-0.000 282 006 337 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 337(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
-0.000 282 006 337(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:

|-0.000 282 006 337| = 0.000 282 006 337


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: 0.000 282 006 337.

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) 0.000 282 006 337 × 2 = 0 + 0.000 564 012 674;
  • 2) 0.000 564 012 674 × 2 = 0 + 0.001 128 025 348;
  • 3) 0.001 128 025 348 × 2 = 0 + 0.002 256 050 696;
  • 4) 0.002 256 050 696 × 2 = 0 + 0.004 512 101 392;
  • 5) 0.004 512 101 392 × 2 = 0 + 0.009 024 202 784;
  • 6) 0.009 024 202 784 × 2 = 0 + 0.018 048 405 568;
  • 7) 0.018 048 405 568 × 2 = 0 + 0.036 096 811 136;
  • 8) 0.036 096 811 136 × 2 = 0 + 0.072 193 622 272;
  • 9) 0.072 193 622 272 × 2 = 0 + 0.144 387 244 544;
  • 10) 0.144 387 244 544 × 2 = 0 + 0.288 774 489 088;
  • 11) 0.288 774 489 088 × 2 = 0 + 0.577 548 978 176;
  • 12) 0.577 548 978 176 × 2 = 1 + 0.155 097 956 352;
  • 13) 0.155 097 956 352 × 2 = 0 + 0.310 195 912 704;
  • 14) 0.310 195 912 704 × 2 = 0 + 0.620 391 825 408;
  • 15) 0.620 391 825 408 × 2 = 1 + 0.240 783 650 816;
  • 16) 0.240 783 650 816 × 2 = 0 + 0.481 567 301 632;
  • 17) 0.481 567 301 632 × 2 = 0 + 0.963 134 603 264;
  • 18) 0.963 134 603 264 × 2 = 1 + 0.926 269 206 528;
  • 19) 0.926 269 206 528 × 2 = 1 + 0.852 538 413 056;
  • 20) 0.852 538 413 056 × 2 = 1 + 0.705 076 826 112;
  • 21) 0.705 076 826 112 × 2 = 1 + 0.410 153 652 224;
  • 22) 0.410 153 652 224 × 2 = 0 + 0.820 307 304 448;
  • 23) 0.820 307 304 448 × 2 = 1 + 0.640 614 608 896;
  • 24) 0.640 614 608 896 × 2 = 1 + 0.281 229 217 792;
  • 25) 0.281 229 217 792 × 2 = 0 + 0.562 458 435 584;
  • 26) 0.562 458 435 584 × 2 = 1 + 0.124 916 871 168;
  • 27) 0.124 916 871 168 × 2 = 0 + 0.249 833 742 336;
  • 28) 0.249 833 742 336 × 2 = 0 + 0.499 667 484 672;
  • 29) 0.499 667 484 672 × 2 = 0 + 0.999 334 969 344;
  • 30) 0.999 334 969 344 × 2 = 1 + 0.998 669 938 688;
  • 31) 0.998 669 938 688 × 2 = 1 + 0.997 339 877 376;
  • 32) 0.997 339 877 376 × 2 = 1 + 0.994 679 754 752;
  • 33) 0.994 679 754 752 × 2 = 1 + 0.989 359 509 504;
  • 34) 0.989 359 509 504 × 2 = 1 + 0.978 719 019 008;
  • 35) 0.978 719 019 008 × 2 = 1 + 0.957 438 038 016;
  • 36) 0.957 438 038 016 × 2 = 1 + 0.914 876 076 032;
  • 37) 0.914 876 076 032 × 2 = 1 + 0.829 752 152 064;
  • 38) 0.829 752 152 064 × 2 = 1 + 0.659 504 304 128;
  • 39) 0.659 504 304 128 × 2 = 1 + 0.319 008 608 256;
  • 40) 0.319 008 608 256 × 2 = 0 + 0.638 017 216 512;
  • 41) 0.638 017 216 512 × 2 = 1 + 0.276 034 433 024;
  • 42) 0.276 034 433 024 × 2 = 0 + 0.552 068 866 048;
  • 43) 0.552 068 866 048 × 2 = 1 + 0.104 137 732 096;
  • 44) 0.104 137 732 096 × 2 = 0 + 0.208 275 464 192;
  • 45) 0.208 275 464 192 × 2 = 0 + 0.416 550 928 384;
  • 46) 0.416 550 928 384 × 2 = 0 + 0.833 101 856 768;
  • 47) 0.833 101 856 768 × 2 = 1 + 0.666 203 713 536;
  • 48) 0.666 203 713 536 × 2 = 1 + 0.332 407 427 072;
  • 49) 0.332 407 427 072 × 2 = 0 + 0.664 814 854 144;
  • 50) 0.664 814 854 144 × 2 = 1 + 0.329 629 708 288;
  • 51) 0.329 629 708 288 × 2 = 0 + 0.659 259 416 576;
  • 52) 0.659 259 416 576 × 2 = 1 + 0.318 518 833 152;
  • 53) 0.318 518 833 152 × 2 = 0 + 0.637 037 666 304;
  • 54) 0.637 037 666 304 × 2 = 1 + 0.274 075 332 608;
  • 55) 0.274 075 332 608 × 2 = 0 + 0.548 150 665 216;
  • 56) 0.548 150 665 216 × 2 = 1 + 0.096 301 330 432;
  • 57) 0.096 301 330 432 × 2 = 0 + 0.192 602 660 864;
  • 58) 0.192 602 660 864 × 2 = 0 + 0.385 205 321 728;
  • 59) 0.385 205 321 728 × 2 = 0 + 0.770 410 643 456;
  • 60) 0.770 410 643 456 × 2 = 1 + 0.540 821 286 912;
  • 61) 0.540 821 286 912 × 2 = 1 + 0.081 642 573 824;
  • 62) 0.081 642 573 824 × 2 = 0 + 0.163 285 147 648;
  • 63) 0.163 285 147 648 × 2 = 0 + 0.326 570 295 296;
  • 64) 0.326 570 295 296 × 2 = 0 + 0.653 140 590 592;

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:


0.000 282 006 337(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1111 1110 1010 0011 0101 0101 0001 1000(2)

6. Positive number before normalization:

0.000 282 006 337(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1111 1110 1010 0011 0101 0101 0001 1000(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:


0.000 282 006 337(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1111 1110 1010 0011 0101 0101 0001 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1111 1110 1010 0011 0101 0101 0001 1000(2) × 20 =

1.0010 0111 1011 0100 0111 1111 1110 1010 0011 0101 0101 0001 1000(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 0111 1111 1110 1010 0011 0101 0101 0001 1000


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 0111 1111 1110 1010 0011 0101 0101 0001 1000 =

0010 0111 1011 0100 0111 1111 1110 1010 0011 0101 0101 0001 1000


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 0111 1111 1110 1010 0011 0101 0101 0001 1000


Decimal number -0.000 282 006 337 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1111 1110 1010 0011 0101 0101 0001 1000

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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