-0.000 282 005 689 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 689(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
-0.000 282 005 689(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:

|-0.000 282 005 689| = 0.000 282 005 689


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: 0.000 282 005 689.

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) 0.000 282 005 689 × 2 = 0 + 0.000 564 011 378;
  • 2) 0.000 564 011 378 × 2 = 0 + 0.001 128 022 756;
  • 3) 0.001 128 022 756 × 2 = 0 + 0.002 256 045 512;
  • 4) 0.002 256 045 512 × 2 = 0 + 0.004 512 091 024;
  • 5) 0.004 512 091 024 × 2 = 0 + 0.009 024 182 048;
  • 6) 0.009 024 182 048 × 2 = 0 + 0.018 048 364 096;
  • 7) 0.018 048 364 096 × 2 = 0 + 0.036 096 728 192;
  • 8) 0.036 096 728 192 × 2 = 0 + 0.072 193 456 384;
  • 9) 0.072 193 456 384 × 2 = 0 + 0.144 386 912 768;
  • 10) 0.144 386 912 768 × 2 = 0 + 0.288 773 825 536;
  • 11) 0.288 773 825 536 × 2 = 0 + 0.577 547 651 072;
  • 12) 0.577 547 651 072 × 2 = 1 + 0.155 095 302 144;
  • 13) 0.155 095 302 144 × 2 = 0 + 0.310 190 604 288;
  • 14) 0.310 190 604 288 × 2 = 0 + 0.620 381 208 576;
  • 15) 0.620 381 208 576 × 2 = 1 + 0.240 762 417 152;
  • 16) 0.240 762 417 152 × 2 = 0 + 0.481 524 834 304;
  • 17) 0.481 524 834 304 × 2 = 0 + 0.963 049 668 608;
  • 18) 0.963 049 668 608 × 2 = 1 + 0.926 099 337 216;
  • 19) 0.926 099 337 216 × 2 = 1 + 0.852 198 674 432;
  • 20) 0.852 198 674 432 × 2 = 1 + 0.704 397 348 864;
  • 21) 0.704 397 348 864 × 2 = 1 + 0.408 794 697 728;
  • 22) 0.408 794 697 728 × 2 = 0 + 0.817 589 395 456;
  • 23) 0.817 589 395 456 × 2 = 1 + 0.635 178 790 912;
  • 24) 0.635 178 790 912 × 2 = 1 + 0.270 357 581 824;
  • 25) 0.270 357 581 824 × 2 = 0 + 0.540 715 163 648;
  • 26) 0.540 715 163 648 × 2 = 1 + 0.081 430 327 296;
  • 27) 0.081 430 327 296 × 2 = 0 + 0.162 860 654 592;
  • 28) 0.162 860 654 592 × 2 = 0 + 0.325 721 309 184;
  • 29) 0.325 721 309 184 × 2 = 0 + 0.651 442 618 368;
  • 30) 0.651 442 618 368 × 2 = 1 + 0.302 885 236 736;
  • 31) 0.302 885 236 736 × 2 = 0 + 0.605 770 473 472;
  • 32) 0.605 770 473 472 × 2 = 1 + 0.211 540 946 944;
  • 33) 0.211 540 946 944 × 2 = 0 + 0.423 081 893 888;
  • 34) 0.423 081 893 888 × 2 = 0 + 0.846 163 787 776;
  • 35) 0.846 163 787 776 × 2 = 1 + 0.692 327 575 552;
  • 36) 0.692 327 575 552 × 2 = 1 + 0.384 655 151 104;
  • 37) 0.384 655 151 104 × 2 = 0 + 0.769 310 302 208;
  • 38) 0.769 310 302 208 × 2 = 1 + 0.538 620 604 416;
  • 39) 0.538 620 604 416 × 2 = 1 + 0.077 241 208 832;
  • 40) 0.077 241 208 832 × 2 = 0 + 0.154 482 417 664;
  • 41) 0.154 482 417 664 × 2 = 0 + 0.308 964 835 328;
  • 42) 0.308 964 835 328 × 2 = 0 + 0.617 929 670 656;
  • 43) 0.617 929 670 656 × 2 = 1 + 0.235 859 341 312;
  • 44) 0.235 859 341 312 × 2 = 0 + 0.471 718 682 624;
  • 45) 0.471 718 682 624 × 2 = 0 + 0.943 437 365 248;
  • 46) 0.943 437 365 248 × 2 = 1 + 0.886 874 730 496;
  • 47) 0.886 874 730 496 × 2 = 1 + 0.773 749 460 992;
  • 48) 0.773 749 460 992 × 2 = 1 + 0.547 498 921 984;
  • 49) 0.547 498 921 984 × 2 = 1 + 0.094 997 843 968;
  • 50) 0.094 997 843 968 × 2 = 0 + 0.189 995 687 936;
  • 51) 0.189 995 687 936 × 2 = 0 + 0.379 991 375 872;
  • 52) 0.379 991 375 872 × 2 = 0 + 0.759 982 751 744;
  • 53) 0.759 982 751 744 × 2 = 1 + 0.519 965 503 488;
  • 54) 0.519 965 503 488 × 2 = 1 + 0.039 931 006 976;
  • 55) 0.039 931 006 976 × 2 = 0 + 0.079 862 013 952;
  • 56) 0.079 862 013 952 × 2 = 0 + 0.159 724 027 904;
  • 57) 0.159 724 027 904 × 2 = 0 + 0.319 448 055 808;
  • 58) 0.319 448 055 808 × 2 = 0 + 0.638 896 111 616;
  • 59) 0.638 896 111 616 × 2 = 1 + 0.277 792 223 232;
  • 60) 0.277 792 223 232 × 2 = 0 + 0.555 584 446 464;
  • 61) 0.555 584 446 464 × 2 = 1 + 0.111 168 892 928;
  • 62) 0.111 168 892 928 × 2 = 0 + 0.222 337 785 856;
  • 63) 0.222 337 785 856 × 2 = 0 + 0.444 675 571 712;
  • 64) 0.444 675 571 712 × 2 = 0 + 0.889 351 143 424;

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:


0.000 282 005 689(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0110 0010 0111 1000 1100 0010 1000(2)

6. Positive number before normalization:

0.000 282 005 689(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0110 0010 0111 1000 1100 0010 1000(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:


0.000 282 005 689(10) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0110 0010 0111 1000 1100 0010 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0101 0011 0110 0010 0111 1000 1100 0010 1000(2) × 20 =

1.0010 0111 1011 0100 0101 0011 0110 0010 0111 1000 1100 0010 1000(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 0011 0110 0010 0111 1000 1100 0010 1000


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 0011 0110 0010 0111 1000 1100 0010 1000 =

0010 0111 1011 0100 0101 0011 0110 0010 0111 1000 1100 0010 1000


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 0011 0110 0010 0111 1000 1100 0010 1000


Decimal number -0.000 282 005 689 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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