-0.000 352 859 482 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 352 859 482(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
-0.000 352 859 482(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:

|-0.000 352 859 482| = 0.000 352 859 482


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: 0.000 352 859 482.

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) 0.000 352 859 482 × 2 = 0 + 0.000 705 718 964;
  • 2) 0.000 705 718 964 × 2 = 0 + 0.001 411 437 928;
  • 3) 0.001 411 437 928 × 2 = 0 + 0.002 822 875 856;
  • 4) 0.002 822 875 856 × 2 = 0 + 0.005 645 751 712;
  • 5) 0.005 645 751 712 × 2 = 0 + 0.011 291 503 424;
  • 6) 0.011 291 503 424 × 2 = 0 + 0.022 583 006 848;
  • 7) 0.022 583 006 848 × 2 = 0 + 0.045 166 013 696;
  • 8) 0.045 166 013 696 × 2 = 0 + 0.090 332 027 392;
  • 9) 0.090 332 027 392 × 2 = 0 + 0.180 664 054 784;
  • 10) 0.180 664 054 784 × 2 = 0 + 0.361 328 109 568;
  • 11) 0.361 328 109 568 × 2 = 0 + 0.722 656 219 136;
  • 12) 0.722 656 219 136 × 2 = 1 + 0.445 312 438 272;
  • 13) 0.445 312 438 272 × 2 = 0 + 0.890 624 876 544;
  • 14) 0.890 624 876 544 × 2 = 1 + 0.781 249 753 088;
  • 15) 0.781 249 753 088 × 2 = 1 + 0.562 499 506 176;
  • 16) 0.562 499 506 176 × 2 = 1 + 0.124 999 012 352;
  • 17) 0.124 999 012 352 × 2 = 0 + 0.249 998 024 704;
  • 18) 0.249 998 024 704 × 2 = 0 + 0.499 996 049 408;
  • 19) 0.499 996 049 408 × 2 = 0 + 0.999 992 098 816;
  • 20) 0.999 992 098 816 × 2 = 1 + 0.999 984 197 632;
  • 21) 0.999 984 197 632 × 2 = 1 + 0.999 968 395 264;
  • 22) 0.999 968 395 264 × 2 = 1 + 0.999 936 790 528;
  • 23) 0.999 936 790 528 × 2 = 1 + 0.999 873 581 056;
  • 24) 0.999 873 581 056 × 2 = 1 + 0.999 747 162 112;
  • 25) 0.999 747 162 112 × 2 = 1 + 0.999 494 324 224;
  • 26) 0.999 494 324 224 × 2 = 1 + 0.998 988 648 448;
  • 27) 0.998 988 648 448 × 2 = 1 + 0.997 977 296 896;
  • 28) 0.997 977 296 896 × 2 = 1 + 0.995 954 593 792;
  • 29) 0.995 954 593 792 × 2 = 1 + 0.991 909 187 584;
  • 30) 0.991 909 187 584 × 2 = 1 + 0.983 818 375 168;
  • 31) 0.983 818 375 168 × 2 = 1 + 0.967 636 750 336;
  • 32) 0.967 636 750 336 × 2 = 1 + 0.935 273 500 672;
  • 33) 0.935 273 500 672 × 2 = 1 + 0.870 547 001 344;
  • 34) 0.870 547 001 344 × 2 = 1 + 0.741 094 002 688;
  • 35) 0.741 094 002 688 × 2 = 1 + 0.482 188 005 376;
  • 36) 0.482 188 005 376 × 2 = 0 + 0.964 376 010 752;
  • 37) 0.964 376 010 752 × 2 = 1 + 0.928 752 021 504;
  • 38) 0.928 752 021 504 × 2 = 1 + 0.857 504 043 008;
  • 39) 0.857 504 043 008 × 2 = 1 + 0.715 008 086 016;
  • 40) 0.715 008 086 016 × 2 = 1 + 0.430 016 172 032;
  • 41) 0.430 016 172 032 × 2 = 0 + 0.860 032 344 064;
  • 42) 0.860 032 344 064 × 2 = 1 + 0.720 064 688 128;
  • 43) 0.720 064 688 128 × 2 = 1 + 0.440 129 376 256;
  • 44) 0.440 129 376 256 × 2 = 0 + 0.880 258 752 512;
  • 45) 0.880 258 752 512 × 2 = 1 + 0.760 517 505 024;
  • 46) 0.760 517 505 024 × 2 = 1 + 0.521 035 010 048;
  • 47) 0.521 035 010 048 × 2 = 1 + 0.042 070 020 096;
  • 48) 0.042 070 020 096 × 2 = 0 + 0.084 140 040 192;
  • 49) 0.084 140 040 192 × 2 = 0 + 0.168 280 080 384;
  • 50) 0.168 280 080 384 × 2 = 0 + 0.336 560 160 768;
  • 51) 0.336 560 160 768 × 2 = 0 + 0.673 120 321 536;
  • 52) 0.673 120 321 536 × 2 = 1 + 0.346 240 643 072;
  • 53) 0.346 240 643 072 × 2 = 0 + 0.692 481 286 144;
  • 54) 0.692 481 286 144 × 2 = 1 + 0.384 962 572 288;
  • 55) 0.384 962 572 288 × 2 = 0 + 0.769 925 144 576;
  • 56) 0.769 925 144 576 × 2 = 1 + 0.539 850 289 152;
  • 57) 0.539 850 289 152 × 2 = 1 + 0.079 700 578 304;
  • 58) 0.079 700 578 304 × 2 = 0 + 0.159 401 156 608;
  • 59) 0.159 401 156 608 × 2 = 0 + 0.318 802 313 216;
  • 60) 0.318 802 313 216 × 2 = 0 + 0.637 604 626 432;
  • 61) 0.637 604 626 432 × 2 = 1 + 0.275 209 252 864;
  • 62) 0.275 209 252 864 × 2 = 0 + 0.550 418 505 728;
  • 63) 0.550 418 505 728 × 2 = 1 + 0.100 837 011 456;
  • 64) 0.100 837 011 456 × 2 = 0 + 0.201 674 022 912;

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:


0.000 352 859 482(10) =

0.0000 0000 0001 0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010(2)

6. Positive number before normalization:

0.000 352 859 482(10) =

0.0000 0000 0001 0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010(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:


0.000 352 859 482(10) =

0.0000 0000 0001 0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010(2) =

0.0000 0000 0001 0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010(2) × 20 =

1.0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010(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):
1.0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010


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. 0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010 =

0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010


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) =
0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010


Decimal number -0.000 352 859 482 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0111 0001 1111 1111 1111 1110 1111 0110 1110 0001 0101 1000 1010

» Convert decimal -0.000 352 859 539 to 64 bit double precision IEEE 754 binary floating point representation standard

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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