-0.008 788 423 612 858 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.008 788 423 612 858(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
-0.008 788 423 612 858(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:

|-0.008 788 423 612 858| = 0.008 788 423 612 858


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: 0.008 788 423 612 858.

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) 0.008 788 423 612 858 × 2 = 0 + 0.017 576 847 225 716;
  • 2) 0.017 576 847 225 716 × 2 = 0 + 0.035 153 694 451 432;
  • 3) 0.035 153 694 451 432 × 2 = 0 + 0.070 307 388 902 864;
  • 4) 0.070 307 388 902 864 × 2 = 0 + 0.140 614 777 805 728;
  • 5) 0.140 614 777 805 728 × 2 = 0 + 0.281 229 555 611 456;
  • 6) 0.281 229 555 611 456 × 2 = 0 + 0.562 459 111 222 912;
  • 7) 0.562 459 111 222 912 × 2 = 1 + 0.124 918 222 445 824;
  • 8) 0.124 918 222 445 824 × 2 = 0 + 0.249 836 444 891 648;
  • 9) 0.249 836 444 891 648 × 2 = 0 + 0.499 672 889 783 296;
  • 10) 0.499 672 889 783 296 × 2 = 0 + 0.999 345 779 566 592;
  • 11) 0.999 345 779 566 592 × 2 = 1 + 0.998 691 559 133 184;
  • 12) 0.998 691 559 133 184 × 2 = 1 + 0.997 383 118 266 368;
  • 13) 0.997 383 118 266 368 × 2 = 1 + 0.994 766 236 532 736;
  • 14) 0.994 766 236 532 736 × 2 = 1 + 0.989 532 473 065 472;
  • 15) 0.989 532 473 065 472 × 2 = 1 + 0.979 064 946 130 944;
  • 16) 0.979 064 946 130 944 × 2 = 1 + 0.958 129 892 261 888;
  • 17) 0.958 129 892 261 888 × 2 = 1 + 0.916 259 784 523 776;
  • 18) 0.916 259 784 523 776 × 2 = 1 + 0.832 519 569 047 552;
  • 19) 0.832 519 569 047 552 × 2 = 1 + 0.665 039 138 095 104;
  • 20) 0.665 039 138 095 104 × 2 = 1 + 0.330 078 276 190 208;
  • 21) 0.330 078 276 190 208 × 2 = 0 + 0.660 156 552 380 416;
  • 22) 0.660 156 552 380 416 × 2 = 1 + 0.320 313 104 760 832;
  • 23) 0.320 313 104 760 832 × 2 = 0 + 0.640 626 209 521 664;
  • 24) 0.640 626 209 521 664 × 2 = 1 + 0.281 252 419 043 328;
  • 25) 0.281 252 419 043 328 × 2 = 0 + 0.562 504 838 086 656;
  • 26) 0.562 504 838 086 656 × 2 = 1 + 0.125 009 676 173 312;
  • 27) 0.125 009 676 173 312 × 2 = 0 + 0.250 019 352 346 624;
  • 28) 0.250 019 352 346 624 × 2 = 0 + 0.500 038 704 693 248;
  • 29) 0.500 038 704 693 248 × 2 = 1 + 0.000 077 409 386 496;
  • 30) 0.000 077 409 386 496 × 2 = 0 + 0.000 154 818 772 992;
  • 31) 0.000 154 818 772 992 × 2 = 0 + 0.000 309 637 545 984;
  • 32) 0.000 309 637 545 984 × 2 = 0 + 0.000 619 275 091 968;
  • 33) 0.000 619 275 091 968 × 2 = 0 + 0.001 238 550 183 936;
  • 34) 0.001 238 550 183 936 × 2 = 0 + 0.002 477 100 367 872;
  • 35) 0.002 477 100 367 872 × 2 = 0 + 0.004 954 200 735 744;
  • 36) 0.004 954 200 735 744 × 2 = 0 + 0.009 908 401 471 488;
  • 37) 0.009 908 401 471 488 × 2 = 0 + 0.019 816 802 942 976;
  • 38) 0.019 816 802 942 976 × 2 = 0 + 0.039 633 605 885 952;
  • 39) 0.039 633 605 885 952 × 2 = 0 + 0.079 267 211 771 904;
  • 40) 0.079 267 211 771 904 × 2 = 0 + 0.158 534 423 543 808;
  • 41) 0.158 534 423 543 808 × 2 = 0 + 0.317 068 847 087 616;
  • 42) 0.317 068 847 087 616 × 2 = 0 + 0.634 137 694 175 232;
  • 43) 0.634 137 694 175 232 × 2 = 1 + 0.268 275 388 350 464;
  • 44) 0.268 275 388 350 464 × 2 = 0 + 0.536 550 776 700 928;
  • 45) 0.536 550 776 700 928 × 2 = 1 + 0.073 101 553 401 856;
  • 46) 0.073 101 553 401 856 × 2 = 0 + 0.146 203 106 803 712;
  • 47) 0.146 203 106 803 712 × 2 = 0 + 0.292 406 213 607 424;
  • 48) 0.292 406 213 607 424 × 2 = 0 + 0.584 812 427 214 848;
  • 49) 0.584 812 427 214 848 × 2 = 1 + 0.169 624 854 429 696;
  • 50) 0.169 624 854 429 696 × 2 = 0 + 0.339 249 708 859 392;
  • 51) 0.339 249 708 859 392 × 2 = 0 + 0.678 499 417 718 784;
  • 52) 0.678 499 417 718 784 × 2 = 1 + 0.356 998 835 437 568;
  • 53) 0.356 998 835 437 568 × 2 = 0 + 0.713 997 670 875 136;
  • 54) 0.713 997 670 875 136 × 2 = 1 + 0.427 995 341 750 272;
  • 55) 0.427 995 341 750 272 × 2 = 0 + 0.855 990 683 500 544;
  • 56) 0.855 990 683 500 544 × 2 = 1 + 0.711 981 367 001 088;
  • 57) 0.711 981 367 001 088 × 2 = 1 + 0.423 962 734 002 176;
  • 58) 0.423 962 734 002 176 × 2 = 0 + 0.847 925 468 004 352;
  • 59) 0.847 925 468 004 352 × 2 = 1 + 0.695 850 936 008 704;

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:


0.008 788 423 612 858(10) =

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

6. Positive number before normalization:

0.008 788 423 612 858(10) =

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


0.008 788 423 612 858(10) =

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

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

1.0001 1111 1111 1010 1010 0100 0000 0000 0001 0100 0100 1010 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 0000 0001 0100 0100 1010 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 0000 0001 0100 0100 1010 1101 =

0001 1111 1111 1010 1010 0100 0000 0000 0001 0100 0100 1010 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 0000 0001 0100 0100 1010 1101


Decimal number -0.008 788 423 612 858 converted to 64 bit double precision IEEE 754 binary floating point representation:

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