-0.000 282 005 914 357 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 357(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
-0.000 282 005 914 357(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:

|-0.000 282 005 914 357| = 0.000 282 005 914 357


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: 0.000 282 005 914 357.

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) 0.000 282 005 914 357 × 2 = 0 + 0.000 564 011 828 714;
  • 2) 0.000 564 011 828 714 × 2 = 0 + 0.001 128 023 657 428;
  • 3) 0.001 128 023 657 428 × 2 = 0 + 0.002 256 047 314 856;
  • 4) 0.002 256 047 314 856 × 2 = 0 + 0.004 512 094 629 712;
  • 5) 0.004 512 094 629 712 × 2 = 0 + 0.009 024 189 259 424;
  • 6) 0.009 024 189 259 424 × 2 = 0 + 0.018 048 378 518 848;
  • 7) 0.018 048 378 518 848 × 2 = 0 + 0.036 096 757 037 696;
  • 8) 0.036 096 757 037 696 × 2 = 0 + 0.072 193 514 075 392;
  • 9) 0.072 193 514 075 392 × 2 = 0 + 0.144 387 028 150 784;
  • 10) 0.144 387 028 150 784 × 2 = 0 + 0.288 774 056 301 568;
  • 11) 0.288 774 056 301 568 × 2 = 0 + 0.577 548 112 603 136;
  • 12) 0.577 548 112 603 136 × 2 = 1 + 0.155 096 225 206 272;
  • 13) 0.155 096 225 206 272 × 2 = 0 + 0.310 192 450 412 544;
  • 14) 0.310 192 450 412 544 × 2 = 0 + 0.620 384 900 825 088;
  • 15) 0.620 384 900 825 088 × 2 = 1 + 0.240 769 801 650 176;
  • 16) 0.240 769 801 650 176 × 2 = 0 + 0.481 539 603 300 352;
  • 17) 0.481 539 603 300 352 × 2 = 0 + 0.963 079 206 600 704;
  • 18) 0.963 079 206 600 704 × 2 = 1 + 0.926 158 413 201 408;
  • 19) 0.926 158 413 201 408 × 2 = 1 + 0.852 316 826 402 816;
  • 20) 0.852 316 826 402 816 × 2 = 1 + 0.704 633 652 805 632;
  • 21) 0.704 633 652 805 632 × 2 = 1 + 0.409 267 305 611 264;
  • 22) 0.409 267 305 611 264 × 2 = 0 + 0.818 534 611 222 528;
  • 23) 0.818 534 611 222 528 × 2 = 1 + 0.637 069 222 445 056;
  • 24) 0.637 069 222 445 056 × 2 = 1 + 0.274 138 444 890 112;
  • 25) 0.274 138 444 890 112 × 2 = 0 + 0.548 276 889 780 224;
  • 26) 0.548 276 889 780 224 × 2 = 1 + 0.096 553 779 560 448;
  • 27) 0.096 553 779 560 448 × 2 = 0 + 0.193 107 559 120 896;
  • 28) 0.193 107 559 120 896 × 2 = 0 + 0.386 215 118 241 792;
  • 29) 0.386 215 118 241 792 × 2 = 0 + 0.772 430 236 483 584;
  • 30) 0.772 430 236 483 584 × 2 = 1 + 0.544 860 472 967 168;
  • 31) 0.544 860 472 967 168 × 2 = 1 + 0.089 720 945 934 336;
  • 32) 0.089 720 945 934 336 × 2 = 0 + 0.179 441 891 868 672;
  • 33) 0.179 441 891 868 672 × 2 = 0 + 0.358 883 783 737 344;
  • 34) 0.358 883 783 737 344 × 2 = 0 + 0.717 767 567 474 688;
  • 35) 0.717 767 567 474 688 × 2 = 1 + 0.435 535 134 949 376;
  • 36) 0.435 535 134 949 376 × 2 = 0 + 0.871 070 269 898 752;
  • 37) 0.871 070 269 898 752 × 2 = 1 + 0.742 140 539 797 504;
  • 38) 0.742 140 539 797 504 × 2 = 1 + 0.484 281 079 595 008;
  • 39) 0.484 281 079 595 008 × 2 = 0 + 0.968 562 159 190 016;
  • 40) 0.968 562 159 190 016 × 2 = 1 + 0.937 124 318 380 032;
  • 41) 0.937 124 318 380 032 × 2 = 1 + 0.874 248 636 760 064;
  • 42) 0.874 248 636 760 064 × 2 = 1 + 0.748 497 273 520 128;
  • 43) 0.748 497 273 520 128 × 2 = 1 + 0.496 994 547 040 256;
  • 44) 0.496 994 547 040 256 × 2 = 0 + 0.993 989 094 080 512;
  • 45) 0.993 989 094 080 512 × 2 = 1 + 0.987 978 188 161 024;
  • 46) 0.987 978 188 161 024 × 2 = 1 + 0.975 956 376 322 048;
  • 47) 0.975 956 376 322 048 × 2 = 1 + 0.951 912 752 644 096;
  • 48) 0.951 912 752 644 096 × 2 = 1 + 0.903 825 505 288 192;
  • 49) 0.903 825 505 288 192 × 2 = 1 + 0.807 651 010 576 384;
  • 50) 0.807 651 010 576 384 × 2 = 1 + 0.615 302 021 152 768;
  • 51) 0.615 302 021 152 768 × 2 = 1 + 0.230 604 042 305 536;
  • 52) 0.230 604 042 305 536 × 2 = 0 + 0.461 208 084 611 072;
  • 53) 0.461 208 084 611 072 × 2 = 0 + 0.922 416 169 222 144;
  • 54) 0.922 416 169 222 144 × 2 = 1 + 0.844 832 338 444 288;
  • 55) 0.844 832 338 444 288 × 2 = 1 + 0.689 664 676 888 576;
  • 56) 0.689 664 676 888 576 × 2 = 1 + 0.379 329 353 777 152;
  • 57) 0.379 329 353 777 152 × 2 = 0 + 0.758 658 707 554 304;
  • 58) 0.758 658 707 554 304 × 2 = 1 + 0.517 317 415 108 608;
  • 59) 0.517 317 415 108 608 × 2 = 1 + 0.034 634 830 217 216;
  • 60) 0.034 634 830 217 216 × 2 = 0 + 0.069 269 660 434 432;
  • 61) 0.069 269 660 434 432 × 2 = 0 + 0.138 539 320 868 864;
  • 62) 0.138 539 320 868 864 × 2 = 0 + 0.277 078 641 737 728;
  • 63) 0.277 078 641 737 728 × 2 = 0 + 0.554 157 283 475 456;
  • 64) 0.554 157 283 475 456 × 2 = 1 + 0.108 314 566 950 912;

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:


0.000 282 005 914 357(10) =

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

6. Positive number before normalization:

0.000 282 005 914 357(10) =

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


0.000 282 005 914 357(10) =

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

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

1.0010 0111 1011 0100 0110 0010 1101 1110 1111 1110 0111 0110 0001(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 1101 1110 1111 1110 0111 0110 0001


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 1101 1110 1111 1110 0111 0110 0001 =

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


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 1101 1110 1111 1110 0111 0110 0001


Decimal number -0.000 282 005 914 357 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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