-0.008 788 423 615 15 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 615 15(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
-0.008 788 423 615 15(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:

|-0.008 788 423 615 15| = 0.008 788 423 615 15


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: 0.008 788 423 615 15.

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) 0.008 788 423 615 15 × 2 = 0 + 0.017 576 847 230 3;
  • 2) 0.017 576 847 230 3 × 2 = 0 + 0.035 153 694 460 6;
  • 3) 0.035 153 694 460 6 × 2 = 0 + 0.070 307 388 921 2;
  • 4) 0.070 307 388 921 2 × 2 = 0 + 0.140 614 777 842 4;
  • 5) 0.140 614 777 842 4 × 2 = 0 + 0.281 229 555 684 8;
  • 6) 0.281 229 555 684 8 × 2 = 0 + 0.562 459 111 369 6;
  • 7) 0.562 459 111 369 6 × 2 = 1 + 0.124 918 222 739 2;
  • 8) 0.124 918 222 739 2 × 2 = 0 + 0.249 836 445 478 4;
  • 9) 0.249 836 445 478 4 × 2 = 0 + 0.499 672 890 956 8;
  • 10) 0.499 672 890 956 8 × 2 = 0 + 0.999 345 781 913 6;
  • 11) 0.999 345 781 913 6 × 2 = 1 + 0.998 691 563 827 2;
  • 12) 0.998 691 563 827 2 × 2 = 1 + 0.997 383 127 654 4;
  • 13) 0.997 383 127 654 4 × 2 = 1 + 0.994 766 255 308 8;
  • 14) 0.994 766 255 308 8 × 2 = 1 + 0.989 532 510 617 6;
  • 15) 0.989 532 510 617 6 × 2 = 1 + 0.979 065 021 235 2;
  • 16) 0.979 065 021 235 2 × 2 = 1 + 0.958 130 042 470 4;
  • 17) 0.958 130 042 470 4 × 2 = 1 + 0.916 260 084 940 8;
  • 18) 0.916 260 084 940 8 × 2 = 1 + 0.832 520 169 881 6;
  • 19) 0.832 520 169 881 6 × 2 = 1 + 0.665 040 339 763 2;
  • 20) 0.665 040 339 763 2 × 2 = 1 + 0.330 080 679 526 4;
  • 21) 0.330 080 679 526 4 × 2 = 0 + 0.660 161 359 052 8;
  • 22) 0.660 161 359 052 8 × 2 = 1 + 0.320 322 718 105 6;
  • 23) 0.320 322 718 105 6 × 2 = 0 + 0.640 645 436 211 2;
  • 24) 0.640 645 436 211 2 × 2 = 1 + 0.281 290 872 422 4;
  • 25) 0.281 290 872 422 4 × 2 = 0 + 0.562 581 744 844 8;
  • 26) 0.562 581 744 844 8 × 2 = 1 + 0.125 163 489 689 6;
  • 27) 0.125 163 489 689 6 × 2 = 0 + 0.250 326 979 379 2;
  • 28) 0.250 326 979 379 2 × 2 = 0 + 0.500 653 958 758 4;
  • 29) 0.500 653 958 758 4 × 2 = 1 + 0.001 307 917 516 8;
  • 30) 0.001 307 917 516 8 × 2 = 0 + 0.002 615 835 033 6;
  • 31) 0.002 615 835 033 6 × 2 = 0 + 0.005 231 670 067 2;
  • 32) 0.005 231 670 067 2 × 2 = 0 + 0.010 463 340 134 4;
  • 33) 0.010 463 340 134 4 × 2 = 0 + 0.020 926 680 268 8;
  • 34) 0.020 926 680 268 8 × 2 = 0 + 0.041 853 360 537 6;
  • 35) 0.041 853 360 537 6 × 2 = 0 + 0.083 706 721 075 2;
  • 36) 0.083 706 721 075 2 × 2 = 0 + 0.167 413 442 150 4;
  • 37) 0.167 413 442 150 4 × 2 = 0 + 0.334 826 884 300 8;
  • 38) 0.334 826 884 300 8 × 2 = 0 + 0.669 653 768 601 6;
  • 39) 0.669 653 768 601 6 × 2 = 1 + 0.339 307 537 203 2;
  • 40) 0.339 307 537 203 2 × 2 = 0 + 0.678 615 074 406 4;
  • 41) 0.678 615 074 406 4 × 2 = 1 + 0.357 230 148 812 8;
  • 42) 0.357 230 148 812 8 × 2 = 0 + 0.714 460 297 625 6;
  • 43) 0.714 460 297 625 6 × 2 = 1 + 0.428 920 595 251 2;
  • 44) 0.428 920 595 251 2 × 2 = 0 + 0.857 841 190 502 4;
  • 45) 0.857 841 190 502 4 × 2 = 1 + 0.715 682 381 004 8;
  • 46) 0.715 682 381 004 8 × 2 = 1 + 0.431 364 762 009 6;
  • 47) 0.431 364 762 009 6 × 2 = 0 + 0.862 729 524 019 2;
  • 48) 0.862 729 524 019 2 × 2 = 1 + 0.725 459 048 038 4;
  • 49) 0.725 459 048 038 4 × 2 = 1 + 0.450 918 096 076 8;
  • 50) 0.450 918 096 076 8 × 2 = 0 + 0.901 836 192 153 6;
  • 51) 0.901 836 192 153 6 × 2 = 1 + 0.803 672 384 307 2;
  • 52) 0.803 672 384 307 2 × 2 = 1 + 0.607 344 768 614 4;
  • 53) 0.607 344 768 614 4 × 2 = 1 + 0.214 689 537 228 8;
  • 54) 0.214 689 537 228 8 × 2 = 0 + 0.429 379 074 457 6;
  • 55) 0.429 379 074 457 6 × 2 = 0 + 0.858 758 148 915 2;
  • 56) 0.858 758 148 915 2 × 2 = 1 + 0.717 516 297 830 4;
  • 57) 0.717 516 297 830 4 × 2 = 1 + 0.435 032 595 660 8;
  • 58) 0.435 032 595 660 8 × 2 = 0 + 0.870 065 191 321 6;
  • 59) 0.870 065 191 321 6 × 2 = 1 + 0.740 130 382 643 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:


0.008 788 423 615 15(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 1010 1101 1011 1001 101(2)

6. Positive number before normalization:

0.008 788 423 615 15(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 1010 1101 1011 1001 101(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:


0.008 788 423 615 15(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 1010 1101 1011 1001 101(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0010 1010 1101 1011 1001 101(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0001 0101 0110 1101 1100 1101(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 0001 0101 0110 1101 1100 1101


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 0001 0101 0110 1101 1100 1101 =

0001 1111 1111 1010 1010 0100 0000 0001 0101 0110 1101 1100 1101


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 0001 0101 0110 1101 1100 1101


Decimal number -0.008 788 423 615 15 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1000 - 0001 1111 1111 1010 1010 0100 0000 0001 0101 0110 1101 1100 1101

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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