-0.000 281 712 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 712(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
-0.000 281 712(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:

|-0.000 281 712| = 0.000 281 712


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: 0.000 281 712.

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) 0.000 281 712 × 2 = 0 + 0.000 563 424;
  • 2) 0.000 563 424 × 2 = 0 + 0.001 126 848;
  • 3) 0.001 126 848 × 2 = 0 + 0.002 253 696;
  • 4) 0.002 253 696 × 2 = 0 + 0.004 507 392;
  • 5) 0.004 507 392 × 2 = 0 + 0.009 014 784;
  • 6) 0.009 014 784 × 2 = 0 + 0.018 029 568;
  • 7) 0.018 029 568 × 2 = 0 + 0.036 059 136;
  • 8) 0.036 059 136 × 2 = 0 + 0.072 118 272;
  • 9) 0.072 118 272 × 2 = 0 + 0.144 236 544;
  • 10) 0.144 236 544 × 2 = 0 + 0.288 473 088;
  • 11) 0.288 473 088 × 2 = 0 + 0.576 946 176;
  • 12) 0.576 946 176 × 2 = 1 + 0.153 892 352;
  • 13) 0.153 892 352 × 2 = 0 + 0.307 784 704;
  • 14) 0.307 784 704 × 2 = 0 + 0.615 569 408;
  • 15) 0.615 569 408 × 2 = 1 + 0.231 138 816;
  • 16) 0.231 138 816 × 2 = 0 + 0.462 277 632;
  • 17) 0.462 277 632 × 2 = 0 + 0.924 555 264;
  • 18) 0.924 555 264 × 2 = 1 + 0.849 110 528;
  • 19) 0.849 110 528 × 2 = 1 + 0.698 221 056;
  • 20) 0.698 221 056 × 2 = 1 + 0.396 442 112;
  • 21) 0.396 442 112 × 2 = 0 + 0.792 884 224;
  • 22) 0.792 884 224 × 2 = 1 + 0.585 768 448;
  • 23) 0.585 768 448 × 2 = 1 + 0.171 536 896;
  • 24) 0.171 536 896 × 2 = 0 + 0.343 073 792;
  • 25) 0.343 073 792 × 2 = 0 + 0.686 147 584;
  • 26) 0.686 147 584 × 2 = 1 + 0.372 295 168;
  • 27) 0.372 295 168 × 2 = 0 + 0.744 590 336;
  • 28) 0.744 590 336 × 2 = 1 + 0.489 180 672;
  • 29) 0.489 180 672 × 2 = 0 + 0.978 361 344;
  • 30) 0.978 361 344 × 2 = 1 + 0.956 722 688;
  • 31) 0.956 722 688 × 2 = 1 + 0.913 445 376;
  • 32) 0.913 445 376 × 2 = 1 + 0.826 890 752;
  • 33) 0.826 890 752 × 2 = 1 + 0.653 781 504;
  • 34) 0.653 781 504 × 2 = 1 + 0.307 563 008;
  • 35) 0.307 563 008 × 2 = 0 + 0.615 126 016;
  • 36) 0.615 126 016 × 2 = 1 + 0.230 252 032;
  • 37) 0.230 252 032 × 2 = 0 + 0.460 504 064;
  • 38) 0.460 504 064 × 2 = 0 + 0.921 008 128;
  • 39) 0.921 008 128 × 2 = 1 + 0.842 016 256;
  • 40) 0.842 016 256 × 2 = 1 + 0.684 032 512;
  • 41) 0.684 032 512 × 2 = 1 + 0.368 065 024;
  • 42) 0.368 065 024 × 2 = 0 + 0.736 130 048;
  • 43) 0.736 130 048 × 2 = 1 + 0.472 260 096;
  • 44) 0.472 260 096 × 2 = 0 + 0.944 520 192;
  • 45) 0.944 520 192 × 2 = 1 + 0.889 040 384;
  • 46) 0.889 040 384 × 2 = 1 + 0.778 080 768;
  • 47) 0.778 080 768 × 2 = 1 + 0.556 161 536;
  • 48) 0.556 161 536 × 2 = 1 + 0.112 323 072;
  • 49) 0.112 323 072 × 2 = 0 + 0.224 646 144;
  • 50) 0.224 646 144 × 2 = 0 + 0.449 292 288;
  • 51) 0.449 292 288 × 2 = 0 + 0.898 584 576;
  • 52) 0.898 584 576 × 2 = 1 + 0.797 169 152;
  • 53) 0.797 169 152 × 2 = 1 + 0.594 338 304;
  • 54) 0.594 338 304 × 2 = 1 + 0.188 676 608;
  • 55) 0.188 676 608 × 2 = 0 + 0.377 353 216;
  • 56) 0.377 353 216 × 2 = 0 + 0.754 706 432;
  • 57) 0.754 706 432 × 2 = 1 + 0.509 412 864;
  • 58) 0.509 412 864 × 2 = 1 + 0.018 825 728;
  • 59) 0.018 825 728 × 2 = 0 + 0.037 651 456;
  • 60) 0.037 651 456 × 2 = 0 + 0.075 302 912;
  • 61) 0.075 302 912 × 2 = 0 + 0.150 605 824;
  • 62) 0.150 605 824 × 2 = 0 + 0.301 211 648;
  • 63) 0.301 211 648 × 2 = 0 + 0.602 423 296;
  • 64) 0.602 423 296 × 2 = 1 + 0.204 846 592;

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:


0.000 281 712(10) =

0.0000 0000 0001 0010 0111 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001(2)

6. Positive number before normalization:

0.000 281 712(10) =

0.0000 0000 0001 0010 0111 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001(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:


0.000 281 712(10) =

0.0000 0000 0001 0010 0111 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001(2) =

0.0000 0000 0001 0010 0111 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001(2) × 20 =

1.0010 0111 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001(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 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001


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 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001 =

0010 0111 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001


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 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001


Decimal number -0.000 281 712 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 0110 0101 0111 1101 0011 1010 1111 0001 1100 1100 0001

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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