| = 0.000 084 993 381 976 750 4


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: 0.000 084 993 381 976 750 4.

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) 0.000 084 993 381 976 750 4 × 2 = 0 + 0.000 169 986 763 953 500 8;
  • 2) 0.000 169 986 763 953 500 8 × 2 = 0 + 0.000 339 973 527 907 001 6;
  • 3) 0.000 339 973 527 907 001 6 × 2 = 0 + 0.000 679 947 055 814 003 2;
  • 4) 0.000 679 947 055 814 003 2 × 2 = 0 + 0.001 359 894 111 628 006 4;
  • 5) 0.001 359 894 111 628 006 4 × 2 = 0 + 0.002 719 788 223 256 012 8;
  • 6) 0.002 719 788 223 256 012 8 × 2 = 0 + 0.005 439 576 446 512 025 6;
  • 7) 0.005 439 576 446 512 025 6 × 2 = 0 + 0.010 879 152 893 024 051 2;
  • 8) 0.010 879 152 893 024 051 2 × 2 = 0 + 0.021 758 305 786 048 102 4;
  • 9) 0.021 758 305 786 048 102 4 × 2 = 0 + 0.043 516 611 572 096 204 8;
  • 10) 0.043 516 611 572 096 204 8 × 2 = 0 + 0.087 033 223 144 192 409 6;
  • 11) 0.087 033 223 144 192 409 6 × 2 = 0 + 0.174 066 446 288 384 819 2;
  • 12) 0.174 066 446 288 384 819 2 × 2 = 0 + 0.348 132 892 576 769 638 4;
  • 13) 0.348 132 892 576 769 638 4 × 2 = 0 + 0.696 265 785 153 539 276 8;
  • 14) 0.696 265 785 153 539 276 8 × 2 = 1 + 0.392 531 570 307 078 553 6;
  • 15) 0.392 531 570 307 078 553 6 × 2 = 0 + 0.785 063 140 614 157 107 2;
  • 16) 0.785 063 140 614 157 107 2 × 2 = 1 + 0.570 126 281 228 314 214 4;
  • 17) 0.570 126 281 228 314 214 4 × 2 = 1 + 0.140 252 562 456 628 428 8;
  • 18) 0.140 252 562 456 628 428 8 × 2 = 0 + 0.280 505 124 913 256 857 6;
  • 19) 0.280 505 124 913 256 857 6 × 2 = 0 + 0.561 010 249 826 513 715 2;
  • 20) 0.561 010 249 826 513 715 2 × 2 = 1 + 0.122 020 499 653 027 430 4;
  • 21) 0.122 020 499 653 027 430 4 × 2 = 0 + 0.244 040 999 306 054 860 8;
  • 22) 0.244 040 999 306 054 860 8 × 2 = 0 + 0.488 081 998 612 109 721 6;
  • 23) 0.488 081 998 612 109 721 6 × 2 = 0 + 0.976 163 997 224 219 443 2;
  • 24) 0.976 163 997 224 219 443 2 × 2 = 1 + 0.952 327 994 448 438 886 4;
  • 25) 0.952 327 994 448 438 886 4 × 2 = 1 + 0.904 655 988 896 877 772 8;
  • 26) 0.904 655 988 896 877 772 8 × 2 = 1 + 0.809 311 977 793 755 545 6;
  • 27) 0.809 311 977 793 755 545 6 × 2 = 1 + 0.618 623 955 587 511 091 2;
  • 28) 0.618 623 955 587 511 091 2 × 2 = 1 + 0.237 247 911 175 022 182 4;
  • 29) 0.237 247 911 175 022 182 4 × 2 = 0 + 0.474 495 822 350 044 364 8;
  • 30) 0.474 495 822 350 044 364 8 × 2 = 0 + 0.948 991 644 700 088 729 6;
  • 31) 0.948 991 644 700 088 729 6 × 2 = 1 + 0.897 983 289 400 177 459 2;
  • 32) 0.897 983 289 400 177 459 2 × 2 = 1 + 0.795 966 578 800 354 918 4;
  • 33) 0.795 966 578 800 354 918 4 × 2 = 1 + 0.591 933 157 600 709 836 8;
  • 34) 0.591 933 157 600 709 836 8 × 2 = 1 + 0.183 866 315 201 419 673 6;
  • 35) 0.183 866 315 201 419 673 6 × 2 = 0 + 0.367 732 630 402 839 347 2;
  • 36) 0.367 732 630 402 839 347 2 × 2 = 0 + 0.735 465 260 805 678 694 4;
  • 37) 0.735 465 260 805 678 694 4 × 2 = 1 + 0.470 930 521 611 357 388 8;
  • 38) 0.470 930 521 611 357 388 8 × 2 = 0 + 0.941 861 043 222 714 777 6;
  • 39) 0.941 861 043 222 714 777 6 × 2 = 1 + 0.883 722 086 445 429 555 2;
  • 40) 0.883 722 086 445 429 555 2 × 2 = 1 + 0.767 444 172 890 859 110 4;
  • 41) 0.767 444 172 890 859 110 4 × 2 = 1 + 0.534 888 345 781 718 220 8;
  • 42) 0.534 888 345 781 718 220 8 × 2 = 1 + 0.069 776 691 563 436 441 6;
  • 43) 0.069 776 691 563 436 441 6 × 2 = 0 + 0.139 553 383 126 872 883 2;
  • 44) 0.139 553 383 126 872 883 2 × 2 = 0 + 0.279 106 766 253 745 766 4;
  • 45) 0.279 106 766 253 745 766 4 × 2 = 0 + 0.558 213 532 507 491 532 8;
  • 46) 0.558 213 532 507 491 532 8 × 2 = 1 + 0.116 427 065 014 983 065 6;
  • 47) 0.116 427 065 014 983 065 6 × 2 = 0 + 0.232 854 130 029 966 131 2;
  • 48) 0.232 854 130 029 966 131 2 × 2 = 0 + 0.465 708 260 059 932 262 4;
  • 49) 0.465 708 260 059 932 262 4 × 2 = 0 + 0.931 416 520 119 864 524 8;
  • 50) 0.931 416 520 119 864 524 8 × 2 = 1 + 0.862 833 040 239 729 049 6;
  • 51) 0.862 833 040 239 729 049 6 × 2 = 1 + 0.725 666 080 479 458 099 2;
  • 52) 0.725 666 080 479 458 099 2 × 2 = 1 + 0.451 332 160 958 916 198 4;
  • 53) 0.451 332 160 958 916 198 4 × 2 = 0 + 0.902 664 321 917 832 396 8;
  • 54) 0.902 664 321 917 832 396 8 × 2 = 1 + 0.805 328 643 835 664 793 6;
  • 55) 0.805 328 643 835 664 793 6 × 2 = 1 + 0.610 657 287 671 329 587 2;
  • 56) 0.610 657 287 671 329 587 2 × 2 = 1 + 0.221 314 575 342 659 174 4;
  • 57) 0.221 314 575 342 659 174 4 × 2 = 0 + 0.442 629 150 685 318 348 8;
  • 58) 0.442 629 150 685 318 348 8 × 2 = 0 + 0.885 258 301 370 636 697 6;
  • 59) 0.885 258 301 370 636 697 6 × 2 = 1 + 0.770 516 602 741 273 395 2;
  • 60) 0.770 516 602 741 273 395 2 × 2 = 1 + 0.541 033 205 482 546 790 4;
  • 61) 0.541 033 205 482 546 790 4 × 2 = 1 + 0.082 066 410 965 093 580 8;
  • 62) 0.082 066 410 965 093 580 8 × 2 = 0 + 0.164 132 821 930 187 161 6;
  • 63) 0.164 132 821 930 187 161 6 × 2 = 0 + 0.328 265 643 860 374 323 2;
  • 64) 0.328 265 643 860 374 323 2 × 2 = 0 + 0.656 531 287 720 748 646 4;
  • 65) 0.656 531 287 720 748 646 4 × 2 = 1 + 0.313 062 575 441 497 292 8;
  • 66) 0.313 062 575 441 497 292 8 × 2 = 0 + 0.626 125 150 882 994 585 6;

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:


0.000 084 993 381 976 750 4(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 0111 0011 1000 10(2)

6. Positive number before normalization:

0.000 084 993 381 976 750 4(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 0111 0011 1000 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 084 993 381 976 750 4(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 0111 0011 1000 10(2) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 0111 0011 1000 10(2) × 20 =

1.0110 0100 0111 1100 1111 0010 1111 0001 0001 1101 1100 1110 0010(2) × 2-14


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): -14

Mantissa (not normalized):
1.0110 0100 0111 1100 1111 0010 1111 0001 0001 1101 1100 1110 0010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-14 + 2(11-1) - 1 =

(-14 + 1 023)(10) =

1 009(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;
  • 1 009 ÷ 2 = 504 + 1;
  • 504 ÷ 2 = 252 + 0;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1009(10) =

011 1111 0001(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. 0110 0100 0111 1100 1111 0010 1111 0001 0001 1101 1100 1110 0010 =

0110 0100 0111 1100 1111 0010 1111 0001 0001 1101 1100 1110 0010


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 0001

Mantissa (52 bits) =
0110 0100 0111 1100 1111 0010 1111 0001 0001 1101 1100 1110 0010


Decimal number -0.000 084 993 381 976 750 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0001 - 0110 0100 0111 1100 1111 0010 1111 0001 0001 1101 1100 1110 0010

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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