-0.000 282 005 914 52 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 52(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 52(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 52| = 0.000 282 005 914 52


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 52.

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 52 × 2 = 0 + 0.000 564 011 829 04;
  • 2) 0.000 564 011 829 04 × 2 = 0 + 0.001 128 023 658 08;
  • 3) 0.001 128 023 658 08 × 2 = 0 + 0.002 256 047 316 16;
  • 4) 0.002 256 047 316 16 × 2 = 0 + 0.004 512 094 632 32;
  • 5) 0.004 512 094 632 32 × 2 = 0 + 0.009 024 189 264 64;
  • 6) 0.009 024 189 264 64 × 2 = 0 + 0.018 048 378 529 28;
  • 7) 0.018 048 378 529 28 × 2 = 0 + 0.036 096 757 058 56;
  • 8) 0.036 096 757 058 56 × 2 = 0 + 0.072 193 514 117 12;
  • 9) 0.072 193 514 117 12 × 2 = 0 + 0.144 387 028 234 24;
  • 10) 0.144 387 028 234 24 × 2 = 0 + 0.288 774 056 468 48;
  • 11) 0.288 774 056 468 48 × 2 = 0 + 0.577 548 112 936 96;
  • 12) 0.577 548 112 936 96 × 2 = 1 + 0.155 096 225 873 92;
  • 13) 0.155 096 225 873 92 × 2 = 0 + 0.310 192 451 747 84;
  • 14) 0.310 192 451 747 84 × 2 = 0 + 0.620 384 903 495 68;
  • 15) 0.620 384 903 495 68 × 2 = 1 + 0.240 769 806 991 36;
  • 16) 0.240 769 806 991 36 × 2 = 0 + 0.481 539 613 982 72;
  • 17) 0.481 539 613 982 72 × 2 = 0 + 0.963 079 227 965 44;
  • 18) 0.963 079 227 965 44 × 2 = 1 + 0.926 158 455 930 88;
  • 19) 0.926 158 455 930 88 × 2 = 1 + 0.852 316 911 861 76;
  • 20) 0.852 316 911 861 76 × 2 = 1 + 0.704 633 823 723 52;
  • 21) 0.704 633 823 723 52 × 2 = 1 + 0.409 267 647 447 04;
  • 22) 0.409 267 647 447 04 × 2 = 0 + 0.818 535 294 894 08;
  • 23) 0.818 535 294 894 08 × 2 = 1 + 0.637 070 589 788 16;
  • 24) 0.637 070 589 788 16 × 2 = 1 + 0.274 141 179 576 32;
  • 25) 0.274 141 179 576 32 × 2 = 0 + 0.548 282 359 152 64;
  • 26) 0.548 282 359 152 64 × 2 = 1 + 0.096 564 718 305 28;
  • 27) 0.096 564 718 305 28 × 2 = 0 + 0.193 129 436 610 56;
  • 28) 0.193 129 436 610 56 × 2 = 0 + 0.386 258 873 221 12;
  • 29) 0.386 258 873 221 12 × 2 = 0 + 0.772 517 746 442 24;
  • 30) 0.772 517 746 442 24 × 2 = 1 + 0.545 035 492 884 48;
  • 31) 0.545 035 492 884 48 × 2 = 1 + 0.090 070 985 768 96;
  • 32) 0.090 070 985 768 96 × 2 = 0 + 0.180 141 971 537 92;
  • 33) 0.180 141 971 537 92 × 2 = 0 + 0.360 283 943 075 84;
  • 34) 0.360 283 943 075 84 × 2 = 0 + 0.720 567 886 151 68;
  • 35) 0.720 567 886 151 68 × 2 = 1 + 0.441 135 772 303 36;
  • 36) 0.441 135 772 303 36 × 2 = 0 + 0.882 271 544 606 72;
  • 37) 0.882 271 544 606 72 × 2 = 1 + 0.764 543 089 213 44;
  • 38) 0.764 543 089 213 44 × 2 = 1 + 0.529 086 178 426 88;
  • 39) 0.529 086 178 426 88 × 2 = 1 + 0.058 172 356 853 76;
  • 40) 0.058 172 356 853 76 × 2 = 0 + 0.116 344 713 707 52;
  • 41) 0.116 344 713 707 52 × 2 = 0 + 0.232 689 427 415 04;
  • 42) 0.232 689 427 415 04 × 2 = 0 + 0.465 378 854 830 08;
  • 43) 0.465 378 854 830 08 × 2 = 0 + 0.930 757 709 660 16;
  • 44) 0.930 757 709 660 16 × 2 = 1 + 0.861 515 419 320 32;
  • 45) 0.861 515 419 320 32 × 2 = 1 + 0.723 030 838 640 64;
  • 46) 0.723 030 838 640 64 × 2 = 1 + 0.446 061 677 281 28;
  • 47) 0.446 061 677 281 28 × 2 = 0 + 0.892 123 354 562 56;
  • 48) 0.892 123 354 562 56 × 2 = 1 + 0.784 246 709 125 12;
  • 49) 0.784 246 709 125 12 × 2 = 1 + 0.568 493 418 250 24;
  • 50) 0.568 493 418 250 24 × 2 = 1 + 0.136 986 836 500 48;
  • 51) 0.136 986 836 500 48 × 2 = 0 + 0.273 973 673 000 96;
  • 52) 0.273 973 673 000 96 × 2 = 0 + 0.547 947 346 001 92;
  • 53) 0.547 947 346 001 92 × 2 = 1 + 0.095 894 692 003 84;
  • 54) 0.095 894 692 003 84 × 2 = 0 + 0.191 789 384 007 68;
  • 55) 0.191 789 384 007 68 × 2 = 0 + 0.383 578 768 015 36;
  • 56) 0.383 578 768 015 36 × 2 = 0 + 0.767 157 536 030 72;
  • 57) 0.767 157 536 030 72 × 2 = 1 + 0.534 315 072 061 44;
  • 58) 0.534 315 072 061 44 × 2 = 1 + 0.068 630 144 122 88;
  • 59) 0.068 630 144 122 88 × 2 = 0 + 0.137 260 288 245 76;
  • 60) 0.137 260 288 245 76 × 2 = 0 + 0.274 520 576 491 52;
  • 61) 0.274 520 576 491 52 × 2 = 0 + 0.549 041 152 983 04;
  • 62) 0.549 041 152 983 04 × 2 = 1 + 0.098 082 305 966 08;
  • 63) 0.098 082 305 966 08 × 2 = 0 + 0.196 164 611 932 16;
  • 64) 0.196 164 611 932 16 × 2 = 0 + 0.392 329 223 864 32;

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 52(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1101 1100 1000 1100 0100(2)

6. Positive number before normalization:

0.000 282 005 914 52(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1101 1100 1000 1100 0100(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 52(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1101 1100 1000 1100 0100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0001 1101 1100 1000 1100 0100(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0001 1101 1100 1000 1100 0100(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 1110 0001 1101 1100 1000 1100 0100


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 1110 0001 1101 1100 1000 1100 0100 =

0010 0111 1011 0100 0110 0010 1110 0001 1101 1100 1000 1100 0100


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 1110 0001 1101 1100 1000 1100 0100


Decimal number -0.000 282 005 914 52 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0001 1101 1100 1000 1100 0100

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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