-0.000 282 356 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 356(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
-0.000 282 356(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:

|-0.000 282 356| = 0.000 282 356


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: 0.000 282 356.

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) 0.000 282 356 × 2 = 0 + 0.000 564 712;
  • 2) 0.000 564 712 × 2 = 0 + 0.001 129 424;
  • 3) 0.001 129 424 × 2 = 0 + 0.002 258 848;
  • 4) 0.002 258 848 × 2 = 0 + 0.004 517 696;
  • 5) 0.004 517 696 × 2 = 0 + 0.009 035 392;
  • 6) 0.009 035 392 × 2 = 0 + 0.018 070 784;
  • 7) 0.018 070 784 × 2 = 0 + 0.036 141 568;
  • 8) 0.036 141 568 × 2 = 0 + 0.072 283 136;
  • 9) 0.072 283 136 × 2 = 0 + 0.144 566 272;
  • 10) 0.144 566 272 × 2 = 0 + 0.289 132 544;
  • 11) 0.289 132 544 × 2 = 0 + 0.578 265 088;
  • 12) 0.578 265 088 × 2 = 1 + 0.156 530 176;
  • 13) 0.156 530 176 × 2 = 0 + 0.313 060 352;
  • 14) 0.313 060 352 × 2 = 0 + 0.626 120 704;
  • 15) 0.626 120 704 × 2 = 1 + 0.252 241 408;
  • 16) 0.252 241 408 × 2 = 0 + 0.504 482 816;
  • 17) 0.504 482 816 × 2 = 1 + 0.008 965 632;
  • 18) 0.008 965 632 × 2 = 0 + 0.017 931 264;
  • 19) 0.017 931 264 × 2 = 0 + 0.035 862 528;
  • 20) 0.035 862 528 × 2 = 0 + 0.071 725 056;
  • 21) 0.071 725 056 × 2 = 0 + 0.143 450 112;
  • 22) 0.143 450 112 × 2 = 0 + 0.286 900 224;
  • 23) 0.286 900 224 × 2 = 0 + 0.573 800 448;
  • 24) 0.573 800 448 × 2 = 1 + 0.147 600 896;
  • 25) 0.147 600 896 × 2 = 0 + 0.295 201 792;
  • 26) 0.295 201 792 × 2 = 0 + 0.590 403 584;
  • 27) 0.590 403 584 × 2 = 1 + 0.180 807 168;
  • 28) 0.180 807 168 × 2 = 0 + 0.361 614 336;
  • 29) 0.361 614 336 × 2 = 0 + 0.723 228 672;
  • 30) 0.723 228 672 × 2 = 1 + 0.446 457 344;
  • 31) 0.446 457 344 × 2 = 0 + 0.892 914 688;
  • 32) 0.892 914 688 × 2 = 1 + 0.785 829 376;
  • 33) 0.785 829 376 × 2 = 1 + 0.571 658 752;
  • 34) 0.571 658 752 × 2 = 1 + 0.143 317 504;
  • 35) 0.143 317 504 × 2 = 0 + 0.286 635 008;
  • 36) 0.286 635 008 × 2 = 0 + 0.573 270 016;
  • 37) 0.573 270 016 × 2 = 1 + 0.146 540 032;
  • 38) 0.146 540 032 × 2 = 0 + 0.293 080 064;
  • 39) 0.293 080 064 × 2 = 0 + 0.586 160 128;
  • 40) 0.586 160 128 × 2 = 1 + 0.172 320 256;
  • 41) 0.172 320 256 × 2 = 0 + 0.344 640 512;
  • 42) 0.344 640 512 × 2 = 0 + 0.689 281 024;
  • 43) 0.689 281 024 × 2 = 1 + 0.378 562 048;
  • 44) 0.378 562 048 × 2 = 0 + 0.757 124 096;
  • 45) 0.757 124 096 × 2 = 1 + 0.514 248 192;
  • 46) 0.514 248 192 × 2 = 1 + 0.028 496 384;
  • 47) 0.028 496 384 × 2 = 0 + 0.056 992 768;
  • 48) 0.056 992 768 × 2 = 0 + 0.113 985 536;
  • 49) 0.113 985 536 × 2 = 0 + 0.227 971 072;
  • 50) 0.227 971 072 × 2 = 0 + 0.455 942 144;
  • 51) 0.455 942 144 × 2 = 0 + 0.911 884 288;
  • 52) 0.911 884 288 × 2 = 1 + 0.823 768 576;
  • 53) 0.823 768 576 × 2 = 1 + 0.647 537 152;
  • 54) 0.647 537 152 × 2 = 1 + 0.295 074 304;
  • 55) 0.295 074 304 × 2 = 0 + 0.590 148 608;
  • 56) 0.590 148 608 × 2 = 1 + 0.180 297 216;
  • 57) 0.180 297 216 × 2 = 0 + 0.360 594 432;
  • 58) 0.360 594 432 × 2 = 0 + 0.721 188 864;
  • 59) 0.721 188 864 × 2 = 1 + 0.442 377 728;
  • 60) 0.442 377 728 × 2 = 0 + 0.884 755 456;
  • 61) 0.884 755 456 × 2 = 1 + 0.769 510 912;
  • 62) 0.769 510 912 × 2 = 1 + 0.539 021 824;
  • 63) 0.539 021 824 × 2 = 1 + 0.078 043 648;
  • 64) 0.078 043 648 × 2 = 0 + 0.156 087 296;

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:


0.000 282 356(10) =

0.0000 0000 0001 0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110(2)

6. Positive number before normalization:

0.000 282 356(10) =

0.0000 0000 0001 0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110(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:


0.000 282 356(10) =

0.0000 0000 0001 0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110(2) =

0.0000 0000 0001 0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110(2) × 20 =

1.0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110(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):
1.0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110


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. 0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110 =

0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110


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) =
0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110


Decimal number -0.000 282 356 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0001 0010 0101 1100 1001 0010 1100 0001 1101 0010 1110

Share

» Convert decimal -0.000 282 374 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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