-0.000 282 408 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 408(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
-0.000 282 408(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:

|-0.000 282 408| = 0.000 282 408


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: 0.000 282 408.

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) 0.000 282 408 × 2 = 0 + 0.000 564 816;
  • 2) 0.000 564 816 × 2 = 0 + 0.001 129 632;
  • 3) 0.001 129 632 × 2 = 0 + 0.002 259 264;
  • 4) 0.002 259 264 × 2 = 0 + 0.004 518 528;
  • 5) 0.004 518 528 × 2 = 0 + 0.009 037 056;
  • 6) 0.009 037 056 × 2 = 0 + 0.018 074 112;
  • 7) 0.018 074 112 × 2 = 0 + 0.036 148 224;
  • 8) 0.036 148 224 × 2 = 0 + 0.072 296 448;
  • 9) 0.072 296 448 × 2 = 0 + 0.144 592 896;
  • 10) 0.144 592 896 × 2 = 0 + 0.289 185 792;
  • 11) 0.289 185 792 × 2 = 0 + 0.578 371 584;
  • 12) 0.578 371 584 × 2 = 1 + 0.156 743 168;
  • 13) 0.156 743 168 × 2 = 0 + 0.313 486 336;
  • 14) 0.313 486 336 × 2 = 0 + 0.626 972 672;
  • 15) 0.626 972 672 × 2 = 1 + 0.253 945 344;
  • 16) 0.253 945 344 × 2 = 0 + 0.507 890 688;
  • 17) 0.507 890 688 × 2 = 1 + 0.015 781 376;
  • 18) 0.015 781 376 × 2 = 0 + 0.031 562 752;
  • 19) 0.031 562 752 × 2 = 0 + 0.063 125 504;
  • 20) 0.063 125 504 × 2 = 0 + 0.126 251 008;
  • 21) 0.126 251 008 × 2 = 0 + 0.252 502 016;
  • 22) 0.252 502 016 × 2 = 0 + 0.505 004 032;
  • 23) 0.505 004 032 × 2 = 1 + 0.010 008 064;
  • 24) 0.010 008 064 × 2 = 0 + 0.020 016 128;
  • 25) 0.020 016 128 × 2 = 0 + 0.040 032 256;
  • 26) 0.040 032 256 × 2 = 0 + 0.080 064 512;
  • 27) 0.080 064 512 × 2 = 0 + 0.160 129 024;
  • 28) 0.160 129 024 × 2 = 0 + 0.320 258 048;
  • 29) 0.320 258 048 × 2 = 0 + 0.640 516 096;
  • 30) 0.640 516 096 × 2 = 1 + 0.281 032 192;
  • 31) 0.281 032 192 × 2 = 0 + 0.562 064 384;
  • 32) 0.562 064 384 × 2 = 1 + 0.124 128 768;
  • 33) 0.124 128 768 × 2 = 0 + 0.248 257 536;
  • 34) 0.248 257 536 × 2 = 0 + 0.496 515 072;
  • 35) 0.496 515 072 × 2 = 0 + 0.993 030 144;
  • 36) 0.993 030 144 × 2 = 1 + 0.986 060 288;
  • 37) 0.986 060 288 × 2 = 1 + 0.972 120 576;
  • 38) 0.972 120 576 × 2 = 1 + 0.944 241 152;
  • 39) 0.944 241 152 × 2 = 1 + 0.888 482 304;
  • 40) 0.888 482 304 × 2 = 1 + 0.776 964 608;
  • 41) 0.776 964 608 × 2 = 1 + 0.553 929 216;
  • 42) 0.553 929 216 × 2 = 1 + 0.107 858 432;
  • 43) 0.107 858 432 × 2 = 0 + 0.215 716 864;
  • 44) 0.215 716 864 × 2 = 0 + 0.431 433 728;
  • 45) 0.431 433 728 × 2 = 0 + 0.862 867 456;
  • 46) 0.862 867 456 × 2 = 1 + 0.725 734 912;
  • 47) 0.725 734 912 × 2 = 1 + 0.451 469 824;
  • 48) 0.451 469 824 × 2 = 0 + 0.902 939 648;
  • 49) 0.902 939 648 × 2 = 1 + 0.805 879 296;
  • 50) 0.805 879 296 × 2 = 1 + 0.611 758 592;
  • 51) 0.611 758 592 × 2 = 1 + 0.223 517 184;
  • 52) 0.223 517 184 × 2 = 0 + 0.447 034 368;
  • 53) 0.447 034 368 × 2 = 0 + 0.894 068 736;
  • 54) 0.894 068 736 × 2 = 1 + 0.788 137 472;
  • 55) 0.788 137 472 × 2 = 1 + 0.576 274 944;
  • 56) 0.576 274 944 × 2 = 1 + 0.152 549 888;
  • 57) 0.152 549 888 × 2 = 0 + 0.305 099 776;
  • 58) 0.305 099 776 × 2 = 0 + 0.610 199 552;
  • 59) 0.610 199 552 × 2 = 1 + 0.220 399 104;
  • 60) 0.220 399 104 × 2 = 0 + 0.440 798 208;
  • 61) 0.440 798 208 × 2 = 0 + 0.881 596 416;
  • 62) 0.881 596 416 × 2 = 1 + 0.763 192 832;
  • 63) 0.763 192 832 × 2 = 1 + 0.526 385 664;
  • 64) 0.526 385 664 × 2 = 1 + 0.052 771 328;

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:


0.000 282 408(10) =

0.0000 0000 0001 0010 1000 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111(2)

6. Positive number before normalization:

0.000 282 408(10) =

0.0000 0000 0001 0010 1000 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111(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:


0.000 282 408(10) =

0.0000 0000 0001 0010 1000 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111(2) =

0.0000 0000 0001 0010 1000 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111(2) × 20 =

1.0010 1000 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111(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 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111


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 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111 =

0010 1000 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111


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 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111


Decimal number -0.000 282 408 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0010 0000 0101 0001 1111 1100 0110 1110 0111 0010 0111

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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