-0.000 282 646 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 646(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
-0.000 282 646(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:

|-0.000 282 646| = 0.000 282 646


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: 0.000 282 646.

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) 0.000 282 646 × 2 = 0 + 0.000 565 292;
  • 2) 0.000 565 292 × 2 = 0 + 0.001 130 584;
  • 3) 0.001 130 584 × 2 = 0 + 0.002 261 168;
  • 4) 0.002 261 168 × 2 = 0 + 0.004 522 336;
  • 5) 0.004 522 336 × 2 = 0 + 0.009 044 672;
  • 6) 0.009 044 672 × 2 = 0 + 0.018 089 344;
  • 7) 0.018 089 344 × 2 = 0 + 0.036 178 688;
  • 8) 0.036 178 688 × 2 = 0 + 0.072 357 376;
  • 9) 0.072 357 376 × 2 = 0 + 0.144 714 752;
  • 10) 0.144 714 752 × 2 = 0 + 0.289 429 504;
  • 11) 0.289 429 504 × 2 = 0 + 0.578 859 008;
  • 12) 0.578 859 008 × 2 = 1 + 0.157 718 016;
  • 13) 0.157 718 016 × 2 = 0 + 0.315 436 032;
  • 14) 0.315 436 032 × 2 = 0 + 0.630 872 064;
  • 15) 0.630 872 064 × 2 = 1 + 0.261 744 128;
  • 16) 0.261 744 128 × 2 = 0 + 0.523 488 256;
  • 17) 0.523 488 256 × 2 = 1 + 0.046 976 512;
  • 18) 0.046 976 512 × 2 = 0 + 0.093 953 024;
  • 19) 0.093 953 024 × 2 = 0 + 0.187 906 048;
  • 20) 0.187 906 048 × 2 = 0 + 0.375 812 096;
  • 21) 0.375 812 096 × 2 = 0 + 0.751 624 192;
  • 22) 0.751 624 192 × 2 = 1 + 0.503 248 384;
  • 23) 0.503 248 384 × 2 = 1 + 0.006 496 768;
  • 24) 0.006 496 768 × 2 = 0 + 0.012 993 536;
  • 25) 0.012 993 536 × 2 = 0 + 0.025 987 072;
  • 26) 0.025 987 072 × 2 = 0 + 0.051 974 144;
  • 27) 0.051 974 144 × 2 = 0 + 0.103 948 288;
  • 28) 0.103 948 288 × 2 = 0 + 0.207 896 576;
  • 29) 0.207 896 576 × 2 = 0 + 0.415 793 152;
  • 30) 0.415 793 152 × 2 = 0 + 0.831 586 304;
  • 31) 0.831 586 304 × 2 = 1 + 0.663 172 608;
  • 32) 0.663 172 608 × 2 = 1 + 0.326 345 216;
  • 33) 0.326 345 216 × 2 = 0 + 0.652 690 432;
  • 34) 0.652 690 432 × 2 = 1 + 0.305 380 864;
  • 35) 0.305 380 864 × 2 = 0 + 0.610 761 728;
  • 36) 0.610 761 728 × 2 = 1 + 0.221 523 456;
  • 37) 0.221 523 456 × 2 = 0 + 0.443 046 912;
  • 38) 0.443 046 912 × 2 = 0 + 0.886 093 824;
  • 39) 0.886 093 824 × 2 = 1 + 0.772 187 648;
  • 40) 0.772 187 648 × 2 = 1 + 0.544 375 296;
  • 41) 0.544 375 296 × 2 = 1 + 0.088 750 592;
  • 42) 0.088 750 592 × 2 = 0 + 0.177 501 184;
  • 43) 0.177 501 184 × 2 = 0 + 0.355 002 368;
  • 44) 0.355 002 368 × 2 = 0 + 0.710 004 736;
  • 45) 0.710 004 736 × 2 = 1 + 0.420 009 472;
  • 46) 0.420 009 472 × 2 = 0 + 0.840 018 944;
  • 47) 0.840 018 944 × 2 = 1 + 0.680 037 888;
  • 48) 0.680 037 888 × 2 = 1 + 0.360 075 776;
  • 49) 0.360 075 776 × 2 = 0 + 0.720 151 552;
  • 50) 0.720 151 552 × 2 = 1 + 0.440 303 104;
  • 51) 0.440 303 104 × 2 = 0 + 0.880 606 208;
  • 52) 0.880 606 208 × 2 = 1 + 0.761 212 416;
  • 53) 0.761 212 416 × 2 = 1 + 0.522 424 832;
  • 54) 0.522 424 832 × 2 = 1 + 0.044 849 664;
  • 55) 0.044 849 664 × 2 = 0 + 0.089 699 328;
  • 56) 0.089 699 328 × 2 = 0 + 0.179 398 656;
  • 57) 0.179 398 656 × 2 = 0 + 0.358 797 312;
  • 58) 0.358 797 312 × 2 = 0 + 0.717 594 624;
  • 59) 0.717 594 624 × 2 = 1 + 0.435 189 248;
  • 60) 0.435 189 248 × 2 = 0 + 0.870 378 496;
  • 61) 0.870 378 496 × 2 = 1 + 0.740 756 992;
  • 62) 0.740 756 992 × 2 = 1 + 0.481 513 984;
  • 63) 0.481 513 984 × 2 = 0 + 0.963 027 968;
  • 64) 0.963 027 968 × 2 = 1 + 0.926 055 936;

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:


0.000 282 646(10) =

0.0000 0000 0001 0010 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101(2)

6. Positive number before normalization:

0.000 282 646(10) =

0.0000 0000 0001 0010 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101(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:


0.000 282 646(10) =

0.0000 0000 0001 0010 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101(2) =

0.0000 0000 0001 0010 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101(2) × 20 =

1.0010 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101(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 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101


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 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101 =

0010 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101


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 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101


Decimal number -0.000 282 646 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0110 0000 0011 0101 0011 1000 1011 0101 1100 0010 1101

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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