-0.008 788 423 612 794 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 612 794(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
-0.008 788 423 612 794(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:

|-0.008 788 423 612 794| = 0.008 788 423 612 794


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: 0.008 788 423 612 794.

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) 0.008 788 423 612 794 × 2 = 0 + 0.017 576 847 225 588;
  • 2) 0.017 576 847 225 588 × 2 = 0 + 0.035 153 694 451 176;
  • 3) 0.035 153 694 451 176 × 2 = 0 + 0.070 307 388 902 352;
  • 4) 0.070 307 388 902 352 × 2 = 0 + 0.140 614 777 804 704;
  • 5) 0.140 614 777 804 704 × 2 = 0 + 0.281 229 555 609 408;
  • 6) 0.281 229 555 609 408 × 2 = 0 + 0.562 459 111 218 816;
  • 7) 0.562 459 111 218 816 × 2 = 1 + 0.124 918 222 437 632;
  • 8) 0.124 918 222 437 632 × 2 = 0 + 0.249 836 444 875 264;
  • 9) 0.249 836 444 875 264 × 2 = 0 + 0.499 672 889 750 528;
  • 10) 0.499 672 889 750 528 × 2 = 0 + 0.999 345 779 501 056;
  • 11) 0.999 345 779 501 056 × 2 = 1 + 0.998 691 559 002 112;
  • 12) 0.998 691 559 002 112 × 2 = 1 + 0.997 383 118 004 224;
  • 13) 0.997 383 118 004 224 × 2 = 1 + 0.994 766 236 008 448;
  • 14) 0.994 766 236 008 448 × 2 = 1 + 0.989 532 472 016 896;
  • 15) 0.989 532 472 016 896 × 2 = 1 + 0.979 064 944 033 792;
  • 16) 0.979 064 944 033 792 × 2 = 1 + 0.958 129 888 067 584;
  • 17) 0.958 129 888 067 584 × 2 = 1 + 0.916 259 776 135 168;
  • 18) 0.916 259 776 135 168 × 2 = 1 + 0.832 519 552 270 336;
  • 19) 0.832 519 552 270 336 × 2 = 1 + 0.665 039 104 540 672;
  • 20) 0.665 039 104 540 672 × 2 = 1 + 0.330 078 209 081 344;
  • 21) 0.330 078 209 081 344 × 2 = 0 + 0.660 156 418 162 688;
  • 22) 0.660 156 418 162 688 × 2 = 1 + 0.320 312 836 325 376;
  • 23) 0.320 312 836 325 376 × 2 = 0 + 0.640 625 672 650 752;
  • 24) 0.640 625 672 650 752 × 2 = 1 + 0.281 251 345 301 504;
  • 25) 0.281 251 345 301 504 × 2 = 0 + 0.562 502 690 603 008;
  • 26) 0.562 502 690 603 008 × 2 = 1 + 0.125 005 381 206 016;
  • 27) 0.125 005 381 206 016 × 2 = 0 + 0.250 010 762 412 032;
  • 28) 0.250 010 762 412 032 × 2 = 0 + 0.500 021 524 824 064;
  • 29) 0.500 021 524 824 064 × 2 = 1 + 0.000 043 049 648 128;
  • 30) 0.000 043 049 648 128 × 2 = 0 + 0.000 086 099 296 256;
  • 31) 0.000 086 099 296 256 × 2 = 0 + 0.000 172 198 592 512;
  • 32) 0.000 172 198 592 512 × 2 = 0 + 0.000 344 397 185 024;
  • 33) 0.000 344 397 185 024 × 2 = 0 + 0.000 688 794 370 048;
  • 34) 0.000 688 794 370 048 × 2 = 0 + 0.001 377 588 740 096;
  • 35) 0.001 377 588 740 096 × 2 = 0 + 0.002 755 177 480 192;
  • 36) 0.002 755 177 480 192 × 2 = 0 + 0.005 510 354 960 384;
  • 37) 0.005 510 354 960 384 × 2 = 0 + 0.011 020 709 920 768;
  • 38) 0.011 020 709 920 768 × 2 = 0 + 0.022 041 419 841 536;
  • 39) 0.022 041 419 841 536 × 2 = 0 + 0.044 082 839 683 072;
  • 40) 0.044 082 839 683 072 × 2 = 0 + 0.088 165 679 366 144;
  • 41) 0.088 165 679 366 144 × 2 = 0 + 0.176 331 358 732 288;
  • 42) 0.176 331 358 732 288 × 2 = 0 + 0.352 662 717 464 576;
  • 43) 0.352 662 717 464 576 × 2 = 0 + 0.705 325 434 929 152;
  • 44) 0.705 325 434 929 152 × 2 = 1 + 0.410 650 869 858 304;
  • 45) 0.410 650 869 858 304 × 2 = 0 + 0.821 301 739 716 608;
  • 46) 0.821 301 739 716 608 × 2 = 1 + 0.642 603 479 433 216;
  • 47) 0.642 603 479 433 216 × 2 = 1 + 0.285 206 958 866 432;
  • 48) 0.285 206 958 866 432 × 2 = 0 + 0.570 413 917 732 864;
  • 49) 0.570 413 917 732 864 × 2 = 1 + 0.140 827 835 465 728;
  • 50) 0.140 827 835 465 728 × 2 = 0 + 0.281 655 670 931 456;
  • 51) 0.281 655 670 931 456 × 2 = 0 + 0.563 311 341 862 912;
  • 52) 0.563 311 341 862 912 × 2 = 1 + 0.126 622 683 725 824;
  • 53) 0.126 622 683 725 824 × 2 = 0 + 0.253 245 367 451 648;
  • 54) 0.253 245 367 451 648 × 2 = 0 + 0.506 490 734 903 296;
  • 55) 0.506 490 734 903 296 × 2 = 1 + 0.012 981 469 806 592;
  • 56) 0.012 981 469 806 592 × 2 = 0 + 0.025 962 939 613 184;
  • 57) 0.025 962 939 613 184 × 2 = 0 + 0.051 925 879 226 368;
  • 58) 0.051 925 879 226 368 × 2 = 0 + 0.103 851 758 452 736;
  • 59) 0.103 851 758 452 736 × 2 = 0 + 0.207 703 516 905 472;

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:


0.008 788 423 612 794(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 0110 1001 0010 000(2)

6. Positive number before normalization:

0.008 788 423 612 794(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 0110 1001 0010 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:


0.008 788 423 612 794(10) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 0110 1001 0010 000(2) =

0.0000 0010 0011 1111 1111 0101 0100 1000 0000 0000 0001 0110 1001 0010 000(2) × 20 =

1.0001 1111 1111 1010 1010 0100 0000 0000 0000 1011 0100 1001 0000(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 0000 1011 0100 1001 0000


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 0000 1011 0100 1001 0000 =

0001 1111 1111 1010 1010 0100 0000 0000 0000 1011 0100 1001 0000


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 0000 1011 0100 1001 0000


Decimal number -0.008 788 423 612 794 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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