-0.008 788 423 614 03 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 614 03(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
-0.008 788 423 614 03(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:

|-0.008 788 423 614 03| = 0.008 788 423 614 03


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: 0.008 788 423 614 03.

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) 0.008 788 423 614 03 × 2 = 0 + 0.017 576 847 228 06;
  • 2) 0.017 576 847 228 06 × 2 = 0 + 0.035 153 694 456 12;
  • 3) 0.035 153 694 456 12 × 2 = 0 + 0.070 307 388 912 24;
  • 4) 0.070 307 388 912 24 × 2 = 0 + 0.140 614 777 824 48;
  • 5) 0.140 614 777 824 48 × 2 = 0 + 0.281 229 555 648 96;
  • 6) 0.281 229 555 648 96 × 2 = 0 + 0.562 459 111 297 92;
  • 7) 0.562 459 111 297 92 × 2 = 1 + 0.124 918 222 595 84;
  • 8) 0.124 918 222 595 84 × 2 = 0 + 0.249 836 445 191 68;
  • 9) 0.249 836 445 191 68 × 2 = 0 + 0.499 672 890 383 36;
  • 10) 0.499 672 890 383 36 × 2 = 0 + 0.999 345 780 766 72;
  • 11) 0.999 345 780 766 72 × 2 = 1 + 0.998 691 561 533 44;
  • 12) 0.998 691 561 533 44 × 2 = 1 + 0.997 383 123 066 88;
  • 13) 0.997 383 123 066 88 × 2 = 1 + 0.994 766 246 133 76;
  • 14) 0.994 766 246 133 76 × 2 = 1 + 0.989 532 492 267 52;
  • 15) 0.989 532 492 267 52 × 2 = 1 + 0.979 064 984 535 04;
  • 16) 0.979 064 984 535 04 × 2 = 1 + 0.958 129 969 070 08;
  • 17) 0.958 129 969 070 08 × 2 = 1 + 0.916 259 938 140 16;
  • 18) 0.916 259 938 140 16 × 2 = 1 + 0.832 519 876 280 32;
  • 19) 0.832 519 876 280 32 × 2 = 1 + 0.665 039 752 560 64;
  • 20) 0.665 039 752 560 64 × 2 = 1 + 0.330 079 505 121 28;
  • 21) 0.330 079 505 121 28 × 2 = 0 + 0.660 159 010 242 56;
  • 22) 0.660 159 010 242 56 × 2 = 1 + 0.320 318 020 485 12;
  • 23) 0.320 318 020 485 12 × 2 = 0 + 0.640 636 040 970 24;
  • 24) 0.640 636 040 970 24 × 2 = 1 + 0.281 272 081 940 48;
  • 25) 0.281 272 081 940 48 × 2 = 0 + 0.562 544 163 880 96;
  • 26) 0.562 544 163 880 96 × 2 = 1 + 0.125 088 327 761 92;
  • 27) 0.125 088 327 761 92 × 2 = 0 + 0.250 176 655 523 84;
  • 28) 0.250 176 655 523 84 × 2 = 0 + 0.500 353 311 047 68;
  • 29) 0.500 353 311 047 68 × 2 = 1 + 0.000 706 622 095 36;
  • 30) 0.000 706 622 095 36 × 2 = 0 + 0.001 413 244 190 72;
  • 31) 0.001 413 244 190 72 × 2 = 0 + 0.002 826 488 381 44;
  • 32) 0.002 826 488 381 44 × 2 = 0 + 0.005 652 976 762 88;
  • 33) 0.005 652 976 762 88 × 2 = 0 + 0.011 305 953 525 76;
  • 34) 0.011 305 953 525 76 × 2 = 0 + 0.022 611 907 051 52;
  • 35) 0.022 611 907 051 52 × 2 = 0 + 0.045 223 814 103 04;
  • 36) 0.045 223 814 103 04 × 2 = 0 + 0.090 447 628 206 08;
  • 37) 0.090 447 628 206 08 × 2 = 0 + 0.180 895 256 412 16;
  • 38) 0.180 895 256 412 16 × 2 = 0 + 0.361 790 512 824 32;
  • 39) 0.361 790 512 824 32 × 2 = 0 + 0.723 581 025 648 64;
  • 40) 0.723 581 025 648 64 × 2 = 1 + 0.447 162 051 297 28;
  • 41) 0.447 162 051 297 28 × 2 = 0 + 0.894 324 102 594 56;
  • 42) 0.894 324 102 594 56 × 2 = 1 + 0.788 648 205 189 12;
  • 43) 0.788 648 205 189 12 × 2 = 1 + 0.577 296 410 378 24;
  • 44) 0.577 296 410 378 24 × 2 = 1 + 0.154 592 820 756 48;
  • 45) 0.154 592 820 756 48 × 2 = 0 + 0.309 185 641 512 96;
  • 46) 0.309 185 641 512 96 × 2 = 0 + 0.618 371 283 025 92;
  • 47) 0.618 371 283 025 92 × 2 = 1 + 0.236 742 566 051 84;
  • 48) 0.236 742 566 051 84 × 2 = 0 + 0.473 485 132 103 68;
  • 49) 0.473 485 132 103 68 × 2 = 0 + 0.946 970 264 207 36;
  • 50) 0.946 970 264 207 36 × 2 = 1 + 0.893 940 528 414 72;
  • 51) 0.893 940 528 414 72 × 2 = 1 + 0.787 881 056 829 44;
  • 52) 0.787 881 056 829 44 × 2 = 1 + 0.575 762 113 658 88;
  • 53) 0.575 762 113 658 88 × 2 = 1 + 0.151 524 227 317 76;
  • 54) 0.151 524 227 317 76 × 2 = 0 + 0.303 048 454 635 52;
  • 55) 0.303 048 454 635 52 × 2 = 0 + 0.606 096 909 271 04;
  • 56) 0.606 096 909 271 04 × 2 = 1 + 0.212 193 818 542 08;
  • 57) 0.212 193 818 542 08 × 2 = 0 + 0.424 387 637 084 16;
  • 58) 0.424 387 637 084 16 × 2 = 0 + 0.848 775 274 168 32;
  • 59) 0.848 775 274 168 32 × 2 = 1 + 0.697 550 548 336 64;

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:


0.008 788 423 614 03(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0111 0010 0111 1001 001(2)

6. Positive number before normalization:

0.008 788 423 614 03(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0111 0010 0111 1001 001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.008 788 423 614 03(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0111 0010 0111 1001 001(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 0111 0010 0111 1001 001(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 1011 1001 0011 1100 1001(2) × 2-7


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): -7

Mantissa (not normalized):
1.0001 1111 1111 1010 1010 0100 0000 0000 1011 1001 0011 1100 1001


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(11-1) - 1 =

(-7 + 1 023)(10) =

1 016(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;
  • 1 016 ÷ 2 = 508 + 0;
  • 508 ÷ 2 = 254 + 0;
  • 254 ÷ 2 = 127 + 0;
  • 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) =

1016(10) =

011 1111 1000(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. 0001 1111 1111 1010 1010 0100 0000 0000 1011 1001 0011 1100 1001 =

0001 1111 1111 1010 1010 0100 0000 0000 1011 1001 0011 1100 1001


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 1000

Mantissa (52 bits) =
0001 1111 1111 1010 1010 0100 0000 0000 1011 1001 0011 1100 1001


Decimal number -0.008 788 423 614 03 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0000 1011 1001 0011 1100 1001

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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