-0.841 470 986 917 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.841 470 986 917 6(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
-0.841 470 986 917 6(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:

|-0.841 470 986 917 6| = 0.841 470 986 917 6


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: 0.841 470 986 917 6.

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) 0.841 470 986 917 6 × 2 = 1 + 0.682 941 973 835 2;
  • 2) 0.682 941 973 835 2 × 2 = 1 + 0.365 883 947 670 4;
  • 3) 0.365 883 947 670 4 × 2 = 0 + 0.731 767 895 340 8;
  • 4) 0.731 767 895 340 8 × 2 = 1 + 0.463 535 790 681 6;
  • 5) 0.463 535 790 681 6 × 2 = 0 + 0.927 071 581 363 2;
  • 6) 0.927 071 581 363 2 × 2 = 1 + 0.854 143 162 726 4;
  • 7) 0.854 143 162 726 4 × 2 = 1 + 0.708 286 325 452 8;
  • 8) 0.708 286 325 452 8 × 2 = 1 + 0.416 572 650 905 6;
  • 9) 0.416 572 650 905 6 × 2 = 0 + 0.833 145 301 811 2;
  • 10) 0.833 145 301 811 2 × 2 = 1 + 0.666 290 603 622 4;
  • 11) 0.666 290 603 622 4 × 2 = 1 + 0.332 581 207 244 8;
  • 12) 0.332 581 207 244 8 × 2 = 0 + 0.665 162 414 489 6;
  • 13) 0.665 162 414 489 6 × 2 = 1 + 0.330 324 828 979 2;
  • 14) 0.330 324 828 979 2 × 2 = 0 + 0.660 649 657 958 4;
  • 15) 0.660 649 657 958 4 × 2 = 1 + 0.321 299 315 916 8;
  • 16) 0.321 299 315 916 8 × 2 = 0 + 0.642 598 631 833 6;
  • 17) 0.642 598 631 833 6 × 2 = 1 + 0.285 197 263 667 2;
  • 18) 0.285 197 263 667 2 × 2 = 0 + 0.570 394 527 334 4;
  • 19) 0.570 394 527 334 4 × 2 = 1 + 0.140 789 054 668 8;
  • 20) 0.140 789 054 668 8 × 2 = 0 + 0.281 578 109 337 6;
  • 21) 0.281 578 109 337 6 × 2 = 0 + 0.563 156 218 675 2;
  • 22) 0.563 156 218 675 2 × 2 = 1 + 0.126 312 437 350 4;
  • 23) 0.126 312 437 350 4 × 2 = 0 + 0.252 624 874 700 8;
  • 24) 0.252 624 874 700 8 × 2 = 0 + 0.505 249 749 401 6;
  • 25) 0.505 249 749 401 6 × 2 = 1 + 0.010 499 498 803 2;
  • 26) 0.010 499 498 803 2 × 2 = 0 + 0.020 998 997 606 4;
  • 27) 0.020 998 997 606 4 × 2 = 0 + 0.041 997 995 212 8;
  • 28) 0.041 997 995 212 8 × 2 = 0 + 0.083 995 990 425 6;
  • 29) 0.083 995 990 425 6 × 2 = 0 + 0.167 991 980 851 2;
  • 30) 0.167 991 980 851 2 × 2 = 0 + 0.335 983 961 702 4;
  • 31) 0.335 983 961 702 4 × 2 = 0 + 0.671 967 923 404 8;
  • 32) 0.671 967 923 404 8 × 2 = 1 + 0.343 935 846 809 6;
  • 33) 0.343 935 846 809 6 × 2 = 0 + 0.687 871 693 619 2;
  • 34) 0.687 871 693 619 2 × 2 = 1 + 0.375 743 387 238 4;
  • 35) 0.375 743 387 238 4 × 2 = 0 + 0.751 486 774 476 8;
  • 36) 0.751 486 774 476 8 × 2 = 1 + 0.502 973 548 953 6;
  • 37) 0.502 973 548 953 6 × 2 = 1 + 0.005 947 097 907 2;
  • 38) 0.005 947 097 907 2 × 2 = 0 + 0.011 894 195 814 4;
  • 39) 0.011 894 195 814 4 × 2 = 0 + 0.023 788 391 628 8;
  • 40) 0.023 788 391 628 8 × 2 = 0 + 0.047 576 783 257 6;
  • 41) 0.047 576 783 257 6 × 2 = 0 + 0.095 153 566 515 2;
  • 42) 0.095 153 566 515 2 × 2 = 0 + 0.190 307 133 030 4;
  • 43) 0.190 307 133 030 4 × 2 = 0 + 0.380 614 266 060 8;
  • 44) 0.380 614 266 060 8 × 2 = 0 + 0.761 228 532 121 6;
  • 45) 0.761 228 532 121 6 × 2 = 1 + 0.522 457 064 243 2;
  • 46) 0.522 457 064 243 2 × 2 = 1 + 0.044 914 128 486 4;
  • 47) 0.044 914 128 486 4 × 2 = 0 + 0.089 828 256 972 8;
  • 48) 0.089 828 256 972 8 × 2 = 0 + 0.179 656 513 945 6;
  • 49) 0.179 656 513 945 6 × 2 = 0 + 0.359 313 027 891 2;
  • 50) 0.359 313 027 891 2 × 2 = 0 + 0.718 626 055 782 4;
  • 51) 0.718 626 055 782 4 × 2 = 1 + 0.437 252 111 564 8;
  • 52) 0.437 252 111 564 8 × 2 = 0 + 0.874 504 223 129 6;
  • 53) 0.874 504 223 129 6 × 2 = 1 + 0.749 008 446 259 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).


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:


0.841 470 986 917 6(10) =

0.1101 0111 0110 1010 1010 0100 1000 0001 0101 1000 0000 1100 0010 1(2)

6. Positive number before normalization:

0.841 470 986 917 6(10) =

0.1101 0111 0110 1010 1010 0100 1000 0001 0101 1000 0000 1100 0010 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.841 470 986 917 6(10) =

0.1101 0111 0110 1010 1010 0100 1000 0001 0101 1000 0000 1100 0010 1(2) =

0.1101 0111 0110 1010 1010 0100 1000 0001 0101 1000 0000 1100 0010 1(2) × 20 =

1.1010 1110 1101 0101 0100 1001 0000 0010 1011 0000 0001 1000 0101(2) × 2-1


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): -1

Mantissa (not normalized):
1.1010 1110 1101 0101 0100 1001 0000 0010 1011 0000 0001 1000 0101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 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) =

1022(10) =

011 1111 1110(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. 1010 1110 1101 0101 0100 1001 0000 0010 1011 0000 0001 1000 0101 =

1010 1110 1101 0101 0100 1001 0000 0010 1011 0000 0001 1000 0101


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 1110

Mantissa (52 bits) =
1010 1110 1101 0101 0100 1001 0000 0010 1011 0000 0001 1000 0101


Decimal number -0.841 470 986 917 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1110 - 1010 1110 1101 0101 0100 1001 0000 0010 1011 0000 0001 1000 0101

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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