-0.000 282 005 726 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 726(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
-0.000 282 005 726(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:

|-0.000 282 005 726| = 0.000 282 005 726


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: 0.000 282 005 726.

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) 0.000 282 005 726 × 2 = 0 + 0.000 564 011 452;
  • 2) 0.000 564 011 452 × 2 = 0 + 0.001 128 022 904;
  • 3) 0.001 128 022 904 × 2 = 0 + 0.002 256 045 808;
  • 4) 0.002 256 045 808 × 2 = 0 + 0.004 512 091 616;
  • 5) 0.004 512 091 616 × 2 = 0 + 0.009 024 183 232;
  • 6) 0.009 024 183 232 × 2 = 0 + 0.018 048 366 464;
  • 7) 0.018 048 366 464 × 2 = 0 + 0.036 096 732 928;
  • 8) 0.036 096 732 928 × 2 = 0 + 0.072 193 465 856;
  • 9) 0.072 193 465 856 × 2 = 0 + 0.144 386 931 712;
  • 10) 0.144 386 931 712 × 2 = 0 + 0.288 773 863 424;
  • 11) 0.288 773 863 424 × 2 = 0 + 0.577 547 726 848;
  • 12) 0.577 547 726 848 × 2 = 1 + 0.155 095 453 696;
  • 13) 0.155 095 453 696 × 2 = 0 + 0.310 190 907 392;
  • 14) 0.310 190 907 392 × 2 = 0 + 0.620 381 814 784;
  • 15) 0.620 381 814 784 × 2 = 1 + 0.240 763 629 568;
  • 16) 0.240 763 629 568 × 2 = 0 + 0.481 527 259 136;
  • 17) 0.481 527 259 136 × 2 = 0 + 0.963 054 518 272;
  • 18) 0.963 054 518 272 × 2 = 1 + 0.926 109 036 544;
  • 19) 0.926 109 036 544 × 2 = 1 + 0.852 218 073 088;
  • 20) 0.852 218 073 088 × 2 = 1 + 0.704 436 146 176;
  • 21) 0.704 436 146 176 × 2 = 1 + 0.408 872 292 352;
  • 22) 0.408 872 292 352 × 2 = 0 + 0.817 744 584 704;
  • 23) 0.817 744 584 704 × 2 = 1 + 0.635 489 169 408;
  • 24) 0.635 489 169 408 × 2 = 1 + 0.270 978 338 816;
  • 25) 0.270 978 338 816 × 2 = 0 + 0.541 956 677 632;
  • 26) 0.541 956 677 632 × 2 = 1 + 0.083 913 355 264;
  • 27) 0.083 913 355 264 × 2 = 0 + 0.167 826 710 528;
  • 28) 0.167 826 710 528 × 2 = 0 + 0.335 653 421 056;
  • 29) 0.335 653 421 056 × 2 = 0 + 0.671 306 842 112;
  • 30) 0.671 306 842 112 × 2 = 1 + 0.342 613 684 224;
  • 31) 0.342 613 684 224 × 2 = 0 + 0.685 227 368 448;
  • 32) 0.685 227 368 448 × 2 = 1 + 0.370 454 736 896;
  • 33) 0.370 454 736 896 × 2 = 0 + 0.740 909 473 792;
  • 34) 0.740 909 473 792 × 2 = 1 + 0.481 818 947 584;
  • 35) 0.481 818 947 584 × 2 = 0 + 0.963 637 895 168;
  • 36) 0.963 637 895 168 × 2 = 1 + 0.927 275 790 336;
  • 37) 0.927 275 790 336 × 2 = 1 + 0.854 551 580 672;
  • 38) 0.854 551 580 672 × 2 = 1 + 0.709 103 161 344;
  • 39) 0.709 103 161 344 × 2 = 1 + 0.418 206 322 688;
  • 40) 0.418 206 322 688 × 2 = 0 + 0.836 412 645 376;
  • 41) 0.836 412 645 376 × 2 = 1 + 0.672 825 290 752;
  • 42) 0.672 825 290 752 × 2 = 1 + 0.345 650 581 504;
  • 43) 0.345 650 581 504 × 2 = 0 + 0.691 301 163 008;
  • 44) 0.691 301 163 008 × 2 = 1 + 0.382 602 326 016;
  • 45) 0.382 602 326 016 × 2 = 0 + 0.765 204 652 032;
  • 46) 0.765 204 652 032 × 2 = 1 + 0.530 409 304 064;
  • 47) 0.530 409 304 064 × 2 = 1 + 0.060 818 608 128;
  • 48) 0.060 818 608 128 × 2 = 0 + 0.121 637 216 256;
  • 49) 0.121 637 216 256 × 2 = 0 + 0.243 274 432 512;
  • 50) 0.243 274 432 512 × 2 = 0 + 0.486 548 865 024;
  • 51) 0.486 548 865 024 × 2 = 0 + 0.973 097 730 048;
  • 52) 0.973 097 730 048 × 2 = 1 + 0.946 195 460 096;
  • 53) 0.946 195 460 096 × 2 = 1 + 0.892 390 920 192;
  • 54) 0.892 390 920 192 × 2 = 1 + 0.784 781 840 384;
  • 55) 0.784 781 840 384 × 2 = 1 + 0.569 563 680 768;
  • 56) 0.569 563 680 768 × 2 = 1 + 0.139 127 361 536;
  • 57) 0.139 127 361 536 × 2 = 0 + 0.278 254 723 072;
  • 58) 0.278 254 723 072 × 2 = 0 + 0.556 509 446 144;
  • 59) 0.556 509 446 144 × 2 = 1 + 0.113 018 892 288;
  • 60) 0.113 018 892 288 × 2 = 0 + 0.226 037 784 576;
  • 61) 0.226 037 784 576 × 2 = 0 + 0.452 075 569 152;
  • 62) 0.452 075 569 152 × 2 = 0 + 0.904 151 138 304;
  • 63) 0.904 151 138 304 × 2 = 1 + 0.808 302 276 608;
  • 64) 0.808 302 276 608 × 2 = 1 + 0.616 604 553 216;

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:


0.000 282 005 726(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 1110 1101 0110 0001 1111 0010 0011(2)

6. Positive number before normalization:

0.000 282 005 726(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 1110 1101 0110 0001 1111 0010 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:


0.000 282 005 726(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 1110 1101 0110 0001 1111 0010 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0101 1110 1101 0110 0001 1111 0010 0011(2) × 20 =

1.0010 0111 1011 0100 0101 0101 1110 1101 0110 0001 1111 0010 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 0101 0101 1110 1101 0110 0001 1111 0010 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 0101 0101 1110 1101 0110 0001 1111 0010 0011 =

0010 0111 1011 0100 0101 0101 1110 1101 0110 0001 1111 0010 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 0101 0101 1110 1101 0110 0001 1111 0010 0011


Decimal number -0.000 282 005 726 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0101 1110 1101 0110 0001 1111 0010 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