-0.000 282 006 044 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 044(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
-0.000 282 006 044(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:

|-0.000 282 006 044| = 0.000 282 006 044


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: 0.000 282 006 044.

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) 0.000 282 006 044 × 2 = 0 + 0.000 564 012 088;
  • 2) 0.000 564 012 088 × 2 = 0 + 0.001 128 024 176;
  • 3) 0.001 128 024 176 × 2 = 0 + 0.002 256 048 352;
  • 4) 0.002 256 048 352 × 2 = 0 + 0.004 512 096 704;
  • 5) 0.004 512 096 704 × 2 = 0 + 0.009 024 193 408;
  • 6) 0.009 024 193 408 × 2 = 0 + 0.018 048 386 816;
  • 7) 0.018 048 386 816 × 2 = 0 + 0.036 096 773 632;
  • 8) 0.036 096 773 632 × 2 = 0 + 0.072 193 547 264;
  • 9) 0.072 193 547 264 × 2 = 0 + 0.144 387 094 528;
  • 10) 0.144 387 094 528 × 2 = 0 + 0.288 774 189 056;
  • 11) 0.288 774 189 056 × 2 = 0 + 0.577 548 378 112;
  • 12) 0.577 548 378 112 × 2 = 1 + 0.155 096 756 224;
  • 13) 0.155 096 756 224 × 2 = 0 + 0.310 193 512 448;
  • 14) 0.310 193 512 448 × 2 = 0 + 0.620 387 024 896;
  • 15) 0.620 387 024 896 × 2 = 1 + 0.240 774 049 792;
  • 16) 0.240 774 049 792 × 2 = 0 + 0.481 548 099 584;
  • 17) 0.481 548 099 584 × 2 = 0 + 0.963 096 199 168;
  • 18) 0.963 096 199 168 × 2 = 1 + 0.926 192 398 336;
  • 19) 0.926 192 398 336 × 2 = 1 + 0.852 384 796 672;
  • 20) 0.852 384 796 672 × 2 = 1 + 0.704 769 593 344;
  • 21) 0.704 769 593 344 × 2 = 1 + 0.409 539 186 688;
  • 22) 0.409 539 186 688 × 2 = 0 + 0.819 078 373 376;
  • 23) 0.819 078 373 376 × 2 = 1 + 0.638 156 746 752;
  • 24) 0.638 156 746 752 × 2 = 1 + 0.276 313 493 504;
  • 25) 0.276 313 493 504 × 2 = 0 + 0.552 626 987 008;
  • 26) 0.552 626 987 008 × 2 = 1 + 0.105 253 974 016;
  • 27) 0.105 253 974 016 × 2 = 0 + 0.210 507 948 032;
  • 28) 0.210 507 948 032 × 2 = 0 + 0.421 015 896 064;
  • 29) 0.421 015 896 064 × 2 = 0 + 0.842 031 792 128;
  • 30) 0.842 031 792 128 × 2 = 1 + 0.684 063 584 256;
  • 31) 0.684 063 584 256 × 2 = 1 + 0.368 127 168 512;
  • 32) 0.368 127 168 512 × 2 = 0 + 0.736 254 337 024;
  • 33) 0.736 254 337 024 × 2 = 1 + 0.472 508 674 048;
  • 34) 0.472 508 674 048 × 2 = 0 + 0.945 017 348 096;
  • 35) 0.945 017 348 096 × 2 = 1 + 0.890 034 696 192;
  • 36) 0.890 034 696 192 × 2 = 1 + 0.780 069 392 384;
  • 37) 0.780 069 392 384 × 2 = 1 + 0.560 138 784 768;
  • 38) 0.560 138 784 768 × 2 = 1 + 0.120 277 569 536;
  • 39) 0.120 277 569 536 × 2 = 0 + 0.240 555 139 072;
  • 40) 0.240 555 139 072 × 2 = 0 + 0.481 110 278 144;
  • 41) 0.481 110 278 144 × 2 = 0 + 0.962 220 556 288;
  • 42) 0.962 220 556 288 × 2 = 1 + 0.924 441 112 576;
  • 43) 0.924 441 112 576 × 2 = 1 + 0.848 882 225 152;
  • 44) 0.848 882 225 152 × 2 = 1 + 0.697 764 450 304;
  • 45) 0.697 764 450 304 × 2 = 1 + 0.395 528 900 608;
  • 46) 0.395 528 900 608 × 2 = 0 + 0.791 057 801 216;
  • 47) 0.791 057 801 216 × 2 = 1 + 0.582 115 602 432;
  • 48) 0.582 115 602 432 × 2 = 1 + 0.164 231 204 864;
  • 49) 0.164 231 204 864 × 2 = 0 + 0.328 462 409 728;
  • 50) 0.328 462 409 728 × 2 = 0 + 0.656 924 819 456;
  • 51) 0.656 924 819 456 × 2 = 1 + 0.313 849 638 912;
  • 52) 0.313 849 638 912 × 2 = 0 + 0.627 699 277 824;
  • 53) 0.627 699 277 824 × 2 = 1 + 0.255 398 555 648;
  • 54) 0.255 398 555 648 × 2 = 0 + 0.510 797 111 296;
  • 55) 0.510 797 111 296 × 2 = 1 + 0.021 594 222 592;
  • 56) 0.021 594 222 592 × 2 = 0 + 0.043 188 445 184;
  • 57) 0.043 188 445 184 × 2 = 0 + 0.086 376 890 368;
  • 58) 0.086 376 890 368 × 2 = 0 + 0.172 753 780 736;
  • 59) 0.172 753 780 736 × 2 = 0 + 0.345 507 561 472;
  • 60) 0.345 507 561 472 × 2 = 0 + 0.691 015 122 944;
  • 61) 0.691 015 122 944 × 2 = 1 + 0.382 030 245 888;
  • 62) 0.382 030 245 888 × 2 = 0 + 0.764 060 491 776;
  • 63) 0.764 060 491 776 × 2 = 1 + 0.528 120 983 552;
  • 64) 0.528 120 983 552 × 2 = 1 + 0.056 241 967 104;

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:


0.000 282 006 044(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 1100 0111 1011 0010 1010 0000 1011(2)

6. Positive number before normalization:

0.000 282 006 044(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 1100 0111 1011 0010 1010 0000 1011(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:


0.000 282 006 044(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 1100 0111 1011 0010 1010 0000 1011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1011 1100 0111 1011 0010 1010 0000 1011(2) × 20 =

1.0010 0111 1011 0100 0110 1011 1100 0111 1011 0010 1010 0000 1011(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 1011 1100 0111 1011 0010 1010 0000 1011


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 1011 1100 0111 1011 0010 1010 0000 1011 =

0010 0111 1011 0100 0110 1011 1100 0111 1011 0010 1010 0000 1011


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 1011 1100 0111 1011 0010 1010 0000 1011


Decimal number -0.000 282 006 044 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1011 1100 0111 1011 0010 1010 0000 1011

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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