-0.000 282 005 914 387 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 387(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
-0.000 282 005 914 387(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:

|-0.000 282 005 914 387| = 0.000 282 005 914 387


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: 0.000 282 005 914 387.

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) 0.000 282 005 914 387 × 2 = 0 + 0.000 564 011 828 774;
  • 2) 0.000 564 011 828 774 × 2 = 0 + 0.001 128 023 657 548;
  • 3) 0.001 128 023 657 548 × 2 = 0 + 0.002 256 047 315 096;
  • 4) 0.002 256 047 315 096 × 2 = 0 + 0.004 512 094 630 192;
  • 5) 0.004 512 094 630 192 × 2 = 0 + 0.009 024 189 260 384;
  • 6) 0.009 024 189 260 384 × 2 = 0 + 0.018 048 378 520 768;
  • 7) 0.018 048 378 520 768 × 2 = 0 + 0.036 096 757 041 536;
  • 8) 0.036 096 757 041 536 × 2 = 0 + 0.072 193 514 083 072;
  • 9) 0.072 193 514 083 072 × 2 = 0 + 0.144 387 028 166 144;
  • 10) 0.144 387 028 166 144 × 2 = 0 + 0.288 774 056 332 288;
  • 11) 0.288 774 056 332 288 × 2 = 0 + 0.577 548 112 664 576;
  • 12) 0.577 548 112 664 576 × 2 = 1 + 0.155 096 225 329 152;
  • 13) 0.155 096 225 329 152 × 2 = 0 + 0.310 192 450 658 304;
  • 14) 0.310 192 450 658 304 × 2 = 0 + 0.620 384 901 316 608;
  • 15) 0.620 384 901 316 608 × 2 = 1 + 0.240 769 802 633 216;
  • 16) 0.240 769 802 633 216 × 2 = 0 + 0.481 539 605 266 432;
  • 17) 0.481 539 605 266 432 × 2 = 0 + 0.963 079 210 532 864;
  • 18) 0.963 079 210 532 864 × 2 = 1 + 0.926 158 421 065 728;
  • 19) 0.926 158 421 065 728 × 2 = 1 + 0.852 316 842 131 456;
  • 20) 0.852 316 842 131 456 × 2 = 1 + 0.704 633 684 262 912;
  • 21) 0.704 633 684 262 912 × 2 = 1 + 0.409 267 368 525 824;
  • 22) 0.409 267 368 525 824 × 2 = 0 + 0.818 534 737 051 648;
  • 23) 0.818 534 737 051 648 × 2 = 1 + 0.637 069 474 103 296;
  • 24) 0.637 069 474 103 296 × 2 = 1 + 0.274 138 948 206 592;
  • 25) 0.274 138 948 206 592 × 2 = 0 + 0.548 277 896 413 184;
  • 26) 0.548 277 896 413 184 × 2 = 1 + 0.096 555 792 826 368;
  • 27) 0.096 555 792 826 368 × 2 = 0 + 0.193 111 585 652 736;
  • 28) 0.193 111 585 652 736 × 2 = 0 + 0.386 223 171 305 472;
  • 29) 0.386 223 171 305 472 × 2 = 0 + 0.772 446 342 610 944;
  • 30) 0.772 446 342 610 944 × 2 = 1 + 0.544 892 685 221 888;
  • 31) 0.544 892 685 221 888 × 2 = 1 + 0.089 785 370 443 776;
  • 32) 0.089 785 370 443 776 × 2 = 0 + 0.179 570 740 887 552;
  • 33) 0.179 570 740 887 552 × 2 = 0 + 0.359 141 481 775 104;
  • 34) 0.359 141 481 775 104 × 2 = 0 + 0.718 282 963 550 208;
  • 35) 0.718 282 963 550 208 × 2 = 1 + 0.436 565 927 100 416;
  • 36) 0.436 565 927 100 416 × 2 = 0 + 0.873 131 854 200 832;
  • 37) 0.873 131 854 200 832 × 2 = 1 + 0.746 263 708 401 664;
  • 38) 0.746 263 708 401 664 × 2 = 1 + 0.492 527 416 803 328;
  • 39) 0.492 527 416 803 328 × 2 = 0 + 0.985 054 833 606 656;
  • 40) 0.985 054 833 606 656 × 2 = 1 + 0.970 109 667 213 312;
  • 41) 0.970 109 667 213 312 × 2 = 1 + 0.940 219 334 426 624;
  • 42) 0.940 219 334 426 624 × 2 = 1 + 0.880 438 668 853 248;
  • 43) 0.880 438 668 853 248 × 2 = 1 + 0.760 877 337 706 496;
  • 44) 0.760 877 337 706 496 × 2 = 1 + 0.521 754 675 412 992;
  • 45) 0.521 754 675 412 992 × 2 = 1 + 0.043 509 350 825 984;
  • 46) 0.043 509 350 825 984 × 2 = 0 + 0.087 018 701 651 968;
  • 47) 0.087 018 701 651 968 × 2 = 0 + 0.174 037 403 303 936;
  • 48) 0.174 037 403 303 936 × 2 = 0 + 0.348 074 806 607 872;
  • 49) 0.348 074 806 607 872 × 2 = 0 + 0.696 149 613 215 744;
  • 50) 0.696 149 613 215 744 × 2 = 1 + 0.392 299 226 431 488;
  • 51) 0.392 299 226 431 488 × 2 = 0 + 0.784 598 452 862 976;
  • 52) 0.784 598 452 862 976 × 2 = 1 + 0.569 196 905 725 952;
  • 53) 0.569 196 905 725 952 × 2 = 1 + 0.138 393 811 451 904;
  • 54) 0.138 393 811 451 904 × 2 = 0 + 0.276 787 622 903 808;
  • 55) 0.276 787 622 903 808 × 2 = 0 + 0.553 575 245 807 616;
  • 56) 0.553 575 245 807 616 × 2 = 1 + 0.107 150 491 615 232;
  • 57) 0.107 150 491 615 232 × 2 = 0 + 0.214 300 983 230 464;
  • 58) 0.214 300 983 230 464 × 2 = 0 + 0.428 601 966 460 928;
  • 59) 0.428 601 966 460 928 × 2 = 0 + 0.857 203 932 921 856;
  • 60) 0.857 203 932 921 856 × 2 = 1 + 0.714 407 865 843 712;
  • 61) 0.714 407 865 843 712 × 2 = 1 + 0.428 815 731 687 424;
  • 62) 0.428 815 731 687 424 × 2 = 0 + 0.857 631 463 374 848;
  • 63) 0.857 631 463 374 848 × 2 = 1 + 0.715 262 926 749 696;
  • 64) 0.715 262 926 749 696 × 2 = 1 + 0.430 525 853 499 392;

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:


0.000 282 005 914 387(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 0101 1001 0001 1011(2)

6. Positive number before normalization:

0.000 282 005 914 387(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 0101 1001 0001 1011(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:


0.000 282 005 914 387(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 0101 1001 0001 1011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 0101 1001 0001 1011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1000 0101 1001 0001 1011(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 1111 1000 0101 1001 0001 1011


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 1111 1000 0101 1001 0001 1011 =

0010 0111 1011 0100 0110 0010 1101 1111 1000 0101 1001 0001 1011


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 1111 1000 0101 1001 0001 1011


Decimal number -0.000 282 005 914 387 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 1000 0101 1001 0001 1011

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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