-0.000 282 005 891 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 891 5(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
-0.000 282 005 891 5(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:

|-0.000 282 005 891 5| = 0.000 282 005 891 5


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: 0.000 282 005 891 5.

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) 0.000 282 005 891 5 × 2 = 0 + 0.000 564 011 783;
  • 2) 0.000 564 011 783 × 2 = 0 + 0.001 128 023 566;
  • 3) 0.001 128 023 566 × 2 = 0 + 0.002 256 047 132;
  • 4) 0.002 256 047 132 × 2 = 0 + 0.004 512 094 264;
  • 5) 0.004 512 094 264 × 2 = 0 + 0.009 024 188 528;
  • 6) 0.009 024 188 528 × 2 = 0 + 0.018 048 377 056;
  • 7) 0.018 048 377 056 × 2 = 0 + 0.036 096 754 112;
  • 8) 0.036 096 754 112 × 2 = 0 + 0.072 193 508 224;
  • 9) 0.072 193 508 224 × 2 = 0 + 0.144 387 016 448;
  • 10) 0.144 387 016 448 × 2 = 0 + 0.288 774 032 896;
  • 11) 0.288 774 032 896 × 2 = 0 + 0.577 548 065 792;
  • 12) 0.577 548 065 792 × 2 = 1 + 0.155 096 131 584;
  • 13) 0.155 096 131 584 × 2 = 0 + 0.310 192 263 168;
  • 14) 0.310 192 263 168 × 2 = 0 + 0.620 384 526 336;
  • 15) 0.620 384 526 336 × 2 = 1 + 0.240 769 052 672;
  • 16) 0.240 769 052 672 × 2 = 0 + 0.481 538 105 344;
  • 17) 0.481 538 105 344 × 2 = 0 + 0.963 076 210 688;
  • 18) 0.963 076 210 688 × 2 = 1 + 0.926 152 421 376;
  • 19) 0.926 152 421 376 × 2 = 1 + 0.852 304 842 752;
  • 20) 0.852 304 842 752 × 2 = 1 + 0.704 609 685 504;
  • 21) 0.704 609 685 504 × 2 = 1 + 0.409 219 371 008;
  • 22) 0.409 219 371 008 × 2 = 0 + 0.818 438 742 016;
  • 23) 0.818 438 742 016 × 2 = 1 + 0.636 877 484 032;
  • 24) 0.636 877 484 032 × 2 = 1 + 0.273 754 968 064;
  • 25) 0.273 754 968 064 × 2 = 0 + 0.547 509 936 128;
  • 26) 0.547 509 936 128 × 2 = 1 + 0.095 019 872 256;
  • 27) 0.095 019 872 256 × 2 = 0 + 0.190 039 744 512;
  • 28) 0.190 039 744 512 × 2 = 0 + 0.380 079 489 024;
  • 29) 0.380 079 489 024 × 2 = 0 + 0.760 158 978 048;
  • 30) 0.760 158 978 048 × 2 = 1 + 0.520 317 956 096;
  • 31) 0.520 317 956 096 × 2 = 1 + 0.040 635 912 192;
  • 32) 0.040 635 912 192 × 2 = 0 + 0.081 271 824 384;
  • 33) 0.081 271 824 384 × 2 = 0 + 0.162 543 648 768;
  • 34) 0.162 543 648 768 × 2 = 0 + 0.325 087 297 536;
  • 35) 0.325 087 297 536 × 2 = 0 + 0.650 174 595 072;
  • 36) 0.650 174 595 072 × 2 = 1 + 0.300 349 190 144;
  • 37) 0.300 349 190 144 × 2 = 0 + 0.600 698 380 288;
  • 38) 0.600 698 380 288 × 2 = 1 + 0.201 396 760 576;
  • 39) 0.201 396 760 576 × 2 = 0 + 0.402 793 521 152;
  • 40) 0.402 793 521 152 × 2 = 0 + 0.805 587 042 304;
  • 41) 0.805 587 042 304 × 2 = 1 + 0.611 174 084 608;
  • 42) 0.611 174 084 608 × 2 = 1 + 0.222 348 169 216;
  • 43) 0.222 348 169 216 × 2 = 0 + 0.444 696 338 432;
  • 44) 0.444 696 338 432 × 2 = 0 + 0.889 392 676 864;
  • 45) 0.889 392 676 864 × 2 = 1 + 0.778 785 353 728;
  • 46) 0.778 785 353 728 × 2 = 1 + 0.557 570 707 456;
  • 47) 0.557 570 707 456 × 2 = 1 + 0.115 141 414 912;
  • 48) 0.115 141 414 912 × 2 = 0 + 0.230 282 829 824;
  • 49) 0.230 282 829 824 × 2 = 0 + 0.460 565 659 648;
  • 50) 0.460 565 659 648 × 2 = 0 + 0.921 131 319 296;
  • 51) 0.921 131 319 296 × 2 = 1 + 0.842 262 638 592;
  • 52) 0.842 262 638 592 × 2 = 1 + 0.684 525 277 184;
  • 53) 0.684 525 277 184 × 2 = 1 + 0.369 050 554 368;
  • 54) 0.369 050 554 368 × 2 = 0 + 0.738 101 108 736;
  • 55) 0.738 101 108 736 × 2 = 1 + 0.476 202 217 472;
  • 56) 0.476 202 217 472 × 2 = 0 + 0.952 404 434 944;
  • 57) 0.952 404 434 944 × 2 = 1 + 0.904 808 869 888;
  • 58) 0.904 808 869 888 × 2 = 1 + 0.809 617 739 776;
  • 59) 0.809 617 739 776 × 2 = 1 + 0.619 235 479 552;
  • 60) 0.619 235 479 552 × 2 = 1 + 0.238 470 959 104;
  • 61) 0.238 470 959 104 × 2 = 0 + 0.476 941 918 208;
  • 62) 0.476 941 918 208 × 2 = 0 + 0.953 883 836 416;
  • 63) 0.953 883 836 416 × 2 = 1 + 0.907 767 672 832;
  • 64) 0.907 767 672 832 × 2 = 1 + 0.815 535 345 664;

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:


0.000 282 005 891 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0100 1100 1110 0011 1010 1111 0011(2)

6. Positive number before normalization:

0.000 282 005 891 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0100 1100 1110 0011 1010 1111 0011(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:


0.000 282 005 891 5(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0100 1100 1110 0011 1010 1111 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0100 1100 1110 0011 1010 1111 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0001 0100 1100 1110 0011 1010 1111 0011(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 0100 1100 1110 0011 1010 1111 0011


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 0100 1100 1110 0011 1010 1111 0011 =

0010 0111 1011 0100 0110 0001 0100 1100 1110 0011 1010 1111 0011


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 0100 1100 1110 0011 1010 1111 0011


Decimal number -0.000 282 005 891 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0100 1100 1110 0011 1010 1111 0011

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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