-0.000 282 005 893 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 893 3(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
-0.000 282 005 893 3(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:

|-0.000 282 005 893 3| = 0.000 282 005 893 3


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: 0.000 282 005 893 3.

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) 0.000 282 005 893 3 × 2 = 0 + 0.000 564 011 786 6;
  • 2) 0.000 564 011 786 6 × 2 = 0 + 0.001 128 023 573 2;
  • 3) 0.001 128 023 573 2 × 2 = 0 + 0.002 256 047 146 4;
  • 4) 0.002 256 047 146 4 × 2 = 0 + 0.004 512 094 292 8;
  • 5) 0.004 512 094 292 8 × 2 = 0 + 0.009 024 188 585 6;
  • 6) 0.009 024 188 585 6 × 2 = 0 + 0.018 048 377 171 2;
  • 7) 0.018 048 377 171 2 × 2 = 0 + 0.036 096 754 342 4;
  • 8) 0.036 096 754 342 4 × 2 = 0 + 0.072 193 508 684 8;
  • 9) 0.072 193 508 684 8 × 2 = 0 + 0.144 387 017 369 6;
  • 10) 0.144 387 017 369 6 × 2 = 0 + 0.288 774 034 739 2;
  • 11) 0.288 774 034 739 2 × 2 = 0 + 0.577 548 069 478 4;
  • 12) 0.577 548 069 478 4 × 2 = 1 + 0.155 096 138 956 8;
  • 13) 0.155 096 138 956 8 × 2 = 0 + 0.310 192 277 913 6;
  • 14) 0.310 192 277 913 6 × 2 = 0 + 0.620 384 555 827 2;
  • 15) 0.620 384 555 827 2 × 2 = 1 + 0.240 769 111 654 4;
  • 16) 0.240 769 111 654 4 × 2 = 0 + 0.481 538 223 308 8;
  • 17) 0.481 538 223 308 8 × 2 = 0 + 0.963 076 446 617 6;
  • 18) 0.963 076 446 617 6 × 2 = 1 + 0.926 152 893 235 2;
  • 19) 0.926 152 893 235 2 × 2 = 1 + 0.852 305 786 470 4;
  • 20) 0.852 305 786 470 4 × 2 = 1 + 0.704 611 572 940 8;
  • 21) 0.704 611 572 940 8 × 2 = 1 + 0.409 223 145 881 6;
  • 22) 0.409 223 145 881 6 × 2 = 0 + 0.818 446 291 763 2;
  • 23) 0.818 446 291 763 2 × 2 = 1 + 0.636 892 583 526 4;
  • 24) 0.636 892 583 526 4 × 2 = 1 + 0.273 785 167 052 8;
  • 25) 0.273 785 167 052 8 × 2 = 0 + 0.547 570 334 105 6;
  • 26) 0.547 570 334 105 6 × 2 = 1 + 0.095 140 668 211 2;
  • 27) 0.095 140 668 211 2 × 2 = 0 + 0.190 281 336 422 4;
  • 28) 0.190 281 336 422 4 × 2 = 0 + 0.380 562 672 844 8;
  • 29) 0.380 562 672 844 8 × 2 = 0 + 0.761 125 345 689 6;
  • 30) 0.761 125 345 689 6 × 2 = 1 + 0.522 250 691 379 2;
  • 31) 0.522 250 691 379 2 × 2 = 1 + 0.044 501 382 758 4;
  • 32) 0.044 501 382 758 4 × 2 = 0 + 0.089 002 765 516 8;
  • 33) 0.089 002 765 516 8 × 2 = 0 + 0.178 005 531 033 6;
  • 34) 0.178 005 531 033 6 × 2 = 0 + 0.356 011 062 067 2;
  • 35) 0.356 011 062 067 2 × 2 = 0 + 0.712 022 124 134 4;
  • 36) 0.712 022 124 134 4 × 2 = 1 + 0.424 044 248 268 8;
  • 37) 0.424 044 248 268 8 × 2 = 0 + 0.848 088 496 537 6;
  • 38) 0.848 088 496 537 6 × 2 = 1 + 0.696 176 993 075 2;
  • 39) 0.696 176 993 075 2 × 2 = 1 + 0.392 353 986 150 4;
  • 40) 0.392 353 986 150 4 × 2 = 0 + 0.784 707 972 300 8;
  • 41) 0.784 707 972 300 8 × 2 = 1 + 0.569 415 944 601 6;
  • 42) 0.569 415 944 601 6 × 2 = 1 + 0.138 831 889 203 2;
  • 43) 0.138 831 889 203 2 × 2 = 0 + 0.277 663 778 406 4;
  • 44) 0.277 663 778 406 4 × 2 = 0 + 0.555 327 556 812 8;
  • 45) 0.555 327 556 812 8 × 2 = 1 + 0.110 655 113 625 6;
  • 46) 0.110 655 113 625 6 × 2 = 0 + 0.221 310 227 251 2;
  • 47) 0.221 310 227 251 2 × 2 = 0 + 0.442 620 454 502 4;
  • 48) 0.442 620 454 502 4 × 2 = 0 + 0.885 240 909 004 8;
  • 49) 0.885 240 909 004 8 × 2 = 1 + 0.770 481 818 009 6;
  • 50) 0.770 481 818 009 6 × 2 = 1 + 0.540 963 636 019 2;
  • 51) 0.540 963 636 019 2 × 2 = 1 + 0.081 927 272 038 4;
  • 52) 0.081 927 272 038 4 × 2 = 0 + 0.163 854 544 076 8;
  • 53) 0.163 854 544 076 8 × 2 = 0 + 0.327 709 088 153 6;
  • 54) 0.327 709 088 153 6 × 2 = 0 + 0.655 418 176 307 2;
  • 55) 0.655 418 176 307 2 × 2 = 1 + 0.310 836 352 614 4;
  • 56) 0.310 836 352 614 4 × 2 = 0 + 0.621 672 705 228 8;
  • 57) 0.621 672 705 228 8 × 2 = 1 + 0.243 345 410 457 6;
  • 58) 0.243 345 410 457 6 × 2 = 0 + 0.486 690 820 915 2;
  • 59) 0.486 690 820 915 2 × 2 = 0 + 0.973 381 641 830 4;
  • 60) 0.973 381 641 830 4 × 2 = 1 + 0.946 763 283 660 8;
  • 61) 0.946 763 283 660 8 × 2 = 1 + 0.893 526 567 321 6;
  • 62) 0.893 526 567 321 6 × 2 = 1 + 0.787 053 134 643 2;
  • 63) 0.787 053 134 643 2 × 2 = 1 + 0.574 106 269 286 4;
  • 64) 0.574 106 269 286 4 × 2 = 1 + 0.148 212 538 572 8;

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:


0.000 282 005 893 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0110 1100 1000 1110 0010 1001 1111(2)

6. Positive number before normalization:

0.000 282 005 893 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0110 1100 1000 1110 0010 1001 1111(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:


0.000 282 005 893 3(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0110 1100 1000 1110 0010 1001 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 0110 1100 1000 1110 0010 1001 1111(2) × 20 =

1.0010 0111 1011 0100 0110 0001 0110 1100 1000 1110 0010 1001 1111(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 0001 0110 1100 1000 1110 0010 1001 1111


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 0001 0110 1100 1000 1110 0010 1001 1111 =

0010 0111 1011 0100 0110 0001 0110 1100 1000 1110 0010 1001 1111


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 0001 0110 1100 1000 1110 0010 1001 1111


Decimal number -0.000 282 005 893 3 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 0110 1100 1000 1110 0010 1001 1111

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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