-0.000 281 839 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 839(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
-0.000 281 839(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:

|-0.000 281 839| = 0.000 281 839


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: 0.000 281 839.

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) 0.000 281 839 × 2 = 0 + 0.000 563 678;
  • 2) 0.000 563 678 × 2 = 0 + 0.001 127 356;
  • 3) 0.001 127 356 × 2 = 0 + 0.002 254 712;
  • 4) 0.002 254 712 × 2 = 0 + 0.004 509 424;
  • 5) 0.004 509 424 × 2 = 0 + 0.009 018 848;
  • 6) 0.009 018 848 × 2 = 0 + 0.018 037 696;
  • 7) 0.018 037 696 × 2 = 0 + 0.036 075 392;
  • 8) 0.036 075 392 × 2 = 0 + 0.072 150 784;
  • 9) 0.072 150 784 × 2 = 0 + 0.144 301 568;
  • 10) 0.144 301 568 × 2 = 0 + 0.288 603 136;
  • 11) 0.288 603 136 × 2 = 0 + 0.577 206 272;
  • 12) 0.577 206 272 × 2 = 1 + 0.154 412 544;
  • 13) 0.154 412 544 × 2 = 0 + 0.308 825 088;
  • 14) 0.308 825 088 × 2 = 0 + 0.617 650 176;
  • 15) 0.617 650 176 × 2 = 1 + 0.235 300 352;
  • 16) 0.235 300 352 × 2 = 0 + 0.470 600 704;
  • 17) 0.470 600 704 × 2 = 0 + 0.941 201 408;
  • 18) 0.941 201 408 × 2 = 1 + 0.882 402 816;
  • 19) 0.882 402 816 × 2 = 1 + 0.764 805 632;
  • 20) 0.764 805 632 × 2 = 1 + 0.529 611 264;
  • 21) 0.529 611 264 × 2 = 1 + 0.059 222 528;
  • 22) 0.059 222 528 × 2 = 0 + 0.118 445 056;
  • 23) 0.118 445 056 × 2 = 0 + 0.236 890 112;
  • 24) 0.236 890 112 × 2 = 0 + 0.473 780 224;
  • 25) 0.473 780 224 × 2 = 0 + 0.947 560 448;
  • 26) 0.947 560 448 × 2 = 1 + 0.895 120 896;
  • 27) 0.895 120 896 × 2 = 1 + 0.790 241 792;
  • 28) 0.790 241 792 × 2 = 1 + 0.580 483 584;
  • 29) 0.580 483 584 × 2 = 1 + 0.160 967 168;
  • 30) 0.160 967 168 × 2 = 0 + 0.321 934 336;
  • 31) 0.321 934 336 × 2 = 0 + 0.643 868 672;
  • 32) 0.643 868 672 × 2 = 1 + 0.287 737 344;
  • 33) 0.287 737 344 × 2 = 0 + 0.575 474 688;
  • 34) 0.575 474 688 × 2 = 1 + 0.150 949 376;
  • 35) 0.150 949 376 × 2 = 0 + 0.301 898 752;
  • 36) 0.301 898 752 × 2 = 0 + 0.603 797 504;
  • 37) 0.603 797 504 × 2 = 1 + 0.207 595 008;
  • 38) 0.207 595 008 × 2 = 0 + 0.415 190 016;
  • 39) 0.415 190 016 × 2 = 0 + 0.830 380 032;
  • 40) 0.830 380 032 × 2 = 1 + 0.660 760 064;
  • 41) 0.660 760 064 × 2 = 1 + 0.321 520 128;
  • 42) 0.321 520 128 × 2 = 0 + 0.643 040 256;
  • 43) 0.643 040 256 × 2 = 1 + 0.286 080 512;
  • 44) 0.286 080 512 × 2 = 0 + 0.572 161 024;
  • 45) 0.572 161 024 × 2 = 1 + 0.144 322 048;
  • 46) 0.144 322 048 × 2 = 0 + 0.288 644 096;
  • 47) 0.288 644 096 × 2 = 0 + 0.577 288 192;
  • 48) 0.577 288 192 × 2 = 1 + 0.154 576 384;
  • 49) 0.154 576 384 × 2 = 0 + 0.309 152 768;
  • 50) 0.309 152 768 × 2 = 0 + 0.618 305 536;
  • 51) 0.618 305 536 × 2 = 1 + 0.236 611 072;
  • 52) 0.236 611 072 × 2 = 0 + 0.473 222 144;
  • 53) 0.473 222 144 × 2 = 0 + 0.946 444 288;
  • 54) 0.946 444 288 × 2 = 1 + 0.892 888 576;
  • 55) 0.892 888 576 × 2 = 1 + 0.785 777 152;
  • 56) 0.785 777 152 × 2 = 1 + 0.571 554 304;
  • 57) 0.571 554 304 × 2 = 1 + 0.143 108 608;
  • 58) 0.143 108 608 × 2 = 0 + 0.286 217 216;
  • 59) 0.286 217 216 × 2 = 0 + 0.572 434 432;
  • 60) 0.572 434 432 × 2 = 1 + 0.144 868 864;
  • 61) 0.144 868 864 × 2 = 0 + 0.289 737 728;
  • 62) 0.289 737 728 × 2 = 0 + 0.579 475 456;
  • 63) 0.579 475 456 × 2 = 1 + 0.158 950 912;
  • 64) 0.158 950 912 × 2 = 0 + 0.317 901 824;

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:


0.000 281 839(10) =

0.0000 0000 0001 0010 0111 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010(2)

6. Positive number before normalization:

0.000 281 839(10) =

0.0000 0000 0001 0010 0111 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010(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:


0.000 281 839(10) =

0.0000 0000 0001 0010 0111 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010(2) =

0.0000 0000 0001 0010 0111 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010(2) × 20 =

1.0010 0111 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010(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 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010


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 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010 =

0010 0111 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010


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 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010


Decimal number -0.000 281 839 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1000 0111 1001 0100 1001 1010 1001 0010 0111 1001 0010

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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