-0.000 282 005 912 61 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 912 61(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
-0.000 282 005 912 61(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:

|-0.000 282 005 912 61| = 0.000 282 005 912 61


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: 0.000 282 005 912 61.

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) 0.000 282 005 912 61 × 2 = 0 + 0.000 564 011 825 22;
  • 2) 0.000 564 011 825 22 × 2 = 0 + 0.001 128 023 650 44;
  • 3) 0.001 128 023 650 44 × 2 = 0 + 0.002 256 047 300 88;
  • 4) 0.002 256 047 300 88 × 2 = 0 + 0.004 512 094 601 76;
  • 5) 0.004 512 094 601 76 × 2 = 0 + 0.009 024 189 203 52;
  • 6) 0.009 024 189 203 52 × 2 = 0 + 0.018 048 378 407 04;
  • 7) 0.018 048 378 407 04 × 2 = 0 + 0.036 096 756 814 08;
  • 8) 0.036 096 756 814 08 × 2 = 0 + 0.072 193 513 628 16;
  • 9) 0.072 193 513 628 16 × 2 = 0 + 0.144 387 027 256 32;
  • 10) 0.144 387 027 256 32 × 2 = 0 + 0.288 774 054 512 64;
  • 11) 0.288 774 054 512 64 × 2 = 0 + 0.577 548 109 025 28;
  • 12) 0.577 548 109 025 28 × 2 = 1 + 0.155 096 218 050 56;
  • 13) 0.155 096 218 050 56 × 2 = 0 + 0.310 192 436 101 12;
  • 14) 0.310 192 436 101 12 × 2 = 0 + 0.620 384 872 202 24;
  • 15) 0.620 384 872 202 24 × 2 = 1 + 0.240 769 744 404 48;
  • 16) 0.240 769 744 404 48 × 2 = 0 + 0.481 539 488 808 96;
  • 17) 0.481 539 488 808 96 × 2 = 0 + 0.963 078 977 617 92;
  • 18) 0.963 078 977 617 92 × 2 = 1 + 0.926 157 955 235 84;
  • 19) 0.926 157 955 235 84 × 2 = 1 + 0.852 315 910 471 68;
  • 20) 0.852 315 910 471 68 × 2 = 1 + 0.704 631 820 943 36;
  • 21) 0.704 631 820 943 36 × 2 = 1 + 0.409 263 641 886 72;
  • 22) 0.409 263 641 886 72 × 2 = 0 + 0.818 527 283 773 44;
  • 23) 0.818 527 283 773 44 × 2 = 1 + 0.637 054 567 546 88;
  • 24) 0.637 054 567 546 88 × 2 = 1 + 0.274 109 135 093 76;
  • 25) 0.274 109 135 093 76 × 2 = 0 + 0.548 218 270 187 52;
  • 26) 0.548 218 270 187 52 × 2 = 1 + 0.096 436 540 375 04;
  • 27) 0.096 436 540 375 04 × 2 = 0 + 0.192 873 080 750 08;
  • 28) 0.192 873 080 750 08 × 2 = 0 + 0.385 746 161 500 16;
  • 29) 0.385 746 161 500 16 × 2 = 0 + 0.771 492 323 000 32;
  • 30) 0.771 492 323 000 32 × 2 = 1 + 0.542 984 646 000 64;
  • 31) 0.542 984 646 000 64 × 2 = 1 + 0.085 969 292 001 28;
  • 32) 0.085 969 292 001 28 × 2 = 0 + 0.171 938 584 002 56;
  • 33) 0.171 938 584 002 56 × 2 = 0 + 0.343 877 168 005 12;
  • 34) 0.343 877 168 005 12 × 2 = 0 + 0.687 754 336 010 24;
  • 35) 0.687 754 336 010 24 × 2 = 1 + 0.375 508 672 020 48;
  • 36) 0.375 508 672 020 48 × 2 = 0 + 0.751 017 344 040 96;
  • 37) 0.751 017 344 040 96 × 2 = 1 + 0.502 034 688 081 92;
  • 38) 0.502 034 688 081 92 × 2 = 1 + 0.004 069 376 163 84;
  • 39) 0.004 069 376 163 84 × 2 = 0 + 0.008 138 752 327 68;
  • 40) 0.008 138 752 327 68 × 2 = 0 + 0.016 277 504 655 36;
  • 41) 0.016 277 504 655 36 × 2 = 0 + 0.032 555 009 310 72;
  • 42) 0.032 555 009 310 72 × 2 = 0 + 0.065 110 018 621 44;
  • 43) 0.065 110 018 621 44 × 2 = 0 + 0.130 220 037 242 88;
  • 44) 0.130 220 037 242 88 × 2 = 0 + 0.260 440 074 485 76;
  • 45) 0.260 440 074 485 76 × 2 = 0 + 0.520 880 148 971 52;
  • 46) 0.520 880 148 971 52 × 2 = 1 + 0.041 760 297 943 04;
  • 47) 0.041 760 297 943 04 × 2 = 0 + 0.083 520 595 886 08;
  • 48) 0.083 520 595 886 08 × 2 = 0 + 0.167 041 191 772 16;
  • 49) 0.167 041 191 772 16 × 2 = 0 + 0.334 082 383 544 32;
  • 50) 0.334 082 383 544 32 × 2 = 0 + 0.668 164 767 088 64;
  • 51) 0.668 164 767 088 64 × 2 = 1 + 0.336 329 534 177 28;
  • 52) 0.336 329 534 177 28 × 2 = 0 + 0.672 659 068 354 56;
  • 53) 0.672 659 068 354 56 × 2 = 1 + 0.345 318 136 709 12;
  • 54) 0.345 318 136 709 12 × 2 = 0 + 0.690 636 273 418 24;
  • 55) 0.690 636 273 418 24 × 2 = 1 + 0.381 272 546 836 48;
  • 56) 0.381 272 546 836 48 × 2 = 0 + 0.762 545 093 672 96;
  • 57) 0.762 545 093 672 96 × 2 = 1 + 0.525 090 187 345 92;
  • 58) 0.525 090 187 345 92 × 2 = 1 + 0.050 180 374 691 84;
  • 59) 0.050 180 374 691 84 × 2 = 0 + 0.100 360 749 383 68;
  • 60) 0.100 360 749 383 68 × 2 = 0 + 0.200 721 498 767 36;
  • 61) 0.200 721 498 767 36 × 2 = 0 + 0.401 442 997 534 72;
  • 62) 0.401 442 997 534 72 × 2 = 0 + 0.802 885 995 069 44;
  • 63) 0.802 885 995 069 44 × 2 = 1 + 0.605 771 990 138 88;
  • 64) 0.605 771 990 138 88 × 2 = 1 + 0.211 543 980 277 76;

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:


0.000 282 005 912 61(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0000 0100 0010 1010 1100 0011(2)

6. Positive number before normalization:

0.000 282 005 912 61(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0000 0100 0010 1010 1100 0011(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:


0.000 282 005 912 61(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0000 0100 0010 1010 1100 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 0000 0100 0010 1010 1100 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 0000 0100 0010 1010 1100 0011(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 1100 0000 0100 0010 1010 1100 0011


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 1100 0000 0100 0010 1010 1100 0011 =

0010 0111 1011 0100 0110 0010 1100 0000 0100 0010 1010 1100 0011


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 1100 0000 0100 0010 1010 1100 0011


Decimal number -0.000 282 005 912 61 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 0000 0100 0010 1010 1100 0011

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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