-0.000 282 005 471 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 471(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
-0.000 282 005 471(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:

|-0.000 282 005 471| = 0.000 282 005 471


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: 0.000 282 005 471.

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) 0.000 282 005 471 × 2 = 0 + 0.000 564 010 942;
  • 2) 0.000 564 010 942 × 2 = 0 + 0.001 128 021 884;
  • 3) 0.001 128 021 884 × 2 = 0 + 0.002 256 043 768;
  • 4) 0.002 256 043 768 × 2 = 0 + 0.004 512 087 536;
  • 5) 0.004 512 087 536 × 2 = 0 + 0.009 024 175 072;
  • 6) 0.009 024 175 072 × 2 = 0 + 0.018 048 350 144;
  • 7) 0.018 048 350 144 × 2 = 0 + 0.036 096 700 288;
  • 8) 0.036 096 700 288 × 2 = 0 + 0.072 193 400 576;
  • 9) 0.072 193 400 576 × 2 = 0 + 0.144 386 801 152;
  • 10) 0.144 386 801 152 × 2 = 0 + 0.288 773 602 304;
  • 11) 0.288 773 602 304 × 2 = 0 + 0.577 547 204 608;
  • 12) 0.577 547 204 608 × 2 = 1 + 0.155 094 409 216;
  • 13) 0.155 094 409 216 × 2 = 0 + 0.310 188 818 432;
  • 14) 0.310 188 818 432 × 2 = 0 + 0.620 377 636 864;
  • 15) 0.620 377 636 864 × 2 = 1 + 0.240 755 273 728;
  • 16) 0.240 755 273 728 × 2 = 0 + 0.481 510 547 456;
  • 17) 0.481 510 547 456 × 2 = 0 + 0.963 021 094 912;
  • 18) 0.963 021 094 912 × 2 = 1 + 0.926 042 189 824;
  • 19) 0.926 042 189 824 × 2 = 1 + 0.852 084 379 648;
  • 20) 0.852 084 379 648 × 2 = 1 + 0.704 168 759 296;
  • 21) 0.704 168 759 296 × 2 = 1 + 0.408 337 518 592;
  • 22) 0.408 337 518 592 × 2 = 0 + 0.816 675 037 184;
  • 23) 0.816 675 037 184 × 2 = 1 + 0.633 350 074 368;
  • 24) 0.633 350 074 368 × 2 = 1 + 0.266 700 148 736;
  • 25) 0.266 700 148 736 × 2 = 0 + 0.533 400 297 472;
  • 26) 0.533 400 297 472 × 2 = 1 + 0.066 800 594 944;
  • 27) 0.066 800 594 944 × 2 = 0 + 0.133 601 189 888;
  • 28) 0.133 601 189 888 × 2 = 0 + 0.267 202 379 776;
  • 29) 0.267 202 379 776 × 2 = 0 + 0.534 404 759 552;
  • 30) 0.534 404 759 552 × 2 = 1 + 0.068 809 519 104;
  • 31) 0.068 809 519 104 × 2 = 0 + 0.137 619 038 208;
  • 32) 0.137 619 038 208 × 2 = 0 + 0.275 238 076 416;
  • 33) 0.275 238 076 416 × 2 = 0 + 0.550 476 152 832;
  • 34) 0.550 476 152 832 × 2 = 1 + 0.100 952 305 664;
  • 35) 0.100 952 305 664 × 2 = 0 + 0.201 904 611 328;
  • 36) 0.201 904 611 328 × 2 = 0 + 0.403 809 222 656;
  • 37) 0.403 809 222 656 × 2 = 0 + 0.807 618 445 312;
  • 38) 0.807 618 445 312 × 2 = 1 + 0.615 236 890 624;
  • 39) 0.615 236 890 624 × 2 = 1 + 0.230 473 781 248;
  • 40) 0.230 473 781 248 × 2 = 0 + 0.460 947 562 496;
  • 41) 0.460 947 562 496 × 2 = 0 + 0.921 895 124 992;
  • 42) 0.921 895 124 992 × 2 = 1 + 0.843 790 249 984;
  • 43) 0.843 790 249 984 × 2 = 1 + 0.687 580 499 968;
  • 44) 0.687 580 499 968 × 2 = 1 + 0.375 160 999 936;
  • 45) 0.375 160 999 936 × 2 = 0 + 0.750 321 999 872;
  • 46) 0.750 321 999 872 × 2 = 1 + 0.500 643 999 744;
  • 47) 0.500 643 999 744 × 2 = 1 + 0.001 287 999 488;
  • 48) 0.001 287 999 488 × 2 = 0 + 0.002 575 998 976;
  • 49) 0.002 575 998 976 × 2 = 0 + 0.005 151 997 952;
  • 50) 0.005 151 997 952 × 2 = 0 + 0.010 303 995 904;
  • 51) 0.010 303 995 904 × 2 = 0 + 0.020 607 991 808;
  • 52) 0.020 607 991 808 × 2 = 0 + 0.041 215 983 616;
  • 53) 0.041 215 983 616 × 2 = 0 + 0.082 431 967 232;
  • 54) 0.082 431 967 232 × 2 = 0 + 0.164 863 934 464;
  • 55) 0.164 863 934 464 × 2 = 0 + 0.329 727 868 928;
  • 56) 0.329 727 868 928 × 2 = 0 + 0.659 455 737 856;
  • 57) 0.659 455 737 856 × 2 = 1 + 0.318 911 475 712;
  • 58) 0.318 911 475 712 × 2 = 0 + 0.637 822 951 424;
  • 59) 0.637 822 951 424 × 2 = 1 + 0.275 645 902 848;
  • 60) 0.275 645 902 848 × 2 = 0 + 0.551 291 805 696;
  • 61) 0.551 291 805 696 × 2 = 1 + 0.102 583 611 392;
  • 62) 0.102 583 611 392 × 2 = 0 + 0.205 167 222 784;
  • 63) 0.205 167 222 784 × 2 = 0 + 0.410 334 445 568;
  • 64) 0.410 334 445 568 × 2 = 0 + 0.820 668 891 136;

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:


0.000 282 005 471(10) =

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

6. Positive number before normalization:

0.000 282 005 471(10) =

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


0.000 282 005 471(10) =

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

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

1.0010 0111 1011 0100 0100 0100 0110 0111 0110 0000 0000 1010 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 0100 0100 0110 0111 0110 0000 0000 1010 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 0100 0100 0110 0111 0110 0000 0000 1010 1000 =

0010 0111 1011 0100 0100 0100 0110 0111 0110 0000 0000 1010 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 0100 0100 0110 0111 0110 0000 0000 1010 1000


Decimal number -0.000 282 005 471 converted to 64 bit double precision IEEE 754 binary floating point representation:

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