-0.000 375 658 488 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 375 658 488 68(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
-0.000 375 658 488 68(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:

|-0.000 375 658 488 68| = 0.000 375 658 488 68


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: 0.000 375 658 488 68.

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) 0.000 375 658 488 68 × 2 = 0 + 0.000 751 316 977 36;
  • 2) 0.000 751 316 977 36 × 2 = 0 + 0.001 502 633 954 72;
  • 3) 0.001 502 633 954 72 × 2 = 0 + 0.003 005 267 909 44;
  • 4) 0.003 005 267 909 44 × 2 = 0 + 0.006 010 535 818 88;
  • 5) 0.006 010 535 818 88 × 2 = 0 + 0.012 021 071 637 76;
  • 6) 0.012 021 071 637 76 × 2 = 0 + 0.024 042 143 275 52;
  • 7) 0.024 042 143 275 52 × 2 = 0 + 0.048 084 286 551 04;
  • 8) 0.048 084 286 551 04 × 2 = 0 + 0.096 168 573 102 08;
  • 9) 0.096 168 573 102 08 × 2 = 0 + 0.192 337 146 204 16;
  • 10) 0.192 337 146 204 16 × 2 = 0 + 0.384 674 292 408 32;
  • 11) 0.384 674 292 408 32 × 2 = 0 + 0.769 348 584 816 64;
  • 12) 0.769 348 584 816 64 × 2 = 1 + 0.538 697 169 633 28;
  • 13) 0.538 697 169 633 28 × 2 = 1 + 0.077 394 339 266 56;
  • 14) 0.077 394 339 266 56 × 2 = 0 + 0.154 788 678 533 12;
  • 15) 0.154 788 678 533 12 × 2 = 0 + 0.309 577 357 066 24;
  • 16) 0.309 577 357 066 24 × 2 = 0 + 0.619 154 714 132 48;
  • 17) 0.619 154 714 132 48 × 2 = 1 + 0.238 309 428 264 96;
  • 18) 0.238 309 428 264 96 × 2 = 0 + 0.476 618 856 529 92;
  • 19) 0.476 618 856 529 92 × 2 = 0 + 0.953 237 713 059 84;
  • 20) 0.953 237 713 059 84 × 2 = 1 + 0.906 475 426 119 68;
  • 21) 0.906 475 426 119 68 × 2 = 1 + 0.812 950 852 239 36;
  • 22) 0.812 950 852 239 36 × 2 = 1 + 0.625 901 704 478 72;
  • 23) 0.625 901 704 478 72 × 2 = 1 + 0.251 803 408 957 44;
  • 24) 0.251 803 408 957 44 × 2 = 0 + 0.503 606 817 914 88;
  • 25) 0.503 606 817 914 88 × 2 = 1 + 0.007 213 635 829 76;
  • 26) 0.007 213 635 829 76 × 2 = 0 + 0.014 427 271 659 52;
  • 27) 0.014 427 271 659 52 × 2 = 0 + 0.028 854 543 319 04;
  • 28) 0.028 854 543 319 04 × 2 = 0 + 0.057 709 086 638 08;
  • 29) 0.057 709 086 638 08 × 2 = 0 + 0.115 418 173 276 16;
  • 30) 0.115 418 173 276 16 × 2 = 0 + 0.230 836 346 552 32;
  • 31) 0.230 836 346 552 32 × 2 = 0 + 0.461 672 693 104 64;
  • 32) 0.461 672 693 104 64 × 2 = 0 + 0.923 345 386 209 28;
  • 33) 0.923 345 386 209 28 × 2 = 1 + 0.846 690 772 418 56;
  • 34) 0.846 690 772 418 56 × 2 = 1 + 0.693 381 544 837 12;
  • 35) 0.693 381 544 837 12 × 2 = 1 + 0.386 763 089 674 24;
  • 36) 0.386 763 089 674 24 × 2 = 0 + 0.773 526 179 348 48;
  • 37) 0.773 526 179 348 48 × 2 = 1 + 0.547 052 358 696 96;
  • 38) 0.547 052 358 696 96 × 2 = 1 + 0.094 104 717 393 92;
  • 39) 0.094 104 717 393 92 × 2 = 0 + 0.188 209 434 787 84;
  • 40) 0.188 209 434 787 84 × 2 = 0 + 0.376 418 869 575 68;
  • 41) 0.376 418 869 575 68 × 2 = 0 + 0.752 837 739 151 36;
  • 42) 0.752 837 739 151 36 × 2 = 1 + 0.505 675 478 302 72;
  • 43) 0.505 675 478 302 72 × 2 = 1 + 0.011 350 956 605 44;
  • 44) 0.011 350 956 605 44 × 2 = 0 + 0.022 701 913 210 88;
  • 45) 0.022 701 913 210 88 × 2 = 0 + 0.045 403 826 421 76;
  • 46) 0.045 403 826 421 76 × 2 = 0 + 0.090 807 652 843 52;
  • 47) 0.090 807 652 843 52 × 2 = 0 + 0.181 615 305 687 04;
  • 48) 0.181 615 305 687 04 × 2 = 0 + 0.363 230 611 374 08;
  • 49) 0.363 230 611 374 08 × 2 = 0 + 0.726 461 222 748 16;
  • 50) 0.726 461 222 748 16 × 2 = 1 + 0.452 922 445 496 32;
  • 51) 0.452 922 445 496 32 × 2 = 0 + 0.905 844 890 992 64;
  • 52) 0.905 844 890 992 64 × 2 = 1 + 0.811 689 781 985 28;
  • 53) 0.811 689 781 985 28 × 2 = 1 + 0.623 379 563 970 56;
  • 54) 0.623 379 563 970 56 × 2 = 1 + 0.246 759 127 941 12;
  • 55) 0.246 759 127 941 12 × 2 = 0 + 0.493 518 255 882 24;
  • 56) 0.493 518 255 882 24 × 2 = 0 + 0.987 036 511 764 48;
  • 57) 0.987 036 511 764 48 × 2 = 1 + 0.974 073 023 528 96;
  • 58) 0.974 073 023 528 96 × 2 = 1 + 0.948 146 047 057 92;
  • 59) 0.948 146 047 057 92 × 2 = 1 + 0.896 292 094 115 84;
  • 60) 0.896 292 094 115 84 × 2 = 1 + 0.792 584 188 231 68;
  • 61) 0.792 584 188 231 68 × 2 = 1 + 0.585 168 376 463 36;
  • 62) 0.585 168 376 463 36 × 2 = 1 + 0.170 336 752 926 72;
  • 63) 0.170 336 752 926 72 × 2 = 0 + 0.340 673 505 853 44;
  • 64) 0.340 673 505 853 44 × 2 = 0 + 0.681 347 011 706 88;

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:


0.000 375 658 488 68(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100(2)

6. Positive number before normalization:

0.000 375 658 488 68(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 375 658 488 68(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100(2) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100(2) × 20 =

1.1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100(2) × 2-12


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): -12

Mantissa (not normalized):
1.1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 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) =

1011(10) =

011 1111 0011(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. 1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100 =

1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100


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 0011

Mantissa (52 bits) =
1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100


Decimal number -0.000 375 658 488 68 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 1000 1001 1110 1000 0000 1110 1100 0110 0000 0101 1100 1111 1100

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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