-0.000 282 005 915 41 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 915 41(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
-0.000 282 005 915 41(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:

|-0.000 282 005 915 41| = 0.000 282 005 915 41


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: 0.000 282 005 915 41.

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) 0.000 282 005 915 41 × 2 = 0 + 0.000 564 011 830 82;
  • 2) 0.000 564 011 830 82 × 2 = 0 + 0.001 128 023 661 64;
  • 3) 0.001 128 023 661 64 × 2 = 0 + 0.002 256 047 323 28;
  • 4) 0.002 256 047 323 28 × 2 = 0 + 0.004 512 094 646 56;
  • 5) 0.004 512 094 646 56 × 2 = 0 + 0.009 024 189 293 12;
  • 6) 0.009 024 189 293 12 × 2 = 0 + 0.018 048 378 586 24;
  • 7) 0.018 048 378 586 24 × 2 = 0 + 0.036 096 757 172 48;
  • 8) 0.036 096 757 172 48 × 2 = 0 + 0.072 193 514 344 96;
  • 9) 0.072 193 514 344 96 × 2 = 0 + 0.144 387 028 689 92;
  • 10) 0.144 387 028 689 92 × 2 = 0 + 0.288 774 057 379 84;
  • 11) 0.288 774 057 379 84 × 2 = 0 + 0.577 548 114 759 68;
  • 12) 0.577 548 114 759 68 × 2 = 1 + 0.155 096 229 519 36;
  • 13) 0.155 096 229 519 36 × 2 = 0 + 0.310 192 459 038 72;
  • 14) 0.310 192 459 038 72 × 2 = 0 + 0.620 384 918 077 44;
  • 15) 0.620 384 918 077 44 × 2 = 1 + 0.240 769 836 154 88;
  • 16) 0.240 769 836 154 88 × 2 = 0 + 0.481 539 672 309 76;
  • 17) 0.481 539 672 309 76 × 2 = 0 + 0.963 079 344 619 52;
  • 18) 0.963 079 344 619 52 × 2 = 1 + 0.926 158 689 239 04;
  • 19) 0.926 158 689 239 04 × 2 = 1 + 0.852 317 378 478 08;
  • 20) 0.852 317 378 478 08 × 2 = 1 + 0.704 634 756 956 16;
  • 21) 0.704 634 756 956 16 × 2 = 1 + 0.409 269 513 912 32;
  • 22) 0.409 269 513 912 32 × 2 = 0 + 0.818 539 027 824 64;
  • 23) 0.818 539 027 824 64 × 2 = 1 + 0.637 078 055 649 28;
  • 24) 0.637 078 055 649 28 × 2 = 1 + 0.274 156 111 298 56;
  • 25) 0.274 156 111 298 56 × 2 = 0 + 0.548 312 222 597 12;
  • 26) 0.548 312 222 597 12 × 2 = 1 + 0.096 624 445 194 24;
  • 27) 0.096 624 445 194 24 × 2 = 0 + 0.193 248 890 388 48;
  • 28) 0.193 248 890 388 48 × 2 = 0 + 0.386 497 780 776 96;
  • 29) 0.386 497 780 776 96 × 2 = 0 + 0.772 995 561 553 92;
  • 30) 0.772 995 561 553 92 × 2 = 1 + 0.545 991 123 107 84;
  • 31) 0.545 991 123 107 84 × 2 = 1 + 0.091 982 246 215 68;
  • 32) 0.091 982 246 215 68 × 2 = 0 + 0.183 964 492 431 36;
  • 33) 0.183 964 492 431 36 × 2 = 0 + 0.367 928 984 862 72;
  • 34) 0.367 928 984 862 72 × 2 = 0 + 0.735 857 969 725 44;
  • 35) 0.735 857 969 725 44 × 2 = 1 + 0.471 715 939 450 88;
  • 36) 0.471 715 939 450 88 × 2 = 0 + 0.943 431 878 901 76;
  • 37) 0.943 431 878 901 76 × 2 = 1 + 0.886 863 757 803 52;
  • 38) 0.886 863 757 803 52 × 2 = 1 + 0.773 727 515 607 04;
  • 39) 0.773 727 515 607 04 × 2 = 1 + 0.547 455 031 214 08;
  • 40) 0.547 455 031 214 08 × 2 = 1 + 0.094 910 062 428 16;
  • 41) 0.094 910 062 428 16 × 2 = 0 + 0.189 820 124 856 32;
  • 42) 0.189 820 124 856 32 × 2 = 0 + 0.379 640 249 712 64;
  • 43) 0.379 640 249 712 64 × 2 = 0 + 0.759 280 499 425 28;
  • 44) 0.759 280 499 425 28 × 2 = 1 + 0.518 560 998 850 56;
  • 45) 0.518 560 998 850 56 × 2 = 1 + 0.037 121 997 701 12;
  • 46) 0.037 121 997 701 12 × 2 = 0 + 0.074 243 995 402 24;
  • 47) 0.074 243 995 402 24 × 2 = 0 + 0.148 487 990 804 48;
  • 48) 0.148 487 990 804 48 × 2 = 0 + 0.296 975 981 608 96;
  • 49) 0.296 975 981 608 96 × 2 = 0 + 0.593 951 963 217 92;
  • 50) 0.593 951 963 217 92 × 2 = 1 + 0.187 903 926 435 84;
  • 51) 0.187 903 926 435 84 × 2 = 0 + 0.375 807 852 871 68;
  • 52) 0.375 807 852 871 68 × 2 = 0 + 0.751 615 705 743 36;
  • 53) 0.751 615 705 743 36 × 2 = 1 + 0.503 231 411 486 72;
  • 54) 0.503 231 411 486 72 × 2 = 1 + 0.006 462 822 973 44;
  • 55) 0.006 462 822 973 44 × 2 = 0 + 0.012 925 645 946 88;
  • 56) 0.012 925 645 946 88 × 2 = 0 + 0.025 851 291 893 76;
  • 57) 0.025 851 291 893 76 × 2 = 0 + 0.051 702 583 787 52;
  • 58) 0.051 702 583 787 52 × 2 = 0 + 0.103 405 167 575 04;
  • 59) 0.103 405 167 575 04 × 2 = 0 + 0.206 810 335 150 08;
  • 60) 0.206 810 335 150 08 × 2 = 0 + 0.413 620 670 300 16;
  • 61) 0.413 620 670 300 16 × 2 = 0 + 0.827 241 340 600 32;
  • 62) 0.827 241 340 600 32 × 2 = 1 + 0.654 482 681 200 64;
  • 63) 0.654 482 681 200 64 × 2 = 1 + 0.308 965 362 401 28;
  • 64) 0.308 965 362 401 28 × 2 = 0 + 0.617 930 724 802 56;

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:


0.000 282 005 915 41(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0001 1000 0100 1100 0000 0110(2)

6. Positive number before normalization:

0.000 282 005 915 41(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0001 1000 0100 1100 0000 0110(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:


0.000 282 005 915 41(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0001 1000 0100 1100 0000 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1111 0001 1000 0100 1100 0000 0110(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1111 0001 1000 0100 1100 0000 0110(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 1111 0001 1000 0100 1100 0000 0110


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 1111 0001 1000 0100 1100 0000 0110 =

0010 0111 1011 0100 0110 0010 1111 0001 1000 0100 1100 0000 0110


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 1111 0001 1000 0100 1100 0000 0110


Decimal number -0.000 282 005 915 41 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1111 0001 1000 0100 1100 0000 0110

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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