-0.000 282 005 651 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 651(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
-0.000 282 005 651(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:

|-0.000 282 005 651| = 0.000 282 005 651


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: 0.000 282 005 651.

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) 0.000 282 005 651 × 2 = 0 + 0.000 564 011 302;
  • 2) 0.000 564 011 302 × 2 = 0 + 0.001 128 022 604;
  • 3) 0.001 128 022 604 × 2 = 0 + 0.002 256 045 208;
  • 4) 0.002 256 045 208 × 2 = 0 + 0.004 512 090 416;
  • 5) 0.004 512 090 416 × 2 = 0 + 0.009 024 180 832;
  • 6) 0.009 024 180 832 × 2 = 0 + 0.018 048 361 664;
  • 7) 0.018 048 361 664 × 2 = 0 + 0.036 096 723 328;
  • 8) 0.036 096 723 328 × 2 = 0 + 0.072 193 446 656;
  • 9) 0.072 193 446 656 × 2 = 0 + 0.144 386 893 312;
  • 10) 0.144 386 893 312 × 2 = 0 + 0.288 773 786 624;
  • 11) 0.288 773 786 624 × 2 = 0 + 0.577 547 573 248;
  • 12) 0.577 547 573 248 × 2 = 1 + 0.155 095 146 496;
  • 13) 0.155 095 146 496 × 2 = 0 + 0.310 190 292 992;
  • 14) 0.310 190 292 992 × 2 = 0 + 0.620 380 585 984;
  • 15) 0.620 380 585 984 × 2 = 1 + 0.240 761 171 968;
  • 16) 0.240 761 171 968 × 2 = 0 + 0.481 522 343 936;
  • 17) 0.481 522 343 936 × 2 = 0 + 0.963 044 687 872;
  • 18) 0.963 044 687 872 × 2 = 1 + 0.926 089 375 744;
  • 19) 0.926 089 375 744 × 2 = 1 + 0.852 178 751 488;
  • 20) 0.852 178 751 488 × 2 = 1 + 0.704 357 502 976;
  • 21) 0.704 357 502 976 × 2 = 1 + 0.408 715 005 952;
  • 22) 0.408 715 005 952 × 2 = 0 + 0.817 430 011 904;
  • 23) 0.817 430 011 904 × 2 = 1 + 0.634 860 023 808;
  • 24) 0.634 860 023 808 × 2 = 1 + 0.269 720 047 616;
  • 25) 0.269 720 047 616 × 2 = 0 + 0.539 440 095 232;
  • 26) 0.539 440 095 232 × 2 = 1 + 0.078 880 190 464;
  • 27) 0.078 880 190 464 × 2 = 0 + 0.157 760 380 928;
  • 28) 0.157 760 380 928 × 2 = 0 + 0.315 520 761 856;
  • 29) 0.315 520 761 856 × 2 = 0 + 0.631 041 523 712;
  • 30) 0.631 041 523 712 × 2 = 1 + 0.262 083 047 424;
  • 31) 0.262 083 047 424 × 2 = 0 + 0.524 166 094 848;
  • 32) 0.524 166 094 848 × 2 = 1 + 0.048 332 189 696;
  • 33) 0.048 332 189 696 × 2 = 0 + 0.096 664 379 392;
  • 34) 0.096 664 379 392 × 2 = 0 + 0.193 328 758 784;
  • 35) 0.193 328 758 784 × 2 = 0 + 0.386 657 517 568;
  • 36) 0.386 657 517 568 × 2 = 0 + 0.773 315 035 136;
  • 37) 0.773 315 035 136 × 2 = 1 + 0.546 630 070 272;
  • 38) 0.546 630 070 272 × 2 = 1 + 0.093 260 140 544;
  • 39) 0.093 260 140 544 × 2 = 0 + 0.186 520 281 088;
  • 40) 0.186 520 281 088 × 2 = 0 + 0.373 040 562 176;
  • 41) 0.373 040 562 176 × 2 = 0 + 0.746 081 124 352;
  • 42) 0.746 081 124 352 × 2 = 1 + 0.492 162 248 704;
  • 43) 0.492 162 248 704 × 2 = 0 + 0.984 324 497 408;
  • 44) 0.984 324 497 408 × 2 = 1 + 0.968 648 994 816;
  • 45) 0.968 648 994 816 × 2 = 1 + 0.937 297 989 632;
  • 46) 0.937 297 989 632 × 2 = 1 + 0.874 595 979 264;
  • 47) 0.874 595 979 264 × 2 = 1 + 0.749 191 958 528;
  • 48) 0.749 191 958 528 × 2 = 1 + 0.498 383 917 056;
  • 49) 0.498 383 917 056 × 2 = 0 + 0.996 767 834 112;
  • 50) 0.996 767 834 112 × 2 = 1 + 0.993 535 668 224;
  • 51) 0.993 535 668 224 × 2 = 1 + 0.987 071 336 448;
  • 52) 0.987 071 336 448 × 2 = 1 + 0.974 142 672 896;
  • 53) 0.974 142 672 896 × 2 = 1 + 0.948 285 345 792;
  • 54) 0.948 285 345 792 × 2 = 1 + 0.896 570 691 584;
  • 55) 0.896 570 691 584 × 2 = 1 + 0.793 141 383 168;
  • 56) 0.793 141 383 168 × 2 = 1 + 0.586 282 766 336;
  • 57) 0.586 282 766 336 × 2 = 1 + 0.172 565 532 672;
  • 58) 0.172 565 532 672 × 2 = 0 + 0.345 131 065 344;
  • 59) 0.345 131 065 344 × 2 = 0 + 0.690 262 130 688;
  • 60) 0.690 262 130 688 × 2 = 1 + 0.380 524 261 376;
  • 61) 0.380 524 261 376 × 2 = 0 + 0.761 048 522 752;
  • 62) 0.761 048 522 752 × 2 = 1 + 0.522 097 045 504;
  • 63) 0.522 097 045 504 × 2 = 1 + 0.044 194 091 008;
  • 64) 0.044 194 091 008 × 2 = 0 + 0.088 388 182 016;

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:


0.000 282 005 651(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0000 1100 0101 1111 0111 1111 1001 0110(2)

6. Positive number before normalization:

0.000 282 005 651(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0000 1100 0101 1111 0111 1111 1001 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:


0.000 282 005 651(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0000 1100 0101 1111 0111 1111 1001 0110(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0000 1100 0101 1111 0111 1111 1001 0110(2) × 20 =

1.0010 0111 1011 0100 0101 0000 1100 0101 1111 0111 1111 1001 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 0101 0000 1100 0101 1111 0111 1111 1001 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 0101 0000 1100 0101 1111 0111 1111 1001 0110 =

0010 0111 1011 0100 0101 0000 1100 0101 1111 0111 1111 1001 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 0101 0000 1100 0101 1111 0111 1111 1001 0110


Decimal number -0.000 282 005 651 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0000 1100 0101 1111 0111 1111 1001 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