-0.000 282 005 962 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 962(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
-0.000 282 005 962(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:

|-0.000 282 005 962| = 0.000 282 005 962


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: 0.000 282 005 962.

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) 0.000 282 005 962 × 2 = 0 + 0.000 564 011 924;
  • 2) 0.000 564 011 924 × 2 = 0 + 0.001 128 023 848;
  • 3) 0.001 128 023 848 × 2 = 0 + 0.002 256 047 696;
  • 4) 0.002 256 047 696 × 2 = 0 + 0.004 512 095 392;
  • 5) 0.004 512 095 392 × 2 = 0 + 0.009 024 190 784;
  • 6) 0.009 024 190 784 × 2 = 0 + 0.018 048 381 568;
  • 7) 0.018 048 381 568 × 2 = 0 + 0.036 096 763 136;
  • 8) 0.036 096 763 136 × 2 = 0 + 0.072 193 526 272;
  • 9) 0.072 193 526 272 × 2 = 0 + 0.144 387 052 544;
  • 10) 0.144 387 052 544 × 2 = 0 + 0.288 774 105 088;
  • 11) 0.288 774 105 088 × 2 = 0 + 0.577 548 210 176;
  • 12) 0.577 548 210 176 × 2 = 1 + 0.155 096 420 352;
  • 13) 0.155 096 420 352 × 2 = 0 + 0.310 192 840 704;
  • 14) 0.310 192 840 704 × 2 = 0 + 0.620 385 681 408;
  • 15) 0.620 385 681 408 × 2 = 1 + 0.240 771 362 816;
  • 16) 0.240 771 362 816 × 2 = 0 + 0.481 542 725 632;
  • 17) 0.481 542 725 632 × 2 = 0 + 0.963 085 451 264;
  • 18) 0.963 085 451 264 × 2 = 1 + 0.926 170 902 528;
  • 19) 0.926 170 902 528 × 2 = 1 + 0.852 341 805 056;
  • 20) 0.852 341 805 056 × 2 = 1 + 0.704 683 610 112;
  • 21) 0.704 683 610 112 × 2 = 1 + 0.409 367 220 224;
  • 22) 0.409 367 220 224 × 2 = 0 + 0.818 734 440 448;
  • 23) 0.818 734 440 448 × 2 = 1 + 0.637 468 880 896;
  • 24) 0.637 468 880 896 × 2 = 1 + 0.274 937 761 792;
  • 25) 0.274 937 761 792 × 2 = 0 + 0.549 875 523 584;
  • 26) 0.549 875 523 584 × 2 = 1 + 0.099 751 047 168;
  • 27) 0.099 751 047 168 × 2 = 0 + 0.199 502 094 336;
  • 28) 0.199 502 094 336 × 2 = 0 + 0.399 004 188 672;
  • 29) 0.399 004 188 672 × 2 = 0 + 0.798 008 377 344;
  • 30) 0.798 008 377 344 × 2 = 1 + 0.596 016 754 688;
  • 31) 0.596 016 754 688 × 2 = 1 + 0.192 033 509 376;
  • 32) 0.192 033 509 376 × 2 = 0 + 0.384 067 018 752;
  • 33) 0.384 067 018 752 × 2 = 0 + 0.768 134 037 504;
  • 34) 0.768 134 037 504 × 2 = 1 + 0.536 268 075 008;
  • 35) 0.536 268 075 008 × 2 = 1 + 0.072 536 150 016;
  • 36) 0.072 536 150 016 × 2 = 0 + 0.145 072 300 032;
  • 37) 0.145 072 300 032 × 2 = 0 + 0.290 144 600 064;
  • 38) 0.290 144 600 064 × 2 = 0 + 0.580 289 200 128;
  • 39) 0.580 289 200 128 × 2 = 1 + 0.160 578 400 256;
  • 40) 0.160 578 400 256 × 2 = 0 + 0.321 156 800 512;
  • 41) 0.321 156 800 512 × 2 = 0 + 0.642 313 601 024;
  • 42) 0.642 313 601 024 × 2 = 1 + 0.284 627 202 048;
  • 43) 0.284 627 202 048 × 2 = 0 + 0.569 254 404 096;
  • 44) 0.569 254 404 096 × 2 = 1 + 0.138 508 808 192;
  • 45) 0.138 508 808 192 × 2 = 0 + 0.277 017 616 384;
  • 46) 0.277 017 616 384 × 2 = 0 + 0.554 035 232 768;
  • 47) 0.554 035 232 768 × 2 = 1 + 0.108 070 465 536;
  • 48) 0.108 070 465 536 × 2 = 0 + 0.216 140 931 072;
  • 49) 0.216 140 931 072 × 2 = 0 + 0.432 281 862 144;
  • 50) 0.432 281 862 144 × 2 = 0 + 0.864 563 724 288;
  • 51) 0.864 563 724 288 × 2 = 1 + 0.729 127 448 576;
  • 52) 0.729 127 448 576 × 2 = 1 + 0.458 254 897 152;
  • 53) 0.458 254 897 152 × 2 = 0 + 0.916 509 794 304;
  • 54) 0.916 509 794 304 × 2 = 1 + 0.833 019 588 608;
  • 55) 0.833 019 588 608 × 2 = 1 + 0.666 039 177 216;
  • 56) 0.666 039 177 216 × 2 = 1 + 0.332 078 354 432;
  • 57) 0.332 078 354 432 × 2 = 0 + 0.664 156 708 864;
  • 58) 0.664 156 708 864 × 2 = 1 + 0.328 313 417 728;
  • 59) 0.328 313 417 728 × 2 = 0 + 0.656 626 835 456;
  • 60) 0.656 626 835 456 × 2 = 1 + 0.313 253 670 912;
  • 61) 0.313 253 670 912 × 2 = 0 + 0.626 507 341 824;
  • 62) 0.626 507 341 824 × 2 = 1 + 0.253 014 683 648;
  • 63) 0.253 014 683 648 × 2 = 0 + 0.506 029 367 296;
  • 64) 0.506 029 367 296 × 2 = 1 + 0.012 058 734 592;

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:


0.000 282 005 962(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 0010 0101 0010 0011 0111 0101 0101(2)

6. Positive number before normalization:

0.000 282 005 962(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 0010 0101 0010 0011 0111 0101 0101(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:


0.000 282 005 962(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 0010 0101 0010 0011 0111 0101 0101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0110 0010 0101 0010 0011 0111 0101 0101(2) × 20 =

1.0010 0111 1011 0100 0110 0110 0010 0101 0010 0011 0111 0101 0101(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 0110 0110 0010 0101 0010 0011 0111 0101 0101


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 0110 0110 0010 0101 0010 0011 0111 0101 0101 =

0010 0111 1011 0100 0110 0110 0010 0101 0010 0011 0111 0101 0101


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 0110 0110 0010 0101 0010 0011 0111 0101 0101


Decimal number -0.000 282 005 962 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0110 0010 0101 0010 0011 0111 0101 0101

Share

» Convert decimal -0.000 282 005 875 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