-0.000 282 005 693 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 693(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
-0.000 282 005 693(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:

|-0.000 282 005 693| = 0.000 282 005 693


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: 0.000 282 005 693.

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) 0.000 282 005 693 × 2 = 0 + 0.000 564 011 386;
  • 2) 0.000 564 011 386 × 2 = 0 + 0.001 128 022 772;
  • 3) 0.001 128 022 772 × 2 = 0 + 0.002 256 045 544;
  • 4) 0.002 256 045 544 × 2 = 0 + 0.004 512 091 088;
  • 5) 0.004 512 091 088 × 2 = 0 + 0.009 024 182 176;
  • 6) 0.009 024 182 176 × 2 = 0 + 0.018 048 364 352;
  • 7) 0.018 048 364 352 × 2 = 0 + 0.036 096 728 704;
  • 8) 0.036 096 728 704 × 2 = 0 + 0.072 193 457 408;
  • 9) 0.072 193 457 408 × 2 = 0 + 0.144 386 914 816;
  • 10) 0.144 386 914 816 × 2 = 0 + 0.288 773 829 632;
  • 11) 0.288 773 829 632 × 2 = 0 + 0.577 547 659 264;
  • 12) 0.577 547 659 264 × 2 = 1 + 0.155 095 318 528;
  • 13) 0.155 095 318 528 × 2 = 0 + 0.310 190 637 056;
  • 14) 0.310 190 637 056 × 2 = 0 + 0.620 381 274 112;
  • 15) 0.620 381 274 112 × 2 = 1 + 0.240 762 548 224;
  • 16) 0.240 762 548 224 × 2 = 0 + 0.481 525 096 448;
  • 17) 0.481 525 096 448 × 2 = 0 + 0.963 050 192 896;
  • 18) 0.963 050 192 896 × 2 = 1 + 0.926 100 385 792;
  • 19) 0.926 100 385 792 × 2 = 1 + 0.852 200 771 584;
  • 20) 0.852 200 771 584 × 2 = 1 + 0.704 401 543 168;
  • 21) 0.704 401 543 168 × 2 = 1 + 0.408 803 086 336;
  • 22) 0.408 803 086 336 × 2 = 0 + 0.817 606 172 672;
  • 23) 0.817 606 172 672 × 2 = 1 + 0.635 212 345 344;
  • 24) 0.635 212 345 344 × 2 = 1 + 0.270 424 690 688;
  • 25) 0.270 424 690 688 × 2 = 0 + 0.540 849 381 376;
  • 26) 0.540 849 381 376 × 2 = 1 + 0.081 698 762 752;
  • 27) 0.081 698 762 752 × 2 = 0 + 0.163 397 525 504;
  • 28) 0.163 397 525 504 × 2 = 0 + 0.326 795 051 008;
  • 29) 0.326 795 051 008 × 2 = 0 + 0.653 590 102 016;
  • 30) 0.653 590 102 016 × 2 = 1 + 0.307 180 204 032;
  • 31) 0.307 180 204 032 × 2 = 0 + 0.614 360 408 064;
  • 32) 0.614 360 408 064 × 2 = 1 + 0.228 720 816 128;
  • 33) 0.228 720 816 128 × 2 = 0 + 0.457 441 632 256;
  • 34) 0.457 441 632 256 × 2 = 0 + 0.914 883 264 512;
  • 35) 0.914 883 264 512 × 2 = 1 + 0.829 766 529 024;
  • 36) 0.829 766 529 024 × 2 = 1 + 0.659 533 058 048;
  • 37) 0.659 533 058 048 × 2 = 1 + 0.319 066 116 096;
  • 38) 0.319 066 116 096 × 2 = 0 + 0.638 132 232 192;
  • 39) 0.638 132 232 192 × 2 = 1 + 0.276 264 464 384;
  • 40) 0.276 264 464 384 × 2 = 0 + 0.552 528 928 768;
  • 41) 0.552 528 928 768 × 2 = 1 + 0.105 057 857 536;
  • 42) 0.105 057 857 536 × 2 = 0 + 0.210 115 715 072;
  • 43) 0.210 115 715 072 × 2 = 0 + 0.420 231 430 144;
  • 44) 0.420 231 430 144 × 2 = 0 + 0.840 462 860 288;
  • 45) 0.840 462 860 288 × 2 = 1 + 0.680 925 720 576;
  • 46) 0.680 925 720 576 × 2 = 1 + 0.361 851 441 152;
  • 47) 0.361 851 441 152 × 2 = 0 + 0.723 702 882 304;
  • 48) 0.723 702 882 304 × 2 = 1 + 0.447 405 764 608;
  • 49) 0.447 405 764 608 × 2 = 0 + 0.894 811 529 216;
  • 50) 0.894 811 529 216 × 2 = 1 + 0.789 623 058 432;
  • 51) 0.789 623 058 432 × 2 = 1 + 0.579 246 116 864;
  • 52) 0.579 246 116 864 × 2 = 1 + 0.158 492 233 728;
  • 53) 0.158 492 233 728 × 2 = 0 + 0.316 984 467 456;
  • 54) 0.316 984 467 456 × 2 = 0 + 0.633 968 934 912;
  • 55) 0.633 968 934 912 × 2 = 1 + 0.267 937 869 824;
  • 56) 0.267 937 869 824 × 2 = 0 + 0.535 875 739 648;
  • 57) 0.535 875 739 648 × 2 = 1 + 0.071 751 479 296;
  • 58) 0.071 751 479 296 × 2 = 0 + 0.143 502 958 592;
  • 59) 0.143 502 958 592 × 2 = 0 + 0.287 005 917 184;
  • 60) 0.287 005 917 184 × 2 = 0 + 0.574 011 834 368;
  • 61) 0.574 011 834 368 × 2 = 1 + 0.148 023 668 736;
  • 62) 0.148 023 668 736 × 2 = 0 + 0.296 047 337 472;
  • 63) 0.296 047 337 472 × 2 = 0 + 0.592 094 674 944;
  • 64) 0.592 094 674 944 × 2 = 1 + 0.184 189 349 888;

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:


0.000 282 005 693(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 1010 1000 1101 0111 0010 1000 1001(2)

6. Positive number before normalization:

0.000 282 005 693(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 1010 1000 1101 0111 0010 1000 1001(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:


0.000 282 005 693(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 1010 1000 1101 0111 0010 1000 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 1010 1000 1101 0111 0010 1000 1001(2) × 20 =

1.0010 0111 1011 0100 0101 0011 1010 1000 1101 0111 0010 1000 1001(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 0101 0011 1010 1000 1101 0111 0010 1000 1001


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 0101 0011 1010 1000 1101 0111 0010 1000 1001 =

0010 0111 1011 0100 0101 0011 1010 1000 1101 0111 0010 1000 1001


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 0101 0011 1010 1000 1101 0111 0010 1000 1001


Decimal number -0.000 282 005 693 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0011 1010 1000 1101 0111 0010 1000 1001

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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