Decimal to 64 Bit IEEE 754 Binary: Convert Number -8 851 100 823 806 338 456 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -8 851 100 823 806 338 456(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-8 851 100 823 806 338 456| = 8 851 100 823 806 338 456


2. 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;
  • 8 851 100 823 806 338 456 ÷ 2 = 4 425 550 411 903 169 228 + 0;
  • 4 425 550 411 903 169 228 ÷ 2 = 2 212 775 205 951 584 614 + 0;
  • 2 212 775 205 951 584 614 ÷ 2 = 1 106 387 602 975 792 307 + 0;
  • 1 106 387 602 975 792 307 ÷ 2 = 553 193 801 487 896 153 + 1;
  • 553 193 801 487 896 153 ÷ 2 = 276 596 900 743 948 076 + 1;
  • 276 596 900 743 948 076 ÷ 2 = 138 298 450 371 974 038 + 0;
  • 138 298 450 371 974 038 ÷ 2 = 69 149 225 185 987 019 + 0;
  • 69 149 225 185 987 019 ÷ 2 = 34 574 612 592 993 509 + 1;
  • 34 574 612 592 993 509 ÷ 2 = 17 287 306 296 496 754 + 1;
  • 17 287 306 296 496 754 ÷ 2 = 8 643 653 148 248 377 + 0;
  • 8 643 653 148 248 377 ÷ 2 = 4 321 826 574 124 188 + 1;
  • 4 321 826 574 124 188 ÷ 2 = 2 160 913 287 062 094 + 0;
  • 2 160 913 287 062 094 ÷ 2 = 1 080 456 643 531 047 + 0;
  • 1 080 456 643 531 047 ÷ 2 = 540 228 321 765 523 + 1;
  • 540 228 321 765 523 ÷ 2 = 270 114 160 882 761 + 1;
  • 270 114 160 882 761 ÷ 2 = 135 057 080 441 380 + 1;
  • 135 057 080 441 380 ÷ 2 = 67 528 540 220 690 + 0;
  • 67 528 540 220 690 ÷ 2 = 33 764 270 110 345 + 0;
  • 33 764 270 110 345 ÷ 2 = 16 882 135 055 172 + 1;
  • 16 882 135 055 172 ÷ 2 = 8 441 067 527 586 + 0;
  • 8 441 067 527 586 ÷ 2 = 4 220 533 763 793 + 0;
  • 4 220 533 763 793 ÷ 2 = 2 110 266 881 896 + 1;
  • 2 110 266 881 896 ÷ 2 = 1 055 133 440 948 + 0;
  • 1 055 133 440 948 ÷ 2 = 527 566 720 474 + 0;
  • 527 566 720 474 ÷ 2 = 263 783 360 237 + 0;
  • 263 783 360 237 ÷ 2 = 131 891 680 118 + 1;
  • 131 891 680 118 ÷ 2 = 65 945 840 059 + 0;
  • 65 945 840 059 ÷ 2 = 32 972 920 029 + 1;
  • 32 972 920 029 ÷ 2 = 16 486 460 014 + 1;
  • 16 486 460 014 ÷ 2 = 8 243 230 007 + 0;
  • 8 243 230 007 ÷ 2 = 4 121 615 003 + 1;
  • 4 121 615 003 ÷ 2 = 2 060 807 501 + 1;
  • 2 060 807 501 ÷ 2 = 1 030 403 750 + 1;
  • 1 030 403 750 ÷ 2 = 515 201 875 + 0;
  • 515 201 875 ÷ 2 = 257 600 937 + 1;
  • 257 600 937 ÷ 2 = 128 800 468 + 1;
  • 128 800 468 ÷ 2 = 64 400 234 + 0;
  • 64 400 234 ÷ 2 = 32 200 117 + 0;
  • 32 200 117 ÷ 2 = 16 100 058 + 1;
  • 16 100 058 ÷ 2 = 8 050 029 + 0;
  • 8 050 029 ÷ 2 = 4 025 014 + 1;
  • 4 025 014 ÷ 2 = 2 012 507 + 0;
  • 2 012 507 ÷ 2 = 1 006 253 + 1;
  • 1 006 253 ÷ 2 = 503 126 + 1;
  • 503 126 ÷ 2 = 251 563 + 0;
  • 251 563 ÷ 2 = 125 781 + 1;
  • 125 781 ÷ 2 = 62 890 + 1;
  • 62 890 ÷ 2 = 31 445 + 0;
  • 31 445 ÷ 2 = 15 722 + 1;
  • 15 722 ÷ 2 = 7 861 + 0;
  • 7 861 ÷ 2 = 3 930 + 1;
  • 3 930 ÷ 2 = 1 965 + 0;
  • 1 965 ÷ 2 = 982 + 1;
  • 982 ÷ 2 = 491 + 0;
  • 491 ÷ 2 = 245 + 1;
  • 245 ÷ 2 = 122 + 1;
  • 122 ÷ 2 = 61 + 0;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the positive number.

Take all the remainders starting from the bottom of the list constructed above.

8 851 100 823 806 338 456(10) =


111 1010 1101 0101 0110 1101 0100 1101 1101 1010 0010 0100 1110 0101 1001 1000(2)


4. Normalize the binary representation of the number.

Shift the decimal mark 62 positions to the left, so that only one non zero digit remains to the left of it:


8 851 100 823 806 338 456(10) =


111 1010 1101 0101 0110 1101 0100 1101 1101 1010 0010 0100 1110 0101 1001 1000(2) =


111 1010 1101 0101 0110 1101 0100 1101 1101 1010 0010 0100 1110 0101 1001 1000(2) × 20 =


1.1110 1011 0101 0101 1011 0101 0011 0111 0110 1000 1001 0011 1001 0110 0110 00(2) × 262


5. 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): 62


Mantissa (not normalized):
1.1110 1011 0101 0101 1011 0101 0011 0111 0110 1000 1001 0011 1001 0110 0110 00


6. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


62 + 2(11-1) - 1 =


(62 + 1 023)(10) =


1 085(10)


7. 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;
  • 1 085 ÷ 2 = 542 + 1;
  • 542 ÷ 2 = 271 + 0;
  • 271 ÷ 2 = 135 + 1;
  • 135 ÷ 2 = 67 + 1;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

8. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1085(10) =


100 0011 1101(2)


9. 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. 1110 1011 0101 0101 1011 0101 0011 0111 0110 1000 1001 0011 1001 01 1001 1000 =


1110 1011 0101 0101 1011 0101 0011 0111 0110 1000 1001 0011 1001


10. 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 0011 1101


Mantissa (52 bits) =
1110 1011 0101 0101 1011 0101 0011 0111 0110 1000 1001 0011 1001


The base ten decimal number -8 851 100 823 806 338 456 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0011 1101 - 1110 1011 0101 0101 1011 0101 0011 0111 0110 1000 1001 0011 1001

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