-0.000 282 005 911 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 911 37(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
-0.000 282 005 911 37(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:

|-0.000 282 005 911 37| = 0.000 282 005 911 37


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: 0.000 282 005 911 37.

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) 0.000 282 005 911 37 × 2 = 0 + 0.000 564 011 822 74;
  • 2) 0.000 564 011 822 74 × 2 = 0 + 0.001 128 023 645 48;
  • 3) 0.001 128 023 645 48 × 2 = 0 + 0.002 256 047 290 96;
  • 4) 0.002 256 047 290 96 × 2 = 0 + 0.004 512 094 581 92;
  • 5) 0.004 512 094 581 92 × 2 = 0 + 0.009 024 189 163 84;
  • 6) 0.009 024 189 163 84 × 2 = 0 + 0.018 048 378 327 68;
  • 7) 0.018 048 378 327 68 × 2 = 0 + 0.036 096 756 655 36;
  • 8) 0.036 096 756 655 36 × 2 = 0 + 0.072 193 513 310 72;
  • 9) 0.072 193 513 310 72 × 2 = 0 + 0.144 387 026 621 44;
  • 10) 0.144 387 026 621 44 × 2 = 0 + 0.288 774 053 242 88;
  • 11) 0.288 774 053 242 88 × 2 = 0 + 0.577 548 106 485 76;
  • 12) 0.577 548 106 485 76 × 2 = 1 + 0.155 096 212 971 52;
  • 13) 0.155 096 212 971 52 × 2 = 0 + 0.310 192 425 943 04;
  • 14) 0.310 192 425 943 04 × 2 = 0 + 0.620 384 851 886 08;
  • 15) 0.620 384 851 886 08 × 2 = 1 + 0.240 769 703 772 16;
  • 16) 0.240 769 703 772 16 × 2 = 0 + 0.481 539 407 544 32;
  • 17) 0.481 539 407 544 32 × 2 = 0 + 0.963 078 815 088 64;
  • 18) 0.963 078 815 088 64 × 2 = 1 + 0.926 157 630 177 28;
  • 19) 0.926 157 630 177 28 × 2 = 1 + 0.852 315 260 354 56;
  • 20) 0.852 315 260 354 56 × 2 = 1 + 0.704 630 520 709 12;
  • 21) 0.704 630 520 709 12 × 2 = 1 + 0.409 261 041 418 24;
  • 22) 0.409 261 041 418 24 × 2 = 0 + 0.818 522 082 836 48;
  • 23) 0.818 522 082 836 48 × 2 = 1 + 0.637 044 165 672 96;
  • 24) 0.637 044 165 672 96 × 2 = 1 + 0.274 088 331 345 92;
  • 25) 0.274 088 331 345 92 × 2 = 0 + 0.548 176 662 691 84;
  • 26) 0.548 176 662 691 84 × 2 = 1 + 0.096 353 325 383 68;
  • 27) 0.096 353 325 383 68 × 2 = 0 + 0.192 706 650 767 36;
  • 28) 0.192 706 650 767 36 × 2 = 0 + 0.385 413 301 534 72;
  • 29) 0.385 413 301 534 72 × 2 = 0 + 0.770 826 603 069 44;
  • 30) 0.770 826 603 069 44 × 2 = 1 + 0.541 653 206 138 88;
  • 31) 0.541 653 206 138 88 × 2 = 1 + 0.083 306 412 277 76;
  • 32) 0.083 306 412 277 76 × 2 = 0 + 0.166 612 824 555 52;
  • 33) 0.166 612 824 555 52 × 2 = 0 + 0.333 225 649 111 04;
  • 34) 0.333 225 649 111 04 × 2 = 0 + 0.666 451 298 222 08;
  • 35) 0.666 451 298 222 08 × 2 = 1 + 0.332 902 596 444 16;
  • 36) 0.332 902 596 444 16 × 2 = 0 + 0.665 805 192 888 32;
  • 37) 0.665 805 192 888 32 × 2 = 1 + 0.331 610 385 776 64;
  • 38) 0.331 610 385 776 64 × 2 = 0 + 0.663 220 771 553 28;
  • 39) 0.663 220 771 553 28 × 2 = 1 + 0.326 441 543 106 56;
  • 40) 0.326 441 543 106 56 × 2 = 0 + 0.652 883 086 213 12;
  • 41) 0.652 883 086 213 12 × 2 = 1 + 0.305 766 172 426 24;
  • 42) 0.305 766 172 426 24 × 2 = 0 + 0.611 532 344 852 48;
  • 43) 0.611 532 344 852 48 × 2 = 1 + 0.223 064 689 704 96;
  • 44) 0.223 064 689 704 96 × 2 = 0 + 0.446 129 379 409 92;
  • 45) 0.446 129 379 409 92 × 2 = 0 + 0.892 258 758 819 84;
  • 46) 0.892 258 758 819 84 × 2 = 1 + 0.784 517 517 639 68;
  • 47) 0.784 517 517 639 68 × 2 = 1 + 0.569 035 035 279 36;
  • 48) 0.569 035 035 279 36 × 2 = 1 + 0.138 070 070 558 72;
  • 49) 0.138 070 070 558 72 × 2 = 0 + 0.276 140 141 117 44;
  • 50) 0.276 140 141 117 44 × 2 = 0 + 0.552 280 282 234 88;
  • 51) 0.552 280 282 234 88 × 2 = 1 + 0.104 560 564 469 76;
  • 52) 0.104 560 564 469 76 × 2 = 0 + 0.209 121 128 939 52;
  • 53) 0.209 121 128 939 52 × 2 = 0 + 0.418 242 257 879 04;
  • 54) 0.418 242 257 879 04 × 2 = 0 + 0.836 484 515 758 08;
  • 55) 0.836 484 515 758 08 × 2 = 1 + 0.672 969 031 516 16;
  • 56) 0.672 969 031 516 16 × 2 = 1 + 0.345 938 063 032 32;
  • 57) 0.345 938 063 032 32 × 2 = 0 + 0.691 876 126 064 64;
  • 58) 0.691 876 126 064 64 × 2 = 1 + 0.383 752 252 129 28;
  • 59) 0.383 752 252 129 28 × 2 = 0 + 0.767 504 504 258 56;
  • 60) 0.767 504 504 258 56 × 2 = 1 + 0.535 009 008 517 12;
  • 61) 0.535 009 008 517 12 × 2 = 1 + 0.070 018 017 034 24;
  • 62) 0.070 018 017 034 24 × 2 = 0 + 0.140 036 034 068 48;
  • 63) 0.140 036 034 068 48 × 2 = 0 + 0.280 072 068 136 96;
  • 64) 0.280 072 068 136 96 × 2 = 0 + 0.560 144 136 273 92;

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:


0.000 282 005 911 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1010 1010 0111 0010 0011 0101 1000(2)

6. Positive number before normalization:

0.000 282 005 911 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1010 1010 0111 0010 0011 0101 1000(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:


0.000 282 005 911 37(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1010 1010 0111 0010 0011 0101 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1010 1010 0111 0010 0011 0101 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1010 1010 0111 0010 0011 0101 1000(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 1010 1010 0111 0010 0011 0101 1000


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 1010 1010 0111 0010 0011 0101 1000 =

0010 0111 1011 0100 0110 0010 1010 1010 0111 0010 0011 0101 1000


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 1010 1010 0111 0010 0011 0101 1000


Decimal number -0.000 282 005 911 37 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1010 1010 0111 0010 0011 0101 1000

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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