-0.000 282 005 737 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 737(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
-0.000 282 005 737(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:

|-0.000 282 005 737| = 0.000 282 005 737


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: 0.000 282 005 737.

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) 0.000 282 005 737 × 2 = 0 + 0.000 564 011 474;
  • 2) 0.000 564 011 474 × 2 = 0 + 0.001 128 022 948;
  • 3) 0.001 128 022 948 × 2 = 0 + 0.002 256 045 896;
  • 4) 0.002 256 045 896 × 2 = 0 + 0.004 512 091 792;
  • 5) 0.004 512 091 792 × 2 = 0 + 0.009 024 183 584;
  • 6) 0.009 024 183 584 × 2 = 0 + 0.018 048 367 168;
  • 7) 0.018 048 367 168 × 2 = 0 + 0.036 096 734 336;
  • 8) 0.036 096 734 336 × 2 = 0 + 0.072 193 468 672;
  • 9) 0.072 193 468 672 × 2 = 0 + 0.144 386 937 344;
  • 10) 0.144 386 937 344 × 2 = 0 + 0.288 773 874 688;
  • 11) 0.288 773 874 688 × 2 = 0 + 0.577 547 749 376;
  • 12) 0.577 547 749 376 × 2 = 1 + 0.155 095 498 752;
  • 13) 0.155 095 498 752 × 2 = 0 + 0.310 190 997 504;
  • 14) 0.310 190 997 504 × 2 = 0 + 0.620 381 995 008;
  • 15) 0.620 381 995 008 × 2 = 1 + 0.240 763 990 016;
  • 16) 0.240 763 990 016 × 2 = 0 + 0.481 527 980 032;
  • 17) 0.481 527 980 032 × 2 = 0 + 0.963 055 960 064;
  • 18) 0.963 055 960 064 × 2 = 1 + 0.926 111 920 128;
  • 19) 0.926 111 920 128 × 2 = 1 + 0.852 223 840 256;
  • 20) 0.852 223 840 256 × 2 = 1 + 0.704 447 680 512;
  • 21) 0.704 447 680 512 × 2 = 1 + 0.408 895 361 024;
  • 22) 0.408 895 361 024 × 2 = 0 + 0.817 790 722 048;
  • 23) 0.817 790 722 048 × 2 = 1 + 0.635 581 444 096;
  • 24) 0.635 581 444 096 × 2 = 1 + 0.271 162 888 192;
  • 25) 0.271 162 888 192 × 2 = 0 + 0.542 325 776 384;
  • 26) 0.542 325 776 384 × 2 = 1 + 0.084 651 552 768;
  • 27) 0.084 651 552 768 × 2 = 0 + 0.169 303 105 536;
  • 28) 0.169 303 105 536 × 2 = 0 + 0.338 606 211 072;
  • 29) 0.338 606 211 072 × 2 = 0 + 0.677 212 422 144;
  • 30) 0.677 212 422 144 × 2 = 1 + 0.354 424 844 288;
  • 31) 0.354 424 844 288 × 2 = 0 + 0.708 849 688 576;
  • 32) 0.708 849 688 576 × 2 = 1 + 0.417 699 377 152;
  • 33) 0.417 699 377 152 × 2 = 0 + 0.835 398 754 304;
  • 34) 0.835 398 754 304 × 2 = 1 + 0.670 797 508 608;
  • 35) 0.670 797 508 608 × 2 = 1 + 0.341 595 017 216;
  • 36) 0.341 595 017 216 × 2 = 0 + 0.683 190 034 432;
  • 37) 0.683 190 034 432 × 2 = 1 + 0.366 380 068 864;
  • 38) 0.366 380 068 864 × 2 = 0 + 0.732 760 137 728;
  • 39) 0.732 760 137 728 × 2 = 1 + 0.465 520 275 456;
  • 40) 0.465 520 275 456 × 2 = 0 + 0.931 040 550 912;
  • 41) 0.931 040 550 912 × 2 = 1 + 0.862 081 101 824;
  • 42) 0.862 081 101 824 × 2 = 1 + 0.724 162 203 648;
  • 43) 0.724 162 203 648 × 2 = 1 + 0.448 324 407 296;
  • 44) 0.448 324 407 296 × 2 = 0 + 0.896 648 814 592;
  • 45) 0.896 648 814 592 × 2 = 1 + 0.793 297 629 184;
  • 46) 0.793 297 629 184 × 2 = 1 + 0.586 595 258 368;
  • 47) 0.586 595 258 368 × 2 = 1 + 0.173 190 516 736;
  • 48) 0.173 190 516 736 × 2 = 0 + 0.346 381 033 472;
  • 49) 0.346 381 033 472 × 2 = 0 + 0.692 762 066 944;
  • 50) 0.692 762 066 944 × 2 = 1 + 0.385 524 133 888;
  • 51) 0.385 524 133 888 × 2 = 0 + 0.771 048 267 776;
  • 52) 0.771 048 267 776 × 2 = 1 + 0.542 096 535 552;
  • 53) 0.542 096 535 552 × 2 = 1 + 0.084 193 071 104;
  • 54) 0.084 193 071 104 × 2 = 0 + 0.168 386 142 208;
  • 55) 0.168 386 142 208 × 2 = 0 + 0.336 772 284 416;
  • 56) 0.336 772 284 416 × 2 = 0 + 0.673 544 568 832;
  • 57) 0.673 544 568 832 × 2 = 1 + 0.347 089 137 664;
  • 58) 0.347 089 137 664 × 2 = 0 + 0.694 178 275 328;
  • 59) 0.694 178 275 328 × 2 = 1 + 0.388 356 550 656;
  • 60) 0.388 356 550 656 × 2 = 0 + 0.776 713 101 312;
  • 61) 0.776 713 101 312 × 2 = 1 + 0.553 426 202 624;
  • 62) 0.553 426 202 624 × 2 = 1 + 0.106 852 405 248;
  • 63) 0.106 852 405 248 × 2 = 0 + 0.213 704 810 496;
  • 64) 0.213 704 810 496 × 2 = 0 + 0.427 409 620 992;

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:


0.000 282 005 737(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 1010 1110 1110 0101 1000 1010 1100(2)

6. Positive number before normalization:

0.000 282 005 737(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 1010 1110 1110 0101 1000 1010 1100(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:


0.000 282 005 737(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 1010 1110 1110 0101 1000 1010 1100(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0110 1010 1110 1110 0101 1000 1010 1100(2) × 20 =

1.0010 0111 1011 0100 0101 0110 1010 1110 1110 0101 1000 1010 1100(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 0110 1010 1110 1110 0101 1000 1010 1100


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 0110 1010 1110 1110 0101 1000 1010 1100 =

0010 0111 1011 0100 0101 0110 1010 1110 1110 0101 1000 1010 1100


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 0110 1010 1110 1110 0101 1000 1010 1100


Decimal number -0.000 282 005 737 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0101 0110 1010 1110 1110 0101 1000 1010 1100

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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