0.000 043 771 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 043 771 4(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
0.000 043 771 4(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 043 771 4.

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) 0.000 043 771 4 × 2 = 0 + 0.000 087 542 8;
  • 2) 0.000 087 542 8 × 2 = 0 + 0.000 175 085 6;
  • 3) 0.000 175 085 6 × 2 = 0 + 0.000 350 171 2;
  • 4) 0.000 350 171 2 × 2 = 0 + 0.000 700 342 4;
  • 5) 0.000 700 342 4 × 2 = 0 + 0.001 400 684 8;
  • 6) 0.001 400 684 8 × 2 = 0 + 0.002 801 369 6;
  • 7) 0.002 801 369 6 × 2 = 0 + 0.005 602 739 2;
  • 8) 0.005 602 739 2 × 2 = 0 + 0.011 205 478 4;
  • 9) 0.011 205 478 4 × 2 = 0 + 0.022 410 956 8;
  • 10) 0.022 410 956 8 × 2 = 0 + 0.044 821 913 6;
  • 11) 0.044 821 913 6 × 2 = 0 + 0.089 643 827 2;
  • 12) 0.089 643 827 2 × 2 = 0 + 0.179 287 654 4;
  • 13) 0.179 287 654 4 × 2 = 0 + 0.358 575 308 8;
  • 14) 0.358 575 308 8 × 2 = 0 + 0.717 150 617 6;
  • 15) 0.717 150 617 6 × 2 = 1 + 0.434 301 235 2;
  • 16) 0.434 301 235 2 × 2 = 0 + 0.868 602 470 4;
  • 17) 0.868 602 470 4 × 2 = 1 + 0.737 204 940 8;
  • 18) 0.737 204 940 8 × 2 = 1 + 0.474 409 881 6;
  • 19) 0.474 409 881 6 × 2 = 0 + 0.948 819 763 2;
  • 20) 0.948 819 763 2 × 2 = 1 + 0.897 639 526 4;
  • 21) 0.897 639 526 4 × 2 = 1 + 0.795 279 052 8;
  • 22) 0.795 279 052 8 × 2 = 1 + 0.590 558 105 6;
  • 23) 0.590 558 105 6 × 2 = 1 + 0.181 116 211 2;
  • 24) 0.181 116 211 2 × 2 = 0 + 0.362 232 422 4;
  • 25) 0.362 232 422 4 × 2 = 0 + 0.724 464 844 8;
  • 26) 0.724 464 844 8 × 2 = 1 + 0.448 929 689 6;
  • 27) 0.448 929 689 6 × 2 = 0 + 0.897 859 379 2;
  • 28) 0.897 859 379 2 × 2 = 1 + 0.795 718 758 4;
  • 29) 0.795 718 758 4 × 2 = 1 + 0.591 437 516 8;
  • 30) 0.591 437 516 8 × 2 = 1 + 0.182 875 033 6;
  • 31) 0.182 875 033 6 × 2 = 0 + 0.365 750 067 2;
  • 32) 0.365 750 067 2 × 2 = 0 + 0.731 500 134 4;
  • 33) 0.731 500 134 4 × 2 = 1 + 0.463 000 268 8;
  • 34) 0.463 000 268 8 × 2 = 0 + 0.926 000 537 6;
  • 35) 0.926 000 537 6 × 2 = 1 + 0.852 001 075 2;
  • 36) 0.852 001 075 2 × 2 = 1 + 0.704 002 150 4;
  • 37) 0.704 002 150 4 × 2 = 1 + 0.408 004 300 8;
  • 38) 0.408 004 300 8 × 2 = 0 + 0.816 008 601 6;
  • 39) 0.816 008 601 6 × 2 = 1 + 0.632 017 203 2;
  • 40) 0.632 017 203 2 × 2 = 1 + 0.264 034 406 4;
  • 41) 0.264 034 406 4 × 2 = 0 + 0.528 068 812 8;
  • 42) 0.528 068 812 8 × 2 = 1 + 0.056 137 625 6;
  • 43) 0.056 137 625 6 × 2 = 0 + 0.112 275 251 2;
  • 44) 0.112 275 251 2 × 2 = 0 + 0.224 550 502 4;
  • 45) 0.224 550 502 4 × 2 = 0 + 0.449 101 004 8;
  • 46) 0.449 101 004 8 × 2 = 0 + 0.898 202 009 6;
  • 47) 0.898 202 009 6 × 2 = 1 + 0.796 404 019 2;
  • 48) 0.796 404 019 2 × 2 = 1 + 0.592 808 038 4;
  • 49) 0.592 808 038 4 × 2 = 1 + 0.185 616 076 8;
  • 50) 0.185 616 076 8 × 2 = 0 + 0.371 232 153 6;
  • 51) 0.371 232 153 6 × 2 = 0 + 0.742 464 307 2;
  • 52) 0.742 464 307 2 × 2 = 1 + 0.484 928 614 4;
  • 53) 0.484 928 614 4 × 2 = 0 + 0.969 857 228 8;
  • 54) 0.969 857 228 8 × 2 = 1 + 0.939 714 457 6;
  • 55) 0.939 714 457 6 × 2 = 1 + 0.879 428 915 2;
  • 56) 0.879 428 915 2 × 2 = 1 + 0.758 857 830 4;
  • 57) 0.758 857 830 4 × 2 = 1 + 0.517 715 660 8;
  • 58) 0.517 715 660 8 × 2 = 1 + 0.035 431 321 6;
  • 59) 0.035 431 321 6 × 2 = 0 + 0.070 862 643 2;
  • 60) 0.070 862 643 2 × 2 = 0 + 0.141 725 286 4;
  • 61) 0.141 725 286 4 × 2 = 0 + 0.283 450 572 8;
  • 62) 0.283 450 572 8 × 2 = 0 + 0.566 901 145 6;
  • 63) 0.566 901 145 6 × 2 = 1 + 0.133 802 291 2;
  • 64) 0.133 802 291 2 × 2 = 0 + 0.267 604 582 4;
  • 65) 0.267 604 582 4 × 2 = 0 + 0.535 209 164 8;
  • 66) 0.535 209 164 8 × 2 = 1 + 0.070 418 329 6;
  • 67) 0.070 418 329 6 × 2 = 0 + 0.140 836 659 2;

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).


4. 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:


0.000 043 771 4(10) =

0.0000 0000 0000 0010 1101 1110 0101 1100 1011 1011 0100 0011 1001 0111 1100 0010 010(2)

5. Positive number before normalization:

0.000 043 771 4(10) =

0.0000 0000 0000 0010 1101 1110 0101 1100 1011 1011 0100 0011 1001 0111 1100 0010 010(2)

6. 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:


0.000 043 771 4(10) =

0.0000 0000 0000 0010 1101 1110 0101 1100 1011 1011 0100 0011 1001 0111 1100 0010 010(2) =

0.0000 0000 0000 0010 1101 1110 0101 1100 1011 1011 0100 0011 1001 0111 1100 0010 010(2) × 20 =

1.0110 1111 0010 1110 0101 1101 1010 0001 1100 1011 1110 0001 0010(2) × 2-15


7. 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 0 (a positive number)

Exponent (unadjusted): -15

Mantissa (not normalized):
1.0110 1111 0010 1110 0101 1101 1010 0001 1100 1011 1110 0001 0010


8. 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)


9. 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;

10. 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)


11. 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. 0110 1111 0010 1110 0101 1101 1010 0001 1100 1011 1110 0001 0010 =

0110 1111 0010 1110 0101 1101 1010 0001 1100 1011 1110 0001 0010


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 0000

Mantissa (52 bits) =
0110 1111 0010 1110 0101 1101 1010 0001 1100 1011 1110 0001 0010


Decimal number 0.000 043 771 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0000 - 0110 1111 0010 1110 0101 1101 1010 0001 1100 1011 1110 0001 0010

» Convert decimal 0.000 043 761 4 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