-0.000 282 006 028 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 028(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
-0.000 282 006 028(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:

|-0.000 282 006 028| = 0.000 282 006 028


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: 0.000 282 006 028.

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) 0.000 282 006 028 × 2 = 0 + 0.000 564 012 056;
  • 2) 0.000 564 012 056 × 2 = 0 + 0.001 128 024 112;
  • 3) 0.001 128 024 112 × 2 = 0 + 0.002 256 048 224;
  • 4) 0.002 256 048 224 × 2 = 0 + 0.004 512 096 448;
  • 5) 0.004 512 096 448 × 2 = 0 + 0.009 024 192 896;
  • 6) 0.009 024 192 896 × 2 = 0 + 0.018 048 385 792;
  • 7) 0.018 048 385 792 × 2 = 0 + 0.036 096 771 584;
  • 8) 0.036 096 771 584 × 2 = 0 + 0.072 193 543 168;
  • 9) 0.072 193 543 168 × 2 = 0 + 0.144 387 086 336;
  • 10) 0.144 387 086 336 × 2 = 0 + 0.288 774 172 672;
  • 11) 0.288 774 172 672 × 2 = 0 + 0.577 548 345 344;
  • 12) 0.577 548 345 344 × 2 = 1 + 0.155 096 690 688;
  • 13) 0.155 096 690 688 × 2 = 0 + 0.310 193 381 376;
  • 14) 0.310 193 381 376 × 2 = 0 + 0.620 386 762 752;
  • 15) 0.620 386 762 752 × 2 = 1 + 0.240 773 525 504;
  • 16) 0.240 773 525 504 × 2 = 0 + 0.481 547 051 008;
  • 17) 0.481 547 051 008 × 2 = 0 + 0.963 094 102 016;
  • 18) 0.963 094 102 016 × 2 = 1 + 0.926 188 204 032;
  • 19) 0.926 188 204 032 × 2 = 1 + 0.852 376 408 064;
  • 20) 0.852 376 408 064 × 2 = 1 + 0.704 752 816 128;
  • 21) 0.704 752 816 128 × 2 = 1 + 0.409 505 632 256;
  • 22) 0.409 505 632 256 × 2 = 0 + 0.819 011 264 512;
  • 23) 0.819 011 264 512 × 2 = 1 + 0.638 022 529 024;
  • 24) 0.638 022 529 024 × 2 = 1 + 0.276 045 058 048;
  • 25) 0.276 045 058 048 × 2 = 0 + 0.552 090 116 096;
  • 26) 0.552 090 116 096 × 2 = 1 + 0.104 180 232 192;
  • 27) 0.104 180 232 192 × 2 = 0 + 0.208 360 464 384;
  • 28) 0.208 360 464 384 × 2 = 0 + 0.416 720 928 768;
  • 29) 0.416 720 928 768 × 2 = 0 + 0.833 441 857 536;
  • 30) 0.833 441 857 536 × 2 = 1 + 0.666 883 715 072;
  • 31) 0.666 883 715 072 × 2 = 1 + 0.333 767 430 144;
  • 32) 0.333 767 430 144 × 2 = 0 + 0.667 534 860 288;
  • 33) 0.667 534 860 288 × 2 = 1 + 0.335 069 720 576;
  • 34) 0.335 069 720 576 × 2 = 0 + 0.670 139 441 152;
  • 35) 0.670 139 441 152 × 2 = 1 + 0.340 278 882 304;
  • 36) 0.340 278 882 304 × 2 = 0 + 0.680 557 764 608;
  • 37) 0.680 557 764 608 × 2 = 1 + 0.361 115 529 216;
  • 38) 0.361 115 529 216 × 2 = 0 + 0.722 231 058 432;
  • 39) 0.722 231 058 432 × 2 = 1 + 0.444 462 116 864;
  • 40) 0.444 462 116 864 × 2 = 0 + 0.888 924 233 728;
  • 41) 0.888 924 233 728 × 2 = 1 + 0.777 848 467 456;
  • 42) 0.777 848 467 456 × 2 = 1 + 0.555 696 934 912;
  • 43) 0.555 696 934 912 × 2 = 1 + 0.111 393 869 824;
  • 44) 0.111 393 869 824 × 2 = 0 + 0.222 787 739 648;
  • 45) 0.222 787 739 648 × 2 = 0 + 0.445 575 479 296;
  • 46) 0.445 575 479 296 × 2 = 0 + 0.891 150 958 592;
  • 47) 0.891 150 958 592 × 2 = 1 + 0.782 301 917 184;
  • 48) 0.782 301 917 184 × 2 = 1 + 0.564 603 834 368;
  • 49) 0.564 603 834 368 × 2 = 1 + 0.129 207 668 736;
  • 50) 0.129 207 668 736 × 2 = 0 + 0.258 415 337 472;
  • 51) 0.258 415 337 472 × 2 = 0 + 0.516 830 674 944;
  • 52) 0.516 830 674 944 × 2 = 1 + 0.033 661 349 888;
  • 53) 0.033 661 349 888 × 2 = 0 + 0.067 322 699 776;
  • 54) 0.067 322 699 776 × 2 = 0 + 0.134 645 399 552;
  • 55) 0.134 645 399 552 × 2 = 0 + 0.269 290 799 104;
  • 56) 0.269 290 799 104 × 2 = 0 + 0.538 581 598 208;
  • 57) 0.538 581 598 208 × 2 = 1 + 0.077 163 196 416;
  • 58) 0.077 163 196 416 × 2 = 0 + 0.154 326 392 832;
  • 59) 0.154 326 392 832 × 2 = 0 + 0.308 652 785 664;
  • 60) 0.308 652 785 664 × 2 = 0 + 0.617 305 571 328;
  • 61) 0.617 305 571 328 × 2 = 1 + 0.234 611 142 656;
  • 62) 0.234 611 142 656 × 2 = 0 + 0.469 222 285 312;
  • 63) 0.469 222 285 312 × 2 = 0 + 0.938 444 570 624;
  • 64) 0.938 444 570 624 × 2 = 1 + 0.876 889 141 248;

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:


0.000 282 006 028(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1010 1010 1110 0011 1001 0000 1000 1001(2)

6. Positive number before normalization:

0.000 282 006 028(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1010 1010 1110 0011 1001 0000 1000 1001(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:


0.000 282 006 028(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1010 1010 1110 0011 1001 0000 1000 1001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1010 1010 1110 0011 1001 0000 1000 1001(2) × 20 =

1.0010 0111 1011 0100 0110 1010 1010 1110 0011 1001 0000 1000 1001(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 1010 1010 1110 0011 1001 0000 1000 1001


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 1010 1010 1110 0011 1001 0000 1000 1001 =

0010 0111 1011 0100 0110 1010 1010 1110 0011 1001 0000 1000 1001


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 1010 1010 1110 0011 1001 0000 1000 1001


Decimal number -0.000 282 006 028 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1010 1010 1110 0011 1001 0000 1000 1001

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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