-0.000 375 658 489 25 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 375 658 489 25(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
-0.000 375 658 489 25(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:

|-0.000 375 658 489 25| = 0.000 375 658 489 25


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: 0.000 375 658 489 25.

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) 0.000 375 658 489 25 × 2 = 0 + 0.000 751 316 978 5;
  • 2) 0.000 751 316 978 5 × 2 = 0 + 0.001 502 633 957;
  • 3) 0.001 502 633 957 × 2 = 0 + 0.003 005 267 914;
  • 4) 0.003 005 267 914 × 2 = 0 + 0.006 010 535 828;
  • 5) 0.006 010 535 828 × 2 = 0 + 0.012 021 071 656;
  • 6) 0.012 021 071 656 × 2 = 0 + 0.024 042 143 312;
  • 7) 0.024 042 143 312 × 2 = 0 + 0.048 084 286 624;
  • 8) 0.048 084 286 624 × 2 = 0 + 0.096 168 573 248;
  • 9) 0.096 168 573 248 × 2 = 0 + 0.192 337 146 496;
  • 10) 0.192 337 146 496 × 2 = 0 + 0.384 674 292 992;
  • 11) 0.384 674 292 992 × 2 = 0 + 0.769 348 585 984;
  • 12) 0.769 348 585 984 × 2 = 1 + 0.538 697 171 968;
  • 13) 0.538 697 171 968 × 2 = 1 + 0.077 394 343 936;
  • 14) 0.077 394 343 936 × 2 = 0 + 0.154 788 687 872;
  • 15) 0.154 788 687 872 × 2 = 0 + 0.309 577 375 744;
  • 16) 0.309 577 375 744 × 2 = 0 + 0.619 154 751 488;
  • 17) 0.619 154 751 488 × 2 = 1 + 0.238 309 502 976;
  • 18) 0.238 309 502 976 × 2 = 0 + 0.476 619 005 952;
  • 19) 0.476 619 005 952 × 2 = 0 + 0.953 238 011 904;
  • 20) 0.953 238 011 904 × 2 = 1 + 0.906 476 023 808;
  • 21) 0.906 476 023 808 × 2 = 1 + 0.812 952 047 616;
  • 22) 0.812 952 047 616 × 2 = 1 + 0.625 904 095 232;
  • 23) 0.625 904 095 232 × 2 = 1 + 0.251 808 190 464;
  • 24) 0.251 808 190 464 × 2 = 0 + 0.503 616 380 928;
  • 25) 0.503 616 380 928 × 2 = 1 + 0.007 232 761 856;
  • 26) 0.007 232 761 856 × 2 = 0 + 0.014 465 523 712;
  • 27) 0.014 465 523 712 × 2 = 0 + 0.028 931 047 424;
  • 28) 0.028 931 047 424 × 2 = 0 + 0.057 862 094 848;
  • 29) 0.057 862 094 848 × 2 = 0 + 0.115 724 189 696;
  • 30) 0.115 724 189 696 × 2 = 0 + 0.231 448 379 392;
  • 31) 0.231 448 379 392 × 2 = 0 + 0.462 896 758 784;
  • 32) 0.462 896 758 784 × 2 = 0 + 0.925 793 517 568;
  • 33) 0.925 793 517 568 × 2 = 1 + 0.851 587 035 136;
  • 34) 0.851 587 035 136 × 2 = 1 + 0.703 174 070 272;
  • 35) 0.703 174 070 272 × 2 = 1 + 0.406 348 140 544;
  • 36) 0.406 348 140 544 × 2 = 0 + 0.812 696 281 088;
  • 37) 0.812 696 281 088 × 2 = 1 + 0.625 392 562 176;
  • 38) 0.625 392 562 176 × 2 = 1 + 0.250 785 124 352;
  • 39) 0.250 785 124 352 × 2 = 0 + 0.501 570 248 704;
  • 40) 0.501 570 248 704 × 2 = 1 + 0.003 140 497 408;
  • 41) 0.003 140 497 408 × 2 = 0 + 0.006 280 994 816;
  • 42) 0.006 280 994 816 × 2 = 0 + 0.012 561 989 632;
  • 43) 0.012 561 989 632 × 2 = 0 + 0.025 123 979 264;
  • 44) 0.025 123 979 264 × 2 = 0 + 0.050 247 958 528;
  • 45) 0.050 247 958 528 × 2 = 0 + 0.100 495 917 056;
  • 46) 0.100 495 917 056 × 2 = 0 + 0.200 991 834 112;
  • 47) 0.200 991 834 112 × 2 = 0 + 0.401 983 668 224;
  • 48) 0.401 983 668 224 × 2 = 0 + 0.803 967 336 448;
  • 49) 0.803 967 336 448 × 2 = 1 + 0.607 934 672 896;
  • 50) 0.607 934 672 896 × 2 = 1 + 0.215 869 345 792;
  • 51) 0.215 869 345 792 × 2 = 0 + 0.431 738 691 584;
  • 52) 0.431 738 691 584 × 2 = 0 + 0.863 477 383 168;
  • 53) 0.863 477 383 168 × 2 = 1 + 0.726 954 766 336;
  • 54) 0.726 954 766 336 × 2 = 1 + 0.453 909 532 672;
  • 55) 0.453 909 532 672 × 2 = 0 + 0.907 819 065 344;
  • 56) 0.907 819 065 344 × 2 = 1 + 0.815 638 130 688;
  • 57) 0.815 638 130 688 × 2 = 1 + 0.631 276 261 376;
  • 58) 0.631 276 261 376 × 2 = 1 + 0.262 552 522 752;
  • 59) 0.262 552 522 752 × 2 = 0 + 0.525 105 045 504;
  • 60) 0.525 105 045 504 × 2 = 1 + 0.050 210 091 008;
  • 61) 0.050 210 091 008 × 2 = 0 + 0.100 420 182 016;
  • 62) 0.100 420 182 016 × 2 = 0 + 0.200 840 364 032;
  • 63) 0.200 840 364 032 × 2 = 0 + 0.401 680 728 064;
  • 64) 0.401 680 728 064 × 2 = 0 + 0.803 361 456 128;

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:


0.000 375 658 489 25(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0000 0000 1100 1101 1101 0000(2)

6. Positive number before normalization:

0.000 375 658 489 25(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0000 0000 1100 1101 1101 0000(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:


0.000 375 658 489 25(10) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0000 0000 1100 1101 1101 0000(2) =

0.0000 0000 0001 1000 1001 1110 1000 0000 1110 1101 0000 0000 1100 1101 1101 0000(2) × 20 =

1.1000 1001 1110 1000 0000 1110 1101 0000 0000 1100 1101 1101 0000(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 1101 0000 0000 1100 1101 1101 0000


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 1101 0000 0000 1100 1101 1101 0000 =

1000 1001 1110 1000 0000 1110 1101 0000 0000 1100 1101 1101 0000


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 1101 0000 0000 1100 1101 1101 0000


Decimal number -0.000 375 658 489 25 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 1000 1001 1110 1000 0000 1110 1101 0000 0000 1100 1101 1101 0000

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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