-0.000 282 005 914 593 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 593(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
-0.000 282 005 914 593(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:

|-0.000 282 005 914 593| = 0.000 282 005 914 593


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: 0.000 282 005 914 593.

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) 0.000 282 005 914 593 × 2 = 0 + 0.000 564 011 829 186;
  • 2) 0.000 564 011 829 186 × 2 = 0 + 0.001 128 023 658 372;
  • 3) 0.001 128 023 658 372 × 2 = 0 + 0.002 256 047 316 744;
  • 4) 0.002 256 047 316 744 × 2 = 0 + 0.004 512 094 633 488;
  • 5) 0.004 512 094 633 488 × 2 = 0 + 0.009 024 189 266 976;
  • 6) 0.009 024 189 266 976 × 2 = 0 + 0.018 048 378 533 952;
  • 7) 0.018 048 378 533 952 × 2 = 0 + 0.036 096 757 067 904;
  • 8) 0.036 096 757 067 904 × 2 = 0 + 0.072 193 514 135 808;
  • 9) 0.072 193 514 135 808 × 2 = 0 + 0.144 387 028 271 616;
  • 10) 0.144 387 028 271 616 × 2 = 0 + 0.288 774 056 543 232;
  • 11) 0.288 774 056 543 232 × 2 = 0 + 0.577 548 113 086 464;
  • 12) 0.577 548 113 086 464 × 2 = 1 + 0.155 096 226 172 928;
  • 13) 0.155 096 226 172 928 × 2 = 0 + 0.310 192 452 345 856;
  • 14) 0.310 192 452 345 856 × 2 = 0 + 0.620 384 904 691 712;
  • 15) 0.620 384 904 691 712 × 2 = 1 + 0.240 769 809 383 424;
  • 16) 0.240 769 809 383 424 × 2 = 0 + 0.481 539 618 766 848;
  • 17) 0.481 539 618 766 848 × 2 = 0 + 0.963 079 237 533 696;
  • 18) 0.963 079 237 533 696 × 2 = 1 + 0.926 158 475 067 392;
  • 19) 0.926 158 475 067 392 × 2 = 1 + 0.852 316 950 134 784;
  • 20) 0.852 316 950 134 784 × 2 = 1 + 0.704 633 900 269 568;
  • 21) 0.704 633 900 269 568 × 2 = 1 + 0.409 267 800 539 136;
  • 22) 0.409 267 800 539 136 × 2 = 0 + 0.818 535 601 078 272;
  • 23) 0.818 535 601 078 272 × 2 = 1 + 0.637 071 202 156 544;
  • 24) 0.637 071 202 156 544 × 2 = 1 + 0.274 142 404 313 088;
  • 25) 0.274 142 404 313 088 × 2 = 0 + 0.548 284 808 626 176;
  • 26) 0.548 284 808 626 176 × 2 = 1 + 0.096 569 617 252 352;
  • 27) 0.096 569 617 252 352 × 2 = 0 + 0.193 139 234 504 704;
  • 28) 0.193 139 234 504 704 × 2 = 0 + 0.386 278 469 009 408;
  • 29) 0.386 278 469 009 408 × 2 = 0 + 0.772 556 938 018 816;
  • 30) 0.772 556 938 018 816 × 2 = 1 + 0.545 113 876 037 632;
  • 31) 0.545 113 876 037 632 × 2 = 1 + 0.090 227 752 075 264;
  • 32) 0.090 227 752 075 264 × 2 = 0 + 0.180 455 504 150 528;
  • 33) 0.180 455 504 150 528 × 2 = 0 + 0.360 911 008 301 056;
  • 34) 0.360 911 008 301 056 × 2 = 0 + 0.721 822 016 602 112;
  • 35) 0.721 822 016 602 112 × 2 = 1 + 0.443 644 033 204 224;
  • 36) 0.443 644 033 204 224 × 2 = 0 + 0.887 288 066 408 448;
  • 37) 0.887 288 066 408 448 × 2 = 1 + 0.774 576 132 816 896;
  • 38) 0.774 576 132 816 896 × 2 = 1 + 0.549 152 265 633 792;
  • 39) 0.549 152 265 633 792 × 2 = 1 + 0.098 304 531 267 584;
  • 40) 0.098 304 531 267 584 × 2 = 0 + 0.196 609 062 535 168;
  • 41) 0.196 609 062 535 168 × 2 = 0 + 0.393 218 125 070 336;
  • 42) 0.393 218 125 070 336 × 2 = 0 + 0.786 436 250 140 672;
  • 43) 0.786 436 250 140 672 × 2 = 1 + 0.572 872 500 281 344;
  • 44) 0.572 872 500 281 344 × 2 = 1 + 0.145 745 000 562 688;
  • 45) 0.145 745 000 562 688 × 2 = 0 + 0.291 490 001 125 376;
  • 46) 0.291 490 001 125 376 × 2 = 0 + 0.582 980 002 250 752;
  • 47) 0.582 980 002 250 752 × 2 = 1 + 0.165 960 004 501 504;
  • 48) 0.165 960 004 501 504 × 2 = 0 + 0.331 920 009 003 008;
  • 49) 0.331 920 009 003 008 × 2 = 0 + 0.663 840 018 006 016;
  • 50) 0.663 840 018 006 016 × 2 = 1 + 0.327 680 036 012 032;
  • 51) 0.327 680 036 012 032 × 2 = 0 + 0.655 360 072 024 064;
  • 52) 0.655 360 072 024 064 × 2 = 1 + 0.310 720 144 048 128;
  • 53) 0.310 720 144 048 128 × 2 = 0 + 0.621 440 288 096 256;
  • 54) 0.621 440 288 096 256 × 2 = 1 + 0.242 880 576 192 512;
  • 55) 0.242 880 576 192 512 × 2 = 0 + 0.485 761 152 385 024;
  • 56) 0.485 761 152 385 024 × 2 = 0 + 0.971 522 304 770 048;
  • 57) 0.971 522 304 770 048 × 2 = 1 + 0.943 044 609 540 096;
  • 58) 0.943 044 609 540 096 × 2 = 1 + 0.886 089 219 080 192;
  • 59) 0.886 089 219 080 192 × 2 = 1 + 0.772 178 438 160 384;
  • 60) 0.772 178 438 160 384 × 2 = 1 + 0.544 356 876 320 768;
  • 61) 0.544 356 876 320 768 × 2 = 1 + 0.088 713 752 641 536;
  • 62) 0.088 713 752 641 536 × 2 = 0 + 0.177 427 505 283 072;
  • 63) 0.177 427 505 283 072 × 2 = 0 + 0.354 855 010 566 144;
  • 64) 0.354 855 010 566 144 × 2 = 0 + 0.709 710 021 132 288;

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:


0.000 282 005 914 593(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0011 0010 0101 0100 1111 1000(2)

6. Positive number before normalization:

0.000 282 005 914 593(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0011 0010 0101 0100 1111 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:


0.000 282 005 914 593(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0011 0010 0101 0100 1111 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0011 0010 0101 0100 1111 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0011 0010 0101 0100 1111 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 0110 0010 1110 0011 0010 0101 0100 1111 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 0110 0010 1110 0011 0010 0101 0100 1111 1000 =

0010 0111 1011 0100 0110 0010 1110 0011 0010 0101 0100 1111 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 0110 0010 1110 0011 0010 0101 0100 1111 1000


Decimal number -0.000 282 005 914 593 converted to 64 bit double precision IEEE 754 binary floating point representation:

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