-0.000 007 820 337 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 007 820 337 6(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
-0.000 007 820 337 6(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:

|-0.000 007 820 337 6| = 0.000 007 820 337 6


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: 0.000 007 820 337 6.

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) 0.000 007 820 337 6 × 2 = 0 + 0.000 015 640 675 2;
  • 2) 0.000 015 640 675 2 × 2 = 0 + 0.000 031 281 350 4;
  • 3) 0.000 031 281 350 4 × 2 = 0 + 0.000 062 562 700 8;
  • 4) 0.000 062 562 700 8 × 2 = 0 + 0.000 125 125 401 6;
  • 5) 0.000 125 125 401 6 × 2 = 0 + 0.000 250 250 803 2;
  • 6) 0.000 250 250 803 2 × 2 = 0 + 0.000 500 501 606 4;
  • 7) 0.000 500 501 606 4 × 2 = 0 + 0.001 001 003 212 8;
  • 8) 0.001 001 003 212 8 × 2 = 0 + 0.002 002 006 425 6;
  • 9) 0.002 002 006 425 6 × 2 = 0 + 0.004 004 012 851 2;
  • 10) 0.004 004 012 851 2 × 2 = 0 + 0.008 008 025 702 4;
  • 11) 0.008 008 025 702 4 × 2 = 0 + 0.016 016 051 404 8;
  • 12) 0.016 016 051 404 8 × 2 = 0 + 0.032 032 102 809 6;
  • 13) 0.032 032 102 809 6 × 2 = 0 + 0.064 064 205 619 2;
  • 14) 0.064 064 205 619 2 × 2 = 0 + 0.128 128 411 238 4;
  • 15) 0.128 128 411 238 4 × 2 = 0 + 0.256 256 822 476 8;
  • 16) 0.256 256 822 476 8 × 2 = 0 + 0.512 513 644 953 6;
  • 17) 0.512 513 644 953 6 × 2 = 1 + 0.025 027 289 907 2;
  • 18) 0.025 027 289 907 2 × 2 = 0 + 0.050 054 579 814 4;
  • 19) 0.050 054 579 814 4 × 2 = 0 + 0.100 109 159 628 8;
  • 20) 0.100 109 159 628 8 × 2 = 0 + 0.200 218 319 257 6;
  • 21) 0.200 218 319 257 6 × 2 = 0 + 0.400 436 638 515 2;
  • 22) 0.400 436 638 515 2 × 2 = 0 + 0.800 873 277 030 4;
  • 23) 0.800 873 277 030 4 × 2 = 1 + 0.601 746 554 060 8;
  • 24) 0.601 746 554 060 8 × 2 = 1 + 0.203 493 108 121 6;
  • 25) 0.203 493 108 121 6 × 2 = 0 + 0.406 986 216 243 2;
  • 26) 0.406 986 216 243 2 × 2 = 0 + 0.813 972 432 486 4;
  • 27) 0.813 972 432 486 4 × 2 = 1 + 0.627 944 864 972 8;
  • 28) 0.627 944 864 972 8 × 2 = 1 + 0.255 889 729 945 6;
  • 29) 0.255 889 729 945 6 × 2 = 0 + 0.511 779 459 891 2;
  • 30) 0.511 779 459 891 2 × 2 = 1 + 0.023 558 919 782 4;
  • 31) 0.023 558 919 782 4 × 2 = 0 + 0.047 117 839 564 8;
  • 32) 0.047 117 839 564 8 × 2 = 0 + 0.094 235 679 129 6;
  • 33) 0.094 235 679 129 6 × 2 = 0 + 0.188 471 358 259 2;
  • 34) 0.188 471 358 259 2 × 2 = 0 + 0.376 942 716 518 4;
  • 35) 0.376 942 716 518 4 × 2 = 0 + 0.753 885 433 036 8;
  • 36) 0.753 885 433 036 8 × 2 = 1 + 0.507 770 866 073 6;
  • 37) 0.507 770 866 073 6 × 2 = 1 + 0.015 541 732 147 2;
  • 38) 0.015 541 732 147 2 × 2 = 0 + 0.031 083 464 294 4;
  • 39) 0.031 083 464 294 4 × 2 = 0 + 0.062 166 928 588 8;
  • 40) 0.062 166 928 588 8 × 2 = 0 + 0.124 333 857 177 6;
  • 41) 0.124 333 857 177 6 × 2 = 0 + 0.248 667 714 355 2;
  • 42) 0.248 667 714 355 2 × 2 = 0 + 0.497 335 428 710 4;
  • 43) 0.497 335 428 710 4 × 2 = 0 + 0.994 670 857 420 8;
  • 44) 0.994 670 857 420 8 × 2 = 1 + 0.989 341 714 841 6;
  • 45) 0.989 341 714 841 6 × 2 = 1 + 0.978 683 429 683 2;
  • 46) 0.978 683 429 683 2 × 2 = 1 + 0.957 366 859 366 4;
  • 47) 0.957 366 859 366 4 × 2 = 1 + 0.914 733 718 732 8;
  • 48) 0.914 733 718 732 8 × 2 = 1 + 0.829 467 437 465 6;
  • 49) 0.829 467 437 465 6 × 2 = 1 + 0.658 934 874 931 2;
  • 50) 0.658 934 874 931 2 × 2 = 1 + 0.317 869 749 862 4;
  • 51) 0.317 869 749 862 4 × 2 = 0 + 0.635 739 499 724 8;
  • 52) 0.635 739 499 724 8 × 2 = 1 + 0.271 478 999 449 6;
  • 53) 0.271 478 999 449 6 × 2 = 0 + 0.542 957 998 899 2;
  • 54) 0.542 957 998 899 2 × 2 = 1 + 0.085 915 997 798 4;
  • 55) 0.085 915 997 798 4 × 2 = 0 + 0.171 831 995 596 8;
  • 56) 0.171 831 995 596 8 × 2 = 0 + 0.343 663 991 193 6;
  • 57) 0.343 663 991 193 6 × 2 = 0 + 0.687 327 982 387 2;
  • 58) 0.687 327 982 387 2 × 2 = 1 + 0.374 655 964 774 4;
  • 59) 0.374 655 964 774 4 × 2 = 0 + 0.749 311 929 548 8;
  • 60) 0.749 311 929 548 8 × 2 = 1 + 0.498 623 859 097 6;
  • 61) 0.498 623 859 097 6 × 2 = 0 + 0.997 247 718 195 2;
  • 62) 0.997 247 718 195 2 × 2 = 1 + 0.994 495 436 390 4;
  • 63) 0.994 495 436 390 4 × 2 = 1 + 0.988 990 872 780 8;
  • 64) 0.988 990 872 780 8 × 2 = 1 + 0.977 981 745 561 6;
  • 65) 0.977 981 745 561 6 × 2 = 1 + 0.955 963 491 123 2;
  • 66) 0.955 963 491 123 2 × 2 = 1 + 0.911 926 982 246 4;
  • 67) 0.911 926 982 246 4 × 2 = 1 + 0.823 853 964 492 8;
  • 68) 0.823 853 964 492 8 × 2 = 1 + 0.647 707 928 985 6;
  • 69) 0.647 707 928 985 6 × 2 = 1 + 0.295 415 857 971 2;

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:


0.000 007 820 337 6(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0001 1000 0001 1111 1101 0100 0101 0111 1111 1(2)

6. Positive number before normalization:

0.000 007 820 337 6(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0001 1000 0001 1111 1101 0100 0101 0111 1111 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 007 820 337 6(10) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0001 1000 0001 1111 1101 0100 0101 0111 1111 1(2) =

0.0000 0000 0000 0000 1000 0011 0011 0100 0001 1000 0001 1111 1101 0100 0101 0111 1111 1(2) × 20 =

1.0000 0110 0110 1000 0011 0000 0011 1111 1010 1000 1010 1111 1111(2) × 2-17


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): -17

Mantissa (not normalized):
1.0000 0110 0110 1000 0011 0000 0011 1111 1010 1000 1010 1111 1111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(11-1) - 1 =

(-17 + 1 023)(10) =

1 006(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;
  • 1 006 ÷ 2 = 503 + 0;
  • 503 ÷ 2 = 251 + 1;
  • 251 ÷ 2 = 125 + 1;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

1006(10) =

011 1110 1110(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. 0000 0110 0110 1000 0011 0000 0011 1111 1010 1000 1010 1111 1111 =

0000 0110 0110 1000 0011 0000 0011 1111 1010 1000 1010 1111 1111


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 1110 1110

Mantissa (52 bits) =
0000 0110 0110 1000 0011 0000 0011 1111 1010 1000 1010 1111 1111


Decimal number -0.000 007 820 337 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1110 - 0000 0110 0110 1000 0011 0000 0011 1111 1010 1000 1010 1111 1111

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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