-0.000 282 005 553 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 553(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
-0.000 282 005 553(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:

|-0.000 282 005 553| = 0.000 282 005 553


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: 0.000 282 005 553.

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) 0.000 282 005 553 × 2 = 0 + 0.000 564 011 106;
  • 2) 0.000 564 011 106 × 2 = 0 + 0.001 128 022 212;
  • 3) 0.001 128 022 212 × 2 = 0 + 0.002 256 044 424;
  • 4) 0.002 256 044 424 × 2 = 0 + 0.004 512 088 848;
  • 5) 0.004 512 088 848 × 2 = 0 + 0.009 024 177 696;
  • 6) 0.009 024 177 696 × 2 = 0 + 0.018 048 355 392;
  • 7) 0.018 048 355 392 × 2 = 0 + 0.036 096 710 784;
  • 8) 0.036 096 710 784 × 2 = 0 + 0.072 193 421 568;
  • 9) 0.072 193 421 568 × 2 = 0 + 0.144 386 843 136;
  • 10) 0.144 386 843 136 × 2 = 0 + 0.288 773 686 272;
  • 11) 0.288 773 686 272 × 2 = 0 + 0.577 547 372 544;
  • 12) 0.577 547 372 544 × 2 = 1 + 0.155 094 745 088;
  • 13) 0.155 094 745 088 × 2 = 0 + 0.310 189 490 176;
  • 14) 0.310 189 490 176 × 2 = 0 + 0.620 378 980 352;
  • 15) 0.620 378 980 352 × 2 = 1 + 0.240 757 960 704;
  • 16) 0.240 757 960 704 × 2 = 0 + 0.481 515 921 408;
  • 17) 0.481 515 921 408 × 2 = 0 + 0.963 031 842 816;
  • 18) 0.963 031 842 816 × 2 = 1 + 0.926 063 685 632;
  • 19) 0.926 063 685 632 × 2 = 1 + 0.852 127 371 264;
  • 20) 0.852 127 371 264 × 2 = 1 + 0.704 254 742 528;
  • 21) 0.704 254 742 528 × 2 = 1 + 0.408 509 485 056;
  • 22) 0.408 509 485 056 × 2 = 0 + 0.817 018 970 112;
  • 23) 0.817 018 970 112 × 2 = 1 + 0.634 037 940 224;
  • 24) 0.634 037 940 224 × 2 = 1 + 0.268 075 880 448;
  • 25) 0.268 075 880 448 × 2 = 0 + 0.536 151 760 896;
  • 26) 0.536 151 760 896 × 2 = 1 + 0.072 303 521 792;
  • 27) 0.072 303 521 792 × 2 = 0 + 0.144 607 043 584;
  • 28) 0.144 607 043 584 × 2 = 0 + 0.289 214 087 168;
  • 29) 0.289 214 087 168 × 2 = 0 + 0.578 428 174 336;
  • 30) 0.578 428 174 336 × 2 = 1 + 0.156 856 348 672;
  • 31) 0.156 856 348 672 × 2 = 0 + 0.313 712 697 344;
  • 32) 0.313 712 697 344 × 2 = 0 + 0.627 425 394 688;
  • 33) 0.627 425 394 688 × 2 = 1 + 0.254 850 789 376;
  • 34) 0.254 850 789 376 × 2 = 0 + 0.509 701 578 752;
  • 35) 0.509 701 578 752 × 2 = 1 + 0.019 403 157 504;
  • 36) 0.019 403 157 504 × 2 = 0 + 0.038 806 315 008;
  • 37) 0.038 806 315 008 × 2 = 0 + 0.077 612 630 016;
  • 38) 0.077 612 630 016 × 2 = 0 + 0.155 225 260 032;
  • 39) 0.155 225 260 032 × 2 = 0 + 0.310 450 520 064;
  • 40) 0.310 450 520 064 × 2 = 0 + 0.620 901 040 128;
  • 41) 0.620 901 040 128 × 2 = 1 + 0.241 802 080 256;
  • 42) 0.241 802 080 256 × 2 = 0 + 0.483 604 160 512;
  • 43) 0.483 604 160 512 × 2 = 0 + 0.967 208 321 024;
  • 44) 0.967 208 321 024 × 2 = 1 + 0.934 416 642 048;
  • 45) 0.934 416 642 048 × 2 = 1 + 0.868 833 284 096;
  • 46) 0.868 833 284 096 × 2 = 1 + 0.737 666 568 192;
  • 47) 0.737 666 568 192 × 2 = 1 + 0.475 333 136 384;
  • 48) 0.475 333 136 384 × 2 = 0 + 0.950 666 272 768;
  • 49) 0.950 666 272 768 × 2 = 1 + 0.901 332 545 536;
  • 50) 0.901 332 545 536 × 2 = 1 + 0.802 665 091 072;
  • 51) 0.802 665 091 072 × 2 = 1 + 0.605 330 182 144;
  • 52) 0.605 330 182 144 × 2 = 1 + 0.210 660 364 288;
  • 53) 0.210 660 364 288 × 2 = 0 + 0.421 320 728 576;
  • 54) 0.421 320 728 576 × 2 = 0 + 0.842 641 457 152;
  • 55) 0.842 641 457 152 × 2 = 1 + 0.685 282 914 304;
  • 56) 0.685 282 914 304 × 2 = 1 + 0.370 565 828 608;
  • 57) 0.370 565 828 608 × 2 = 0 + 0.741 131 657 216;
  • 58) 0.741 131 657 216 × 2 = 1 + 0.482 263 314 432;
  • 59) 0.482 263 314 432 × 2 = 0 + 0.964 526 628 864;
  • 60) 0.964 526 628 864 × 2 = 1 + 0.929 053 257 728;
  • 61) 0.929 053 257 728 × 2 = 1 + 0.858 106 515 456;
  • 62) 0.858 106 515 456 × 2 = 1 + 0.716 213 030 912;
  • 63) 0.716 213 030 912 × 2 = 1 + 0.432 426 061 824;
  • 64) 0.432 426 061 824 × 2 = 0 + 0.864 852 123 648;

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:


0.000 282 005 553(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 1110(2)

6. Positive number before normalization:

0.000 282 005 553(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 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:


0.000 282 005 553(10) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 1110(2) × 20 =

1.0010 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 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 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 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 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 1110 =

0010 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 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 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 1110


Decimal number -0.000 282 005 553 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0100 1010 0000 1001 1110 1111 0011 0101 1110

Share

» Convert decimal -0.000 282 005 527 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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