-0.000 282 005 896 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 896(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
-0.000 282 005 896(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:

|-0.000 282 005 896| = 0.000 282 005 896


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: 0.000 282 005 896.

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) 0.000 282 005 896 × 2 = 0 + 0.000 564 011 792;
  • 2) 0.000 564 011 792 × 2 = 0 + 0.001 128 023 584;
  • 3) 0.001 128 023 584 × 2 = 0 + 0.002 256 047 168;
  • 4) 0.002 256 047 168 × 2 = 0 + 0.004 512 094 336;
  • 5) 0.004 512 094 336 × 2 = 0 + 0.009 024 188 672;
  • 6) 0.009 024 188 672 × 2 = 0 + 0.018 048 377 344;
  • 7) 0.018 048 377 344 × 2 = 0 + 0.036 096 754 688;
  • 8) 0.036 096 754 688 × 2 = 0 + 0.072 193 509 376;
  • 9) 0.072 193 509 376 × 2 = 0 + 0.144 387 018 752;
  • 10) 0.144 387 018 752 × 2 = 0 + 0.288 774 037 504;
  • 11) 0.288 774 037 504 × 2 = 0 + 0.577 548 075 008;
  • 12) 0.577 548 075 008 × 2 = 1 + 0.155 096 150 016;
  • 13) 0.155 096 150 016 × 2 = 0 + 0.310 192 300 032;
  • 14) 0.310 192 300 032 × 2 = 0 + 0.620 384 600 064;
  • 15) 0.620 384 600 064 × 2 = 1 + 0.240 769 200 128;
  • 16) 0.240 769 200 128 × 2 = 0 + 0.481 538 400 256;
  • 17) 0.481 538 400 256 × 2 = 0 + 0.963 076 800 512;
  • 18) 0.963 076 800 512 × 2 = 1 + 0.926 153 601 024;
  • 19) 0.926 153 601 024 × 2 = 1 + 0.852 307 202 048;
  • 20) 0.852 307 202 048 × 2 = 1 + 0.704 614 404 096;
  • 21) 0.704 614 404 096 × 2 = 1 + 0.409 228 808 192;
  • 22) 0.409 228 808 192 × 2 = 0 + 0.818 457 616 384;
  • 23) 0.818 457 616 384 × 2 = 1 + 0.636 915 232 768;
  • 24) 0.636 915 232 768 × 2 = 1 + 0.273 830 465 536;
  • 25) 0.273 830 465 536 × 2 = 0 + 0.547 660 931 072;
  • 26) 0.547 660 931 072 × 2 = 1 + 0.095 321 862 144;
  • 27) 0.095 321 862 144 × 2 = 0 + 0.190 643 724 288;
  • 28) 0.190 643 724 288 × 2 = 0 + 0.381 287 448 576;
  • 29) 0.381 287 448 576 × 2 = 0 + 0.762 574 897 152;
  • 30) 0.762 574 897 152 × 2 = 1 + 0.525 149 794 304;
  • 31) 0.525 149 794 304 × 2 = 1 + 0.050 299 588 608;
  • 32) 0.050 299 588 608 × 2 = 0 + 0.100 599 177 216;
  • 33) 0.100 599 177 216 × 2 = 0 + 0.201 198 354 432;
  • 34) 0.201 198 354 432 × 2 = 0 + 0.402 396 708 864;
  • 35) 0.402 396 708 864 × 2 = 0 + 0.804 793 417 728;
  • 36) 0.804 793 417 728 × 2 = 1 + 0.609 586 835 456;
  • 37) 0.609 586 835 456 × 2 = 1 + 0.219 173 670 912;
  • 38) 0.219 173 670 912 × 2 = 0 + 0.438 347 341 824;
  • 39) 0.438 347 341 824 × 2 = 0 + 0.876 694 683 648;
  • 40) 0.876 694 683 648 × 2 = 1 + 0.753 389 367 296;
  • 41) 0.753 389 367 296 × 2 = 1 + 0.506 778 734 592;
  • 42) 0.506 778 734 592 × 2 = 1 + 0.013 557 469 184;
  • 43) 0.013 557 469 184 × 2 = 0 + 0.027 114 938 368;
  • 44) 0.027 114 938 368 × 2 = 0 + 0.054 229 876 736;
  • 45) 0.054 229 876 736 × 2 = 0 + 0.108 459 753 472;
  • 46) 0.108 459 753 472 × 2 = 0 + 0.216 919 506 944;
  • 47) 0.216 919 506 944 × 2 = 0 + 0.433 839 013 888;
  • 48) 0.433 839 013 888 × 2 = 0 + 0.867 678 027 776;
  • 49) 0.867 678 027 776 × 2 = 1 + 0.735 356 055 552;
  • 50) 0.735 356 055 552 × 2 = 1 + 0.470 712 111 104;
  • 51) 0.470 712 111 104 × 2 = 0 + 0.941 424 222 208;
  • 52) 0.941 424 222 208 × 2 = 1 + 0.882 848 444 416;
  • 53) 0.882 848 444 416 × 2 = 1 + 0.765 696 888 832;
  • 54) 0.765 696 888 832 × 2 = 1 + 0.531 393 777 664;
  • 55) 0.531 393 777 664 × 2 = 1 + 0.062 787 555 328;
  • 56) 0.062 787 555 328 × 2 = 0 + 0.125 575 110 656;
  • 57) 0.125 575 110 656 × 2 = 0 + 0.251 150 221 312;
  • 58) 0.251 150 221 312 × 2 = 0 + 0.502 300 442 624;
  • 59) 0.502 300 442 624 × 2 = 1 + 0.004 600 885 248;
  • 60) 0.004 600 885 248 × 2 = 0 + 0.009 201 770 496;
  • 61) 0.009 201 770 496 × 2 = 0 + 0.018 403 540 992;
  • 62) 0.018 403 540 992 × 2 = 0 + 0.036 807 081 984;
  • 63) 0.036 807 081 984 × 2 = 0 + 0.073 614 163 968;
  • 64) 0.073 614 163 968 × 2 = 0 + 0.147 228 327 936;

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:


0.000 282 005 896(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1001 1100 0000 1101 1110 0010 0000(2)

6. Positive number before normalization:

0.000 282 005 896(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1001 1100 0000 1101 1110 0010 0000(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:


0.000 282 005 896(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1001 1100 0000 1101 1110 0010 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1001 1100 0000 1101 1110 0010 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1001 1100 0000 1101 1110 0010 0000(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 1100 0000 1101 1110 0010 0000


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 1100 0000 1101 1110 0010 0000 =

0010 0111 1011 0100 0110 0001 1001 1100 0000 1101 1110 0010 0000


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 1100 0000 1101 1110 0010 0000


Decimal number -0.000 282 005 896 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 1001 1100 0000 1101 1110 0010 0000

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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