-0.000 282 005 912 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 912 7(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
-0.000 282 005 912 7(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:

|-0.000 282 005 912 7| = 0.000 282 005 912 7


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: 0.000 282 005 912 7.

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) 0.000 282 005 912 7 × 2 = 0 + 0.000 564 011 825 4;
  • 2) 0.000 564 011 825 4 × 2 = 0 + 0.001 128 023 650 8;
  • 3) 0.001 128 023 650 8 × 2 = 0 + 0.002 256 047 301 6;
  • 4) 0.002 256 047 301 6 × 2 = 0 + 0.004 512 094 603 2;
  • 5) 0.004 512 094 603 2 × 2 = 0 + 0.009 024 189 206 4;
  • 6) 0.009 024 189 206 4 × 2 = 0 + 0.018 048 378 412 8;
  • 7) 0.018 048 378 412 8 × 2 = 0 + 0.036 096 756 825 6;
  • 8) 0.036 096 756 825 6 × 2 = 0 + 0.072 193 513 651 2;
  • 9) 0.072 193 513 651 2 × 2 = 0 + 0.144 387 027 302 4;
  • 10) 0.144 387 027 302 4 × 2 = 0 + 0.288 774 054 604 8;
  • 11) 0.288 774 054 604 8 × 2 = 0 + 0.577 548 109 209 6;
  • 12) 0.577 548 109 209 6 × 2 = 1 + 0.155 096 218 419 2;
  • 13) 0.155 096 218 419 2 × 2 = 0 + 0.310 192 436 838 4;
  • 14) 0.310 192 436 838 4 × 2 = 0 + 0.620 384 873 676 8;
  • 15) 0.620 384 873 676 8 × 2 = 1 + 0.240 769 747 353 6;
  • 16) 0.240 769 747 353 6 × 2 = 0 + 0.481 539 494 707 2;
  • 17) 0.481 539 494 707 2 × 2 = 0 + 0.963 078 989 414 4;
  • 18) 0.963 078 989 414 4 × 2 = 1 + 0.926 157 978 828 8;
  • 19) 0.926 157 978 828 8 × 2 = 1 + 0.852 315 957 657 6;
  • 20) 0.852 315 957 657 6 × 2 = 1 + 0.704 631 915 315 2;
  • 21) 0.704 631 915 315 2 × 2 = 1 + 0.409 263 830 630 4;
  • 22) 0.409 263 830 630 4 × 2 = 0 + 0.818 527 661 260 8;
  • 23) 0.818 527 661 260 8 × 2 = 1 + 0.637 055 322 521 6;
  • 24) 0.637 055 322 521 6 × 2 = 1 + 0.274 110 645 043 2;
  • 25) 0.274 110 645 043 2 × 2 = 0 + 0.548 221 290 086 4;
  • 26) 0.548 221 290 086 4 × 2 = 1 + 0.096 442 580 172 8;
  • 27) 0.096 442 580 172 8 × 2 = 0 + 0.192 885 160 345 6;
  • 28) 0.192 885 160 345 6 × 2 = 0 + 0.385 770 320 691 2;
  • 29) 0.385 770 320 691 2 × 2 = 0 + 0.771 540 641 382 4;
  • 30) 0.771 540 641 382 4 × 2 = 1 + 0.543 081 282 764 8;
  • 31) 0.543 081 282 764 8 × 2 = 1 + 0.086 162 565 529 6;
  • 32) 0.086 162 565 529 6 × 2 = 0 + 0.172 325 131 059 2;
  • 33) 0.172 325 131 059 2 × 2 = 0 + 0.344 650 262 118 4;
  • 34) 0.344 650 262 118 4 × 2 = 0 + 0.689 300 524 236 8;
  • 35) 0.689 300 524 236 8 × 2 = 1 + 0.378 601 048 473 6;
  • 36) 0.378 601 048 473 6 × 2 = 0 + 0.757 202 096 947 2;
  • 37) 0.757 202 096 947 2 × 2 = 1 + 0.514 404 193 894 4;
  • 38) 0.514 404 193 894 4 × 2 = 1 + 0.028 808 387 788 8;
  • 39) 0.028 808 387 788 8 × 2 = 0 + 0.057 616 775 577 6;
  • 40) 0.057 616 775 577 6 × 2 = 0 + 0.115 233 551 155 2;
  • 41) 0.115 233 551 155 2 × 2 = 0 + 0.230 467 102 310 4;
  • 42) 0.230 467 102 310 4 × 2 = 0 + 0.460 934 204 620 8;
  • 43) 0.460 934 204 620 8 × 2 = 0 + 0.921 868 409 241 6;
  • 44) 0.921 868 409 241 6 × 2 = 1 + 0.843 736 818 483 2;
  • 45) 0.843 736 818 483 2 × 2 = 1 + 0.687 473 636 966 4;
  • 46) 0.687 473 636 966 4 × 2 = 1 + 0.374 947 273 932 8;
  • 47) 0.374 947 273 932 8 × 2 = 0 + 0.749 894 547 865 6;
  • 48) 0.749 894 547 865 6 × 2 = 1 + 0.499 789 095 731 2;
  • 49) 0.499 789 095 731 2 × 2 = 0 + 0.999 578 191 462 4;
  • 50) 0.999 578 191 462 4 × 2 = 1 + 0.999 156 382 924 8;
  • 51) 0.999 156 382 924 8 × 2 = 1 + 0.998 312 765 849 6;
  • 52) 0.998 312 765 849 6 × 2 = 1 + 0.996 625 531 699 2;
  • 53) 0.996 625 531 699 2 × 2 = 1 + 0.993 251 063 398 4;
  • 54) 0.993 251 063 398 4 × 2 = 1 + 0.986 502 126 796 8;
  • 55) 0.986 502 126 796 8 × 2 = 1 + 0.973 004 253 593 6;
  • 56) 0.973 004 253 593 6 × 2 = 1 + 0.946 008 507 187 2;
  • 57) 0.946 008 507 187 2 × 2 = 1 + 0.892 017 014 374 4;
  • 58) 0.892 017 014 374 4 × 2 = 1 + 0.784 034 028 748 8;
  • 59) 0.784 034 028 748 8 × 2 = 1 + 0.568 068 057 497 6;
  • 60) 0.568 068 057 497 6 × 2 = 1 + 0.136 136 114 995 2;
  • 61) 0.136 136 114 995 2 × 2 = 0 + 0.272 272 229 990 4;
  • 62) 0.272 272 229 990 4 × 2 = 0 + 0.544 544 459 980 8;
  • 63) 0.544 544 459 980 8 × 2 = 1 + 0.089 088 919 961 6;
  • 64) 0.089 088 919 961 6 × 2 = 0 + 0.178 177 839 923 2;

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:


0.000 282 005 912 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0001 1101 0111 1111 1111 0010(2)

6. Positive number before normalization:

0.000 282 005 912 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0001 1101 0111 1111 1111 0010(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:


0.000 282 005 912 7(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0001 1101 0111 1111 1111 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0001 1101 0111 1111 1111 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 0001 1101 0111 1111 1111 0010(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 0010 1100 0001 1101 0111 1111 1111 0010


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 0010 1100 0001 1101 0111 1111 1111 0010 =

0010 0111 1011 0100 0110 0010 1100 0001 1101 0111 1111 1111 0010


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 0010 1100 0001 1101 0111 1111 1111 0010


Decimal number -0.000 282 005 912 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 0001 1101 0111 1111 1111 0010

Share

» Convert decimal -0.000 282 005 919 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