Convert -442.034 455 41 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

-442.034 455 41(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. Start with the positive version of the number:

|-442.034 455 41| = 442.034 455 41

2. First, convert to the binary (base 2) the integer part: 442.
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;
  • 442 ÷ 2 = 221 + 0;
  • 221 ÷ 2 = 110 + 1;
  • 110 ÷ 2 = 55 + 0;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.


442(10) =


1 1011 1010(2)


4. Convert to the binary (base 2) the fractional part: 0.034 455 41.

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) 0.034 455 41 × 2 = 0 + 0.068 910 82;
  • 2) 0.068 910 82 × 2 = 0 + 0.137 821 64;
  • 3) 0.137 821 64 × 2 = 0 + 0.275 643 28;
  • 4) 0.275 643 28 × 2 = 0 + 0.551 286 56;
  • 5) 0.551 286 56 × 2 = 1 + 0.102 573 12;
  • 6) 0.102 573 12 × 2 = 0 + 0.205 146 24;
  • 7) 0.205 146 24 × 2 = 0 + 0.410 292 48;
  • 8) 0.410 292 48 × 2 = 0 + 0.820 584 96;
  • 9) 0.820 584 96 × 2 = 1 + 0.641 169 92;
  • 10) 0.641 169 92 × 2 = 1 + 0.282 339 84;
  • 11) 0.282 339 84 × 2 = 0 + 0.564 679 68;
  • 12) 0.564 679 68 × 2 = 1 + 0.129 359 36;
  • 13) 0.129 359 36 × 2 = 0 + 0.258 718 72;
  • 14) 0.258 718 72 × 2 = 0 + 0.517 437 44;
  • 15) 0.517 437 44 × 2 = 1 + 0.034 874 88;
  • 16) 0.034 874 88 × 2 = 0 + 0.069 749 76;
  • 17) 0.069 749 76 × 2 = 0 + 0.139 499 52;
  • 18) 0.139 499 52 × 2 = 0 + 0.278 999 04;
  • 19) 0.278 999 04 × 2 = 0 + 0.557 998 08;
  • 20) 0.557 998 08 × 2 = 1 + 0.115 996 16;
  • 21) 0.115 996 16 × 2 = 0 + 0.231 992 32;
  • 22) 0.231 992 32 × 2 = 0 + 0.463 984 64;
  • 23) 0.463 984 64 × 2 = 0 + 0.927 969 28;
  • 24) 0.927 969 28 × 2 = 1 + 0.855 938 56;
  • 25) 0.855 938 56 × 2 = 1 + 0.711 877 12;
  • 26) 0.711 877 12 × 2 = 1 + 0.423 754 24;
  • 27) 0.423 754 24 × 2 = 0 + 0.847 508 48;
  • 28) 0.847 508 48 × 2 = 1 + 0.695 016 96;
  • 29) 0.695 016 96 × 2 = 1 + 0.390 033 92;
  • 30) 0.390 033 92 × 2 = 0 + 0.780 067 84;
  • 31) 0.780 067 84 × 2 = 1 + 0.560 135 68;
  • 32) 0.560 135 68 × 2 = 1 + 0.120 271 36;
  • 33) 0.120 271 36 × 2 = 0 + 0.240 542 72;
  • 34) 0.240 542 72 × 2 = 0 + 0.481 085 44;
  • 35) 0.481 085 44 × 2 = 0 + 0.962 170 88;
  • 36) 0.962 170 88 × 2 = 1 + 0.924 341 76;
  • 37) 0.924 341 76 × 2 = 1 + 0.848 683 52;
  • 38) 0.848 683 52 × 2 = 1 + 0.697 367 04;
  • 39) 0.697 367 04 × 2 = 1 + 0.394 734 08;
  • 40) 0.394 734 08 × 2 = 0 + 0.789 468 16;
  • 41) 0.789 468 16 × 2 = 1 + 0.578 936 32;
  • 42) 0.578 936 32 × 2 = 1 + 0.157 872 64;
  • 43) 0.157 872 64 × 2 = 0 + 0.315 745 28;
  • 44) 0.315 745 28 × 2 = 0 + 0.631 490 56;
  • 45) 0.631 490 56 × 2 = 1 + 0.262 981 12;
  • 46) 0.262 981 12 × 2 = 0 + 0.525 962 24;
  • 47) 0.525 962 24 × 2 = 1 + 0.051 924 48;
  • 48) 0.051 924 48 × 2 = 0 + 0.103 848 96;
  • 49) 0.103 848 96 × 2 = 0 + 0.207 697 92;
  • 50) 0.207 697 92 × 2 = 0 + 0.415 395 84;
  • 51) 0.415 395 84 × 2 = 0 + 0.830 791 68;
  • 52) 0.830 791 68 × 2 = 1 + 0.661 583 36;
  • 53) 0.661 583 36 × 2 = 1 + 0.323 166 72;

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...)


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:


0.034 455 41(10) =


0.0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100 1010 0001 1(2)


6. Positive number before normalization:

442.034 455 41(10) =


1 1011 1010.0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100 1010 0001 1(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 8 positions to the left so that only one non zero digit remains to the left of it:


442.034 455 41(10) =


1 1011 1010.0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100 1010 0001 1(2) =


1 1011 1010.0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100 1010 0001 1(2) × 20 =


1.1011 1010 0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100 1010 0001 1(2) × 28


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): 8


Mantissa (not normalized):
1.1011 1010 0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100 1010 0001 1


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


8 + 2(11-1) - 1 =


(8 + 1 023)(10) =


1 031(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;
  • 1 031 ÷ 2 = 515 + 1;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


1031(10) =


100 0000 0111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 1011 1010 0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100 1 0100 0011 =


1011 1010 0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100


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) =
100 0000 0111


Mantissa (52 bits) =
1011 1010 0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100


Number -442.034 455 41 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
1 - 100 0000 0111 - 1011 1010 0000 1000 1101 0010 0001 0001 1101 1011 0001 1110 1100

(64 bits IEEE 754)

More operations of this kind:

-442.034 455 42 = ? ... -442.034 455 4 = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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