-0.000 038 124 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 038 124(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
-0.000 038 124(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:

|-0.000 038 124| = 0.000 038 124


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: 0.000 038 124.

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) 0.000 038 124 × 2 = 0 + 0.000 076 248;
  • 2) 0.000 076 248 × 2 = 0 + 0.000 152 496;
  • 3) 0.000 152 496 × 2 = 0 + 0.000 304 992;
  • 4) 0.000 304 992 × 2 = 0 + 0.000 609 984;
  • 5) 0.000 609 984 × 2 = 0 + 0.001 219 968;
  • 6) 0.001 219 968 × 2 = 0 + 0.002 439 936;
  • 7) 0.002 439 936 × 2 = 0 + 0.004 879 872;
  • 8) 0.004 879 872 × 2 = 0 + 0.009 759 744;
  • 9) 0.009 759 744 × 2 = 0 + 0.019 519 488;
  • 10) 0.019 519 488 × 2 = 0 + 0.039 038 976;
  • 11) 0.039 038 976 × 2 = 0 + 0.078 077 952;
  • 12) 0.078 077 952 × 2 = 0 + 0.156 155 904;
  • 13) 0.156 155 904 × 2 = 0 + 0.312 311 808;
  • 14) 0.312 311 808 × 2 = 0 + 0.624 623 616;
  • 15) 0.624 623 616 × 2 = 1 + 0.249 247 232;
  • 16) 0.249 247 232 × 2 = 0 + 0.498 494 464;
  • 17) 0.498 494 464 × 2 = 0 + 0.996 988 928;
  • 18) 0.996 988 928 × 2 = 1 + 0.993 977 856;
  • 19) 0.993 977 856 × 2 = 1 + 0.987 955 712;
  • 20) 0.987 955 712 × 2 = 1 + 0.975 911 424;
  • 21) 0.975 911 424 × 2 = 1 + 0.951 822 848;
  • 22) 0.951 822 848 × 2 = 1 + 0.903 645 696;
  • 23) 0.903 645 696 × 2 = 1 + 0.807 291 392;
  • 24) 0.807 291 392 × 2 = 1 + 0.614 582 784;
  • 25) 0.614 582 784 × 2 = 1 + 0.229 165 568;
  • 26) 0.229 165 568 × 2 = 0 + 0.458 331 136;
  • 27) 0.458 331 136 × 2 = 0 + 0.916 662 272;
  • 28) 0.916 662 272 × 2 = 1 + 0.833 324 544;
  • 29) 0.833 324 544 × 2 = 1 + 0.666 649 088;
  • 30) 0.666 649 088 × 2 = 1 + 0.333 298 176;
  • 31) 0.333 298 176 × 2 = 0 + 0.666 596 352;
  • 32) 0.666 596 352 × 2 = 1 + 0.333 192 704;
  • 33) 0.333 192 704 × 2 = 0 + 0.666 385 408;
  • 34) 0.666 385 408 × 2 = 1 + 0.332 770 816;
  • 35) 0.332 770 816 × 2 = 0 + 0.665 541 632;
  • 36) 0.665 541 632 × 2 = 1 + 0.331 083 264;
  • 37) 0.331 083 264 × 2 = 0 + 0.662 166 528;
  • 38) 0.662 166 528 × 2 = 1 + 0.324 333 056;
  • 39) 0.324 333 056 × 2 = 0 + 0.648 666 112;
  • 40) 0.648 666 112 × 2 = 1 + 0.297 332 224;
  • 41) 0.297 332 224 × 2 = 0 + 0.594 664 448;
  • 42) 0.594 664 448 × 2 = 1 + 0.189 328 896;
  • 43) 0.189 328 896 × 2 = 0 + 0.378 657 792;
  • 44) 0.378 657 792 × 2 = 0 + 0.757 315 584;
  • 45) 0.757 315 584 × 2 = 1 + 0.514 631 168;
  • 46) 0.514 631 168 × 2 = 1 + 0.029 262 336;
  • 47) 0.029 262 336 × 2 = 0 + 0.058 524 672;
  • 48) 0.058 524 672 × 2 = 0 + 0.117 049 344;
  • 49) 0.117 049 344 × 2 = 0 + 0.234 098 688;
  • 50) 0.234 098 688 × 2 = 0 + 0.468 197 376;
  • 51) 0.468 197 376 × 2 = 0 + 0.936 394 752;
  • 52) 0.936 394 752 × 2 = 1 + 0.872 789 504;
  • 53) 0.872 789 504 × 2 = 1 + 0.745 579 008;
  • 54) 0.745 579 008 × 2 = 1 + 0.491 158 016;
  • 55) 0.491 158 016 × 2 = 0 + 0.982 316 032;
  • 56) 0.982 316 032 × 2 = 1 + 0.964 632 064;
  • 57) 0.964 632 064 × 2 = 1 + 0.929 264 128;
  • 58) 0.929 264 128 × 2 = 1 + 0.858 528 256;
  • 59) 0.858 528 256 × 2 = 1 + 0.717 056 512;
  • 60) 0.717 056 512 × 2 = 1 + 0.434 113 024;
  • 61) 0.434 113 024 × 2 = 0 + 0.868 226 048;
  • 62) 0.868 226 048 × 2 = 1 + 0.736 452 096;
  • 63) 0.736 452 096 × 2 = 1 + 0.472 904 192;
  • 64) 0.472 904 192 × 2 = 0 + 0.945 808 384;
  • 65) 0.945 808 384 × 2 = 1 + 0.891 616 768;
  • 66) 0.891 616 768 × 2 = 1 + 0.783 233 536;
  • 67) 0.783 233 536 × 2 = 1 + 0.566 467 072;

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:


0.000 038 124(10) =

0.0000 0000 0000 0010 0111 1111 1001 1101 0101 0101 0100 1100 0001 1101 1111 0110 111(2)

6. Positive number before normalization:

0.000 038 124(10) =

0.0000 0000 0000 0010 0111 1111 1001 1101 0101 0101 0100 1100 0001 1101 1111 0110 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 038 124(10) =

0.0000 0000 0000 0010 0111 1111 1001 1101 0101 0101 0100 1100 0001 1101 1111 0110 111(2) =

0.0000 0000 0000 0010 0111 1111 1001 1101 0101 0101 0100 1100 0001 1101 1111 0110 111(2) × 20 =

1.0011 1111 1100 1110 1010 1010 1010 0110 0000 1110 1111 1011 0111(2) × 2-15


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): -15

Mantissa (not normalized):
1.0011 1111 1100 1110 1010 1010 1010 0110 0000 1110 1111 1011 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-15 + 2(11-1) - 1 =

(-15 + 1 023)(10) =

1 008(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;
  • 1 008 ÷ 2 = 504 + 0;
  • 504 ÷ 2 = 252 + 0;
  • 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) =

1008(10) =

011 1111 0000(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. 0011 1111 1100 1110 1010 1010 1010 0110 0000 1110 1111 1011 0111 =

0011 1111 1100 1110 1010 1010 1010 0110 0000 1110 1111 1011 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 0000

Mantissa (52 bits) =
0011 1111 1100 1110 1010 1010 1010 0110 0000 1110 1111 1011 0111


Decimal number -0.000 038 124 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0000 - 0011 1111 1100 1110 1010 1010 1010 0110 0000 1110 1111 1011 0111

» Convert decimal -0.000 038 215 to 64 bit double precision IEEE 754 binary floating point representation standard

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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