-0.000 282 005 953 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 953 1(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
-0.000 282 005 953 1(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:

|-0.000 282 005 953 1| = 0.000 282 005 953 1


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: 0.000 282 005 953 1.

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) 0.000 282 005 953 1 × 2 = 0 + 0.000 564 011 906 2;
  • 2) 0.000 564 011 906 2 × 2 = 0 + 0.001 128 023 812 4;
  • 3) 0.001 128 023 812 4 × 2 = 0 + 0.002 256 047 624 8;
  • 4) 0.002 256 047 624 8 × 2 = 0 + 0.004 512 095 249 6;
  • 5) 0.004 512 095 249 6 × 2 = 0 + 0.009 024 190 499 2;
  • 6) 0.009 024 190 499 2 × 2 = 0 + 0.018 048 380 998 4;
  • 7) 0.018 048 380 998 4 × 2 = 0 + 0.036 096 761 996 8;
  • 8) 0.036 096 761 996 8 × 2 = 0 + 0.072 193 523 993 6;
  • 9) 0.072 193 523 993 6 × 2 = 0 + 0.144 387 047 987 2;
  • 10) 0.144 387 047 987 2 × 2 = 0 + 0.288 774 095 974 4;
  • 11) 0.288 774 095 974 4 × 2 = 0 + 0.577 548 191 948 8;
  • 12) 0.577 548 191 948 8 × 2 = 1 + 0.155 096 383 897 6;
  • 13) 0.155 096 383 897 6 × 2 = 0 + 0.310 192 767 795 2;
  • 14) 0.310 192 767 795 2 × 2 = 0 + 0.620 385 535 590 4;
  • 15) 0.620 385 535 590 4 × 2 = 1 + 0.240 771 071 180 8;
  • 16) 0.240 771 071 180 8 × 2 = 0 + 0.481 542 142 361 6;
  • 17) 0.481 542 142 361 6 × 2 = 0 + 0.963 084 284 723 2;
  • 18) 0.963 084 284 723 2 × 2 = 1 + 0.926 168 569 446 4;
  • 19) 0.926 168 569 446 4 × 2 = 1 + 0.852 337 138 892 8;
  • 20) 0.852 337 138 892 8 × 2 = 1 + 0.704 674 277 785 6;
  • 21) 0.704 674 277 785 6 × 2 = 1 + 0.409 348 555 571 2;
  • 22) 0.409 348 555 571 2 × 2 = 0 + 0.818 697 111 142 4;
  • 23) 0.818 697 111 142 4 × 2 = 1 + 0.637 394 222 284 8;
  • 24) 0.637 394 222 284 8 × 2 = 1 + 0.274 788 444 569 6;
  • 25) 0.274 788 444 569 6 × 2 = 0 + 0.549 576 889 139 2;
  • 26) 0.549 576 889 139 2 × 2 = 1 + 0.099 153 778 278 4;
  • 27) 0.099 153 778 278 4 × 2 = 0 + 0.198 307 556 556 8;
  • 28) 0.198 307 556 556 8 × 2 = 0 + 0.396 615 113 113 6;
  • 29) 0.396 615 113 113 6 × 2 = 0 + 0.793 230 226 227 2;
  • 30) 0.793 230 226 227 2 × 2 = 1 + 0.586 460 452 454 4;
  • 31) 0.586 460 452 454 4 × 2 = 1 + 0.172 920 904 908 8;
  • 32) 0.172 920 904 908 8 × 2 = 0 + 0.345 841 809 817 6;
  • 33) 0.345 841 809 817 6 × 2 = 0 + 0.691 683 619 635 2;
  • 34) 0.691 683 619 635 2 × 2 = 1 + 0.383 367 239 270 4;
  • 35) 0.383 367 239 270 4 × 2 = 0 + 0.766 734 478 540 8;
  • 36) 0.766 734 478 540 8 × 2 = 1 + 0.533 468 957 081 6;
  • 37) 0.533 468 957 081 6 × 2 = 1 + 0.066 937 914 163 2;
  • 38) 0.066 937 914 163 2 × 2 = 0 + 0.133 875 828 326 4;
  • 39) 0.133 875 828 326 4 × 2 = 0 + 0.267 751 656 652 8;
  • 40) 0.267 751 656 652 8 × 2 = 0 + 0.535 503 313 305 6;
  • 41) 0.535 503 313 305 6 × 2 = 1 + 0.071 006 626 611 2;
  • 42) 0.071 006 626 611 2 × 2 = 0 + 0.142 013 253 222 4;
  • 43) 0.142 013 253 222 4 × 2 = 0 + 0.284 026 506 444 8;
  • 44) 0.284 026 506 444 8 × 2 = 0 + 0.568 053 012 889 6;
  • 45) 0.568 053 012 889 6 × 2 = 1 + 0.136 106 025 779 2;
  • 46) 0.136 106 025 779 2 × 2 = 0 + 0.272 212 051 558 4;
  • 47) 0.272 212 051 558 4 × 2 = 0 + 0.544 424 103 116 8;
  • 48) 0.544 424 103 116 8 × 2 = 1 + 0.088 848 206 233 6;
  • 49) 0.088 848 206 233 6 × 2 = 0 + 0.177 696 412 467 2;
  • 50) 0.177 696 412 467 2 × 2 = 0 + 0.355 392 824 934 4;
  • 51) 0.355 392 824 934 4 × 2 = 0 + 0.710 785 649 868 8;
  • 52) 0.710 785 649 868 8 × 2 = 1 + 0.421 571 299 737 6;
  • 53) 0.421 571 299 737 6 × 2 = 0 + 0.843 142 599 475 2;
  • 54) 0.843 142 599 475 2 × 2 = 1 + 0.686 285 198 950 4;
  • 55) 0.686 285 198 950 4 × 2 = 1 + 0.372 570 397 900 8;
  • 56) 0.372 570 397 900 8 × 2 = 0 + 0.745 140 795 801 6;
  • 57) 0.745 140 795 801 6 × 2 = 1 + 0.490 281 591 603 2;
  • 58) 0.490 281 591 603 2 × 2 = 0 + 0.980 563 183 206 4;
  • 59) 0.980 563 183 206 4 × 2 = 1 + 0.961 126 366 412 8;
  • 60) 0.961 126 366 412 8 × 2 = 1 + 0.922 252 732 825 6;
  • 61) 0.922 252 732 825 6 × 2 = 1 + 0.844 505 465 651 2;
  • 62) 0.844 505 465 651 2 × 2 = 1 + 0.689 010 931 302 4;
  • 63) 0.689 010 931 302 4 × 2 = 1 + 0.378 021 862 604 8;
  • 64) 0.378 021 862 604 8 × 2 = 0 + 0.756 043 725 209 6;

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:


0.000 282 005 953 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1000 1000 1001 0001 0110 1011 1110(2)

6. Positive number before normalization:

0.000 282 005 953 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1000 1000 1001 0001 0110 1011 1110(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:


0.000 282 005 953 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1000 1000 1001 0001 0110 1011 1110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0101 1000 1000 1001 0001 0110 1011 1110(2) × 20 =

1.0010 0111 1011 0100 0110 0101 1000 1000 1001 0001 0110 1011 1110(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 0101 1000 1000 1001 0001 0110 1011 1110


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 0101 1000 1000 1001 0001 0110 1011 1110 =

0010 0111 1011 0100 0110 0101 1000 1000 1001 0001 0110 1011 1110


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 0101 1000 1000 1001 0001 0110 1011 1110


Decimal number -0.000 282 005 953 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0101 1000 1000 1001 0001 0110 1011 1110

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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