-0.008 788 423 614 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 614 2(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 2(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 2| = 0.008 788 423 614 2


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 2.

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 2 × 2 = 0 + 0.017 576 847 228 4;
  • 2) 0.017 576 847 228 4 × 2 = 0 + 0.035 153 694 456 8;
  • 3) 0.035 153 694 456 8 × 2 = 0 + 0.070 307 388 913 6;
  • 4) 0.070 307 388 913 6 × 2 = 0 + 0.140 614 777 827 2;
  • 5) 0.140 614 777 827 2 × 2 = 0 + 0.281 229 555 654 4;
  • 6) 0.281 229 555 654 4 × 2 = 0 + 0.562 459 111 308 8;
  • 7) 0.562 459 111 308 8 × 2 = 1 + 0.124 918 222 617 6;
  • 8) 0.124 918 222 617 6 × 2 = 0 + 0.249 836 445 235 2;
  • 9) 0.249 836 445 235 2 × 2 = 0 + 0.499 672 890 470 4;
  • 10) 0.499 672 890 470 4 × 2 = 0 + 0.999 345 780 940 8;
  • 11) 0.999 345 780 940 8 × 2 = 1 + 0.998 691 561 881 6;
  • 12) 0.998 691 561 881 6 × 2 = 1 + 0.997 383 123 763 2;
  • 13) 0.997 383 123 763 2 × 2 = 1 + 0.994 766 247 526 4;
  • 14) 0.994 766 247 526 4 × 2 = 1 + 0.989 532 495 052 8;
  • 15) 0.989 532 495 052 8 × 2 = 1 + 0.979 064 990 105 6;
  • 16) 0.979 064 990 105 6 × 2 = 1 + 0.958 129 980 211 2;
  • 17) 0.958 129 980 211 2 × 2 = 1 + 0.916 259 960 422 4;
  • 18) 0.916 259 960 422 4 × 2 = 1 + 0.832 519 920 844 8;
  • 19) 0.832 519 920 844 8 × 2 = 1 + 0.665 039 841 689 6;
  • 20) 0.665 039 841 689 6 × 2 = 1 + 0.330 079 683 379 2;
  • 21) 0.330 079 683 379 2 × 2 = 0 + 0.660 159 366 758 4;
  • 22) 0.660 159 366 758 4 × 2 = 1 + 0.320 318 733 516 8;
  • 23) 0.320 318 733 516 8 × 2 = 0 + 0.640 637 467 033 6;
  • 24) 0.640 637 467 033 6 × 2 = 1 + 0.281 274 934 067 2;
  • 25) 0.281 274 934 067 2 × 2 = 0 + 0.562 549 868 134 4;
  • 26) 0.562 549 868 134 4 × 2 = 1 + 0.125 099 736 268 8;
  • 27) 0.125 099 736 268 8 × 2 = 0 + 0.250 199 472 537 6;
  • 28) 0.250 199 472 537 6 × 2 = 0 + 0.500 398 945 075 2;
  • 29) 0.500 398 945 075 2 × 2 = 1 + 0.000 797 890 150 4;
  • 30) 0.000 797 890 150 4 × 2 = 0 + 0.001 595 780 300 8;
  • 31) 0.001 595 780 300 8 × 2 = 0 + 0.003 191 560 601 6;
  • 32) 0.003 191 560 601 6 × 2 = 0 + 0.006 383 121 203 2;
  • 33) 0.006 383 121 203 2 × 2 = 0 + 0.012 766 242 406 4;
  • 34) 0.012 766 242 406 4 × 2 = 0 + 0.025 532 484 812 8;
  • 35) 0.025 532 484 812 8 × 2 = 0 + 0.051 064 969 625 6;
  • 36) 0.051 064 969 625 6 × 2 = 0 + 0.102 129 939 251 2;
  • 37) 0.102 129 939 251 2 × 2 = 0 + 0.204 259 878 502 4;
  • 38) 0.204 259 878 502 4 × 2 = 0 + 0.408 519 757 004 8;
  • 39) 0.408 519 757 004 8 × 2 = 0 + 0.817 039 514 009 6;
  • 40) 0.817 039 514 009 6 × 2 = 1 + 0.634 079 028 019 2;
  • 41) 0.634 079 028 019 2 × 2 = 1 + 0.268 158 056 038 4;
  • 42) 0.268 158 056 038 4 × 2 = 0 + 0.536 316 112 076 8;
  • 43) 0.536 316 112 076 8 × 2 = 1 + 0.072 632 224 153 6;
  • 44) 0.072 632 224 153 6 × 2 = 0 + 0.145 264 448 307 2;
  • 45) 0.145 264 448 307 2 × 2 = 0 + 0.290 528 896 614 4;
  • 46) 0.290 528 896 614 4 × 2 = 0 + 0.581 057 793 228 8;
  • 47) 0.581 057 793 228 8 × 2 = 1 + 0.162 115 586 457 6;
  • 48) 0.162 115 586 457 6 × 2 = 0 + 0.324 231 172 915 2;
  • 49) 0.324 231 172 915 2 × 2 = 0 + 0.648 462 345 830 4;
  • 50) 0.648 462 345 830 4 × 2 = 1 + 0.296 924 691 660 8;
  • 51) 0.296 924 691 660 8 × 2 = 0 + 0.593 849 383 321 6;
  • 52) 0.593 849 383 321 6 × 2 = 1 + 0.187 698 766 643 2;
  • 53) 0.187 698 766 643 2 × 2 = 0 + 0.375 397 533 286 4;
  • 54) 0.375 397 533 286 4 × 2 = 0 + 0.750 795 066 572 8;
  • 55) 0.750 795 066 572 8 × 2 = 1 + 0.501 590 133 145 6;
  • 56) 0.501 590 133 145 6 × 2 = 1 + 0.003 180 266 291 2;
  • 57) 0.003 180 266 291 2 × 2 = 0 + 0.006 360 532 582 4;
  • 58) 0.006 360 532 582 4 × 2 = 0 + 0.012 721 065 164 8;
  • 59) 0.012 721 065 164 8 × 2 = 0 + 0.025 442 130 329 6;

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 2(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 1010 0010 0101 0011 000(2)

6. Positive number before normalization:

0.008 788 423 614 2(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 1010 0010 0101 0011 000(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 2(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 1010 0010 0101 0011 000(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0001 1010 0010 0101 0011 000(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 1101 0001 0010 1001 1000(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 1101 0001 0010 1001 1000


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 1101 0001 0010 1001 1000 =

0001 1111 1111 1010 1010 0100 0000 0000 1101 0001 0010 1001 1000


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 1101 0001 0010 1001 1000


Decimal number -0.008 788 423 614 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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