-0.000 282 005 895 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 895 8(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
-0.000 282 005 895 8(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:

|-0.000 282 005 895 8| = 0.000 282 005 895 8


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: 0.000 282 005 895 8.

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) 0.000 282 005 895 8 × 2 = 0 + 0.000 564 011 791 6;
  • 2) 0.000 564 011 791 6 × 2 = 0 + 0.001 128 023 583 2;
  • 3) 0.001 128 023 583 2 × 2 = 0 + 0.002 256 047 166 4;
  • 4) 0.002 256 047 166 4 × 2 = 0 + 0.004 512 094 332 8;
  • 5) 0.004 512 094 332 8 × 2 = 0 + 0.009 024 188 665 6;
  • 6) 0.009 024 188 665 6 × 2 = 0 + 0.018 048 377 331 2;
  • 7) 0.018 048 377 331 2 × 2 = 0 + 0.036 096 754 662 4;
  • 8) 0.036 096 754 662 4 × 2 = 0 + 0.072 193 509 324 8;
  • 9) 0.072 193 509 324 8 × 2 = 0 + 0.144 387 018 649 6;
  • 10) 0.144 387 018 649 6 × 2 = 0 + 0.288 774 037 299 2;
  • 11) 0.288 774 037 299 2 × 2 = 0 + 0.577 548 074 598 4;
  • 12) 0.577 548 074 598 4 × 2 = 1 + 0.155 096 149 196 8;
  • 13) 0.155 096 149 196 8 × 2 = 0 + 0.310 192 298 393 6;
  • 14) 0.310 192 298 393 6 × 2 = 0 + 0.620 384 596 787 2;
  • 15) 0.620 384 596 787 2 × 2 = 1 + 0.240 769 193 574 4;
  • 16) 0.240 769 193 574 4 × 2 = 0 + 0.481 538 387 148 8;
  • 17) 0.481 538 387 148 8 × 2 = 0 + 0.963 076 774 297 6;
  • 18) 0.963 076 774 297 6 × 2 = 1 + 0.926 153 548 595 2;
  • 19) 0.926 153 548 595 2 × 2 = 1 + 0.852 307 097 190 4;
  • 20) 0.852 307 097 190 4 × 2 = 1 + 0.704 614 194 380 8;
  • 21) 0.704 614 194 380 8 × 2 = 1 + 0.409 228 388 761 6;
  • 22) 0.409 228 388 761 6 × 2 = 0 + 0.818 456 777 523 2;
  • 23) 0.818 456 777 523 2 × 2 = 1 + 0.636 913 555 046 4;
  • 24) 0.636 913 555 046 4 × 2 = 1 + 0.273 827 110 092 8;
  • 25) 0.273 827 110 092 8 × 2 = 0 + 0.547 654 220 185 6;
  • 26) 0.547 654 220 185 6 × 2 = 1 + 0.095 308 440 371 2;
  • 27) 0.095 308 440 371 2 × 2 = 0 + 0.190 616 880 742 4;
  • 28) 0.190 616 880 742 4 × 2 = 0 + 0.381 233 761 484 8;
  • 29) 0.381 233 761 484 8 × 2 = 0 + 0.762 467 522 969 6;
  • 30) 0.762 467 522 969 6 × 2 = 1 + 0.524 935 045 939 2;
  • 31) 0.524 935 045 939 2 × 2 = 1 + 0.049 870 091 878 4;
  • 32) 0.049 870 091 878 4 × 2 = 0 + 0.099 740 183 756 8;
  • 33) 0.099 740 183 756 8 × 2 = 0 + 0.199 480 367 513 6;
  • 34) 0.199 480 367 513 6 × 2 = 0 + 0.398 960 735 027 2;
  • 35) 0.398 960 735 027 2 × 2 = 0 + 0.797 921 470 054 4;
  • 36) 0.797 921 470 054 4 × 2 = 1 + 0.595 842 940 108 8;
  • 37) 0.595 842 940 108 8 × 2 = 1 + 0.191 685 880 217 6;
  • 38) 0.191 685 880 217 6 × 2 = 0 + 0.383 371 760 435 2;
  • 39) 0.383 371 760 435 2 × 2 = 0 + 0.766 743 520 870 4;
  • 40) 0.766 743 520 870 4 × 2 = 1 + 0.533 487 041 740 8;
  • 41) 0.533 487 041 740 8 × 2 = 1 + 0.066 974 083 481 6;
  • 42) 0.066 974 083 481 6 × 2 = 0 + 0.133 948 166 963 2;
  • 43) 0.133 948 166 963 2 × 2 = 0 + 0.267 896 333 926 4;
  • 44) 0.267 896 333 926 4 × 2 = 0 + 0.535 792 667 852 8;
  • 45) 0.535 792 667 852 8 × 2 = 1 + 0.071 585 335 705 6;
  • 46) 0.071 585 335 705 6 × 2 = 0 + 0.143 170 671 411 2;
  • 47) 0.143 170 671 411 2 × 2 = 0 + 0.286 341 342 822 4;
  • 48) 0.286 341 342 822 4 × 2 = 0 + 0.572 682 685 644 8;
  • 49) 0.572 682 685 644 8 × 2 = 1 + 0.145 365 371 289 6;
  • 50) 0.145 365 371 289 6 × 2 = 0 + 0.290 730 742 579 2;
  • 51) 0.290 730 742 579 2 × 2 = 0 + 0.581 461 485 158 4;
  • 52) 0.581 461 485 158 4 × 2 = 1 + 0.162 922 970 316 8;
  • 53) 0.162 922 970 316 8 × 2 = 0 + 0.325 845 940 633 6;
  • 54) 0.325 845 940 633 6 × 2 = 0 + 0.651 691 881 267 2;
  • 55) 0.651 691 881 267 2 × 2 = 1 + 0.303 383 762 534 4;
  • 56) 0.303 383 762 534 4 × 2 = 0 + 0.606 767 525 068 8;
  • 57) 0.606 767 525 068 8 × 2 = 1 + 0.213 535 050 137 6;
  • 58) 0.213 535 050 137 6 × 2 = 0 + 0.427 070 100 275 2;
  • 59) 0.427 070 100 275 2 × 2 = 0 + 0.854 140 200 550 4;
  • 60) 0.854 140 200 550 4 × 2 = 1 + 0.708 280 401 100 8;
  • 61) 0.708 280 401 100 8 × 2 = 1 + 0.416 560 802 201 6;
  • 62) 0.416 560 802 201 6 × 2 = 0 + 0.833 121 604 403 2;
  • 63) 0.833 121 604 403 2 × 2 = 1 + 0.666 243 208 806 4;
  • 64) 0.666 243 208 806 4 × 2 = 1 + 0.332 486 417 612 8;

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:


0.000 282 005 895 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1001 1000 1000 1001 0010 1001 1011(2)

6. Positive number before normalization:

0.000 282 005 895 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1001 1000 1000 1001 0010 1001 1011(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:


0.000 282 005 895 8(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1001 1000 1000 1001 0010 1001 1011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1001 1000 1000 1001 0010 1001 1011(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1001 1000 1000 1001 0010 1001 1011(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 0001 1001 1000 1000 1001 0010 1001 1011


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 0001 1001 1000 1000 1001 0010 1001 1011 =

0010 0111 1011 0100 0110 0001 1001 1000 1000 1001 0010 1001 1011


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 0001 1001 1000 1000 1001 0010 1001 1011


Decimal number -0.000 282 005 895 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 1001 1000 1000 1001 0010 1001 1011

Share

» Convert decimal -0.000 282 005 888 7 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 Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 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