-0.000 084 993 381 976 777 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 084 993 381 976 777 99(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
-0.000 084 993 381 976 777 99(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:

|-0.000 084 993 381 976 777 99| = 0.000 084 993 381 976 777 99


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: 0.000 084 993 381 976 777 99.

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) 0.000 084 993 381 976 777 99 × 2 = 0 + 0.000 169 986 763 953 555 98;
  • 2) 0.000 169 986 763 953 555 98 × 2 = 0 + 0.000 339 973 527 907 111 96;
  • 3) 0.000 339 973 527 907 111 96 × 2 = 0 + 0.000 679 947 055 814 223 92;
  • 4) 0.000 679 947 055 814 223 92 × 2 = 0 + 0.001 359 894 111 628 447 84;
  • 5) 0.001 359 894 111 628 447 84 × 2 = 0 + 0.002 719 788 223 256 895 68;
  • 6) 0.002 719 788 223 256 895 68 × 2 = 0 + 0.005 439 576 446 513 791 36;
  • 7) 0.005 439 576 446 513 791 36 × 2 = 0 + 0.010 879 152 893 027 582 72;
  • 8) 0.010 879 152 893 027 582 72 × 2 = 0 + 0.021 758 305 786 055 165 44;
  • 9) 0.021 758 305 786 055 165 44 × 2 = 0 + 0.043 516 611 572 110 330 88;
  • 10) 0.043 516 611 572 110 330 88 × 2 = 0 + 0.087 033 223 144 220 661 76;
  • 11) 0.087 033 223 144 220 661 76 × 2 = 0 + 0.174 066 446 288 441 323 52;
  • 12) 0.174 066 446 288 441 323 52 × 2 = 0 + 0.348 132 892 576 882 647 04;
  • 13) 0.348 132 892 576 882 647 04 × 2 = 0 + 0.696 265 785 153 765 294 08;
  • 14) 0.696 265 785 153 765 294 08 × 2 = 1 + 0.392 531 570 307 530 588 16;
  • 15) 0.392 531 570 307 530 588 16 × 2 = 0 + 0.785 063 140 615 061 176 32;
  • 16) 0.785 063 140 615 061 176 32 × 2 = 1 + 0.570 126 281 230 122 352 64;
  • 17) 0.570 126 281 230 122 352 64 × 2 = 1 + 0.140 252 562 460 244 705 28;
  • 18) 0.140 252 562 460 244 705 28 × 2 = 0 + 0.280 505 124 920 489 410 56;
  • 19) 0.280 505 124 920 489 410 56 × 2 = 0 + 0.561 010 249 840 978 821 12;
  • 20) 0.561 010 249 840 978 821 12 × 2 = 1 + 0.122 020 499 681 957 642 24;
  • 21) 0.122 020 499 681 957 642 24 × 2 = 0 + 0.244 040 999 363 915 284 48;
  • 22) 0.244 040 999 363 915 284 48 × 2 = 0 + 0.488 081 998 727 830 568 96;
  • 23) 0.488 081 998 727 830 568 96 × 2 = 0 + 0.976 163 997 455 661 137 92;
  • 24) 0.976 163 997 455 661 137 92 × 2 = 1 + 0.952 327 994 911 322 275 84;
  • 25) 0.952 327 994 911 322 275 84 × 2 = 1 + 0.904 655 989 822 644 551 68;
  • 26) 0.904 655 989 822 644 551 68 × 2 = 1 + 0.809 311 979 645 289 103 36;
  • 27) 0.809 311 979 645 289 103 36 × 2 = 1 + 0.618 623 959 290 578 206 72;
  • 28) 0.618 623 959 290 578 206 72 × 2 = 1 + 0.237 247 918 581 156 413 44;
  • 29) 0.237 247 918 581 156 413 44 × 2 = 0 + 0.474 495 837 162 312 826 88;
  • 30) 0.474 495 837 162 312 826 88 × 2 = 0 + 0.948 991 674 324 625 653 76;
  • 31) 0.948 991 674 324 625 653 76 × 2 = 1 + 0.897 983 348 649 251 307 52;
  • 32) 0.897 983 348 649 251 307 52 × 2 = 1 + 0.795 966 697 298 502 615 04;
  • 33) 0.795 966 697 298 502 615 04 × 2 = 1 + 0.591 933 394 597 005 230 08;
  • 34) 0.591 933 394 597 005 230 08 × 2 = 1 + 0.183 866 789 194 010 460 16;
  • 35) 0.183 866 789 194 010 460 16 × 2 = 0 + 0.367 733 578 388 020 920 32;
  • 36) 0.367 733 578 388 020 920 32 × 2 = 0 + 0.735 467 156 776 041 840 64;
  • 37) 0.735 467 156 776 041 840 64 × 2 = 1 + 0.470 934 313 552 083 681 28;
  • 38) 0.470 934 313 552 083 681 28 × 2 = 0 + 0.941 868 627 104 167 362 56;
  • 39) 0.941 868 627 104 167 362 56 × 2 = 1 + 0.883 737 254 208 334 725 12;
  • 40) 0.883 737 254 208 334 725 12 × 2 = 1 + 0.767 474 508 416 669 450 24;
  • 41) 0.767 474 508 416 669 450 24 × 2 = 1 + 0.534 949 016 833 338 900 48;
  • 42) 0.534 949 016 833 338 900 48 × 2 = 1 + 0.069 898 033 666 677 800 96;
  • 43) 0.069 898 033 666 677 800 96 × 2 = 0 + 0.139 796 067 333 355 601 92;
  • 44) 0.139 796 067 333 355 601 92 × 2 = 0 + 0.279 592 134 666 711 203 84;
  • 45) 0.279 592 134 666 711 203 84 × 2 = 0 + 0.559 184 269 333 422 407 68;
  • 46) 0.559 184 269 333 422 407 68 × 2 = 1 + 0.118 368 538 666 844 815 36;
  • 47) 0.118 368 538 666 844 815 36 × 2 = 0 + 0.236 737 077 333 689 630 72;
  • 48) 0.236 737 077 333 689 630 72 × 2 = 0 + 0.473 474 154 667 379 261 44;
  • 49) 0.473 474 154 667 379 261 44 × 2 = 0 + 0.946 948 309 334 758 522 88;
  • 50) 0.946 948 309 334 758 522 88 × 2 = 1 + 0.893 896 618 669 517 045 76;
  • 51) 0.893 896 618 669 517 045 76 × 2 = 1 + 0.787 793 237 339 034 091 52;
  • 52) 0.787 793 237 339 034 091 52 × 2 = 1 + 0.575 586 474 678 068 183 04;
  • 53) 0.575 586 474 678 068 183 04 × 2 = 1 + 0.151 172 949 356 136 366 08;
  • 54) 0.151 172 949 356 136 366 08 × 2 = 0 + 0.302 345 898 712 272 732 16;
  • 55) 0.302 345 898 712 272 732 16 × 2 = 0 + 0.604 691 797 424 545 464 32;
  • 56) 0.604 691 797 424 545 464 32 × 2 = 1 + 0.209 383 594 849 090 928 64;
  • 57) 0.209 383 594 849 090 928 64 × 2 = 0 + 0.418 767 189 698 181 857 28;
  • 58) 0.418 767 189 698 181 857 28 × 2 = 0 + 0.837 534 379 396 363 714 56;
  • 59) 0.837 534 379 396 363 714 56 × 2 = 1 + 0.675 068 758 792 727 429 12;
  • 60) 0.675 068 758 792 727 429 12 × 2 = 1 + 0.350 137 517 585 454 858 24;
  • 61) 0.350 137 517 585 454 858 24 × 2 = 0 + 0.700 275 035 170 909 716 48;
  • 62) 0.700 275 035 170 909 716 48 × 2 = 1 + 0.400 550 070 341 819 432 96;
  • 63) 0.400 550 070 341 819 432 96 × 2 = 0 + 0.801 100 140 683 638 865 92;
  • 64) 0.801 100 140 683 638 865 92 × 2 = 1 + 0.602 200 281 367 277 731 84;
  • 65) 0.602 200 281 367 277 731 84 × 2 = 1 + 0.204 400 562 734 555 463 68;
  • 66) 0.204 400 562 734 555 463 68 × 2 = 0 + 0.408 801 125 469 110 927 36;

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:


0.000 084 993 381 976 777 99(10) =

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

6. Positive number before normalization:

0.000 084 993 381 976 777 99(10) =

0.0000 0000 0000 0101 1001 0001 1111 0011 1100 1011 1100 0100 0111 1001 0011 0101 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:


0.000 084 993 381 976 777 99(10) =

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

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

1.0110 0100 0111 1100 1111 0010 1111 0001 0001 1110 0100 1101 0110(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 1110 0100 1101 0110


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 1110 0100 1101 0110 =

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


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 1110 0100 1101 0110


Decimal number -0.000 084 993 381 976 777 99 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 1110 0100 1101 0110

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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