6.999 999 999 999 999 555 912 99 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.999 999 999 999 999 555 912 99(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
6.999 999 999 999 999 555 912 99(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. First, convert to binary (in base 2) the integer part: 6.
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;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. 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.

6(10) =

110(2)


3. Convert to binary (base 2) the fractional part: 0.999 999 999 999 999 555 912 99.

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) 0.999 999 999 999 999 555 912 99 × 2 = 1 + 0.999 999 999 999 999 111 825 98;
  • 2) 0.999 999 999 999 999 111 825 98 × 2 = 1 + 0.999 999 999 999 998 223 651 96;
  • 3) 0.999 999 999 999 998 223 651 96 × 2 = 1 + 0.999 999 999 999 996 447 303 92;
  • 4) 0.999 999 999 999 996 447 303 92 × 2 = 1 + 0.999 999 999 999 992 894 607 84;
  • 5) 0.999 999 999 999 992 894 607 84 × 2 = 1 + 0.999 999 999 999 985 789 215 68;
  • 6) 0.999 999 999 999 985 789 215 68 × 2 = 1 + 0.999 999 999 999 971 578 431 36;
  • 7) 0.999 999 999 999 971 578 431 36 × 2 = 1 + 0.999 999 999 999 943 156 862 72;
  • 8) 0.999 999 999 999 943 156 862 72 × 2 = 1 + 0.999 999 999 999 886 313 725 44;
  • 9) 0.999 999 999 999 886 313 725 44 × 2 = 1 + 0.999 999 999 999 772 627 450 88;
  • 10) 0.999 999 999 999 772 627 450 88 × 2 = 1 + 0.999 999 999 999 545 254 901 76;
  • 11) 0.999 999 999 999 545 254 901 76 × 2 = 1 + 0.999 999 999 999 090 509 803 52;
  • 12) 0.999 999 999 999 090 509 803 52 × 2 = 1 + 0.999 999 999 998 181 019 607 04;
  • 13) 0.999 999 999 998 181 019 607 04 × 2 = 1 + 0.999 999 999 996 362 039 214 08;
  • 14) 0.999 999 999 996 362 039 214 08 × 2 = 1 + 0.999 999 999 992 724 078 428 16;
  • 15) 0.999 999 999 992 724 078 428 16 × 2 = 1 + 0.999 999 999 985 448 156 856 32;
  • 16) 0.999 999 999 985 448 156 856 32 × 2 = 1 + 0.999 999 999 970 896 313 712 64;
  • 17) 0.999 999 999 970 896 313 712 64 × 2 = 1 + 0.999 999 999 941 792 627 425 28;
  • 18) 0.999 999 999 941 792 627 425 28 × 2 = 1 + 0.999 999 999 883 585 254 850 56;
  • 19) 0.999 999 999 883 585 254 850 56 × 2 = 1 + 0.999 999 999 767 170 509 701 12;
  • 20) 0.999 999 999 767 170 509 701 12 × 2 = 1 + 0.999 999 999 534 341 019 402 24;
  • 21) 0.999 999 999 534 341 019 402 24 × 2 = 1 + 0.999 999 999 068 682 038 804 48;
  • 22) 0.999 999 999 068 682 038 804 48 × 2 = 1 + 0.999 999 998 137 364 077 608 96;
  • 23) 0.999 999 998 137 364 077 608 96 × 2 = 1 + 0.999 999 996 274 728 155 217 92;
  • 24) 0.999 999 996 274 728 155 217 92 × 2 = 1 + 0.999 999 992 549 456 310 435 84;
  • 25) 0.999 999 992 549 456 310 435 84 × 2 = 1 + 0.999 999 985 098 912 620 871 68;
  • 26) 0.999 999 985 098 912 620 871 68 × 2 = 1 + 0.999 999 970 197 825 241 743 36;
  • 27) 0.999 999 970 197 825 241 743 36 × 2 = 1 + 0.999 999 940 395 650 483 486 72;
  • 28) 0.999 999 940 395 650 483 486 72 × 2 = 1 + 0.999 999 880 791 300 966 973 44;
  • 29) 0.999 999 880 791 300 966 973 44 × 2 = 1 + 0.999 999 761 582 601 933 946 88;
  • 30) 0.999 999 761 582 601 933 946 88 × 2 = 1 + 0.999 999 523 165 203 867 893 76;
  • 31) 0.999 999 523 165 203 867 893 76 × 2 = 1 + 0.999 999 046 330 407 735 787 52;
  • 32) 0.999 999 046 330 407 735 787 52 × 2 = 1 + 0.999 998 092 660 815 471 575 04;
  • 33) 0.999 998 092 660 815 471 575 04 × 2 = 1 + 0.999 996 185 321 630 943 150 08;
  • 34) 0.999 996 185 321 630 943 150 08 × 2 = 1 + 0.999 992 370 643 261 886 300 16;
  • 35) 0.999 992 370 643 261 886 300 16 × 2 = 1 + 0.999 984 741 286 523 772 600 32;
  • 36) 0.999 984 741 286 523 772 600 32 × 2 = 1 + 0.999 969 482 573 047 545 200 64;
  • 37) 0.999 969 482 573 047 545 200 64 × 2 = 1 + 0.999 938 965 146 095 090 401 28;
  • 38) 0.999 938 965 146 095 090 401 28 × 2 = 1 + 0.999 877 930 292 190 180 802 56;
  • 39) 0.999 877 930 292 190 180 802 56 × 2 = 1 + 0.999 755 860 584 380 361 605 12;
  • 40) 0.999 755 860 584 380 361 605 12 × 2 = 1 + 0.999 511 721 168 760 723 210 24;
  • 41) 0.999 511 721 168 760 723 210 24 × 2 = 1 + 0.999 023 442 337 521 446 420 48;
  • 42) 0.999 023 442 337 521 446 420 48 × 2 = 1 + 0.998 046 884 675 042 892 840 96;
  • 43) 0.998 046 884 675 042 892 840 96 × 2 = 1 + 0.996 093 769 350 085 785 681 92;
  • 44) 0.996 093 769 350 085 785 681 92 × 2 = 1 + 0.992 187 538 700 171 571 363 84;
  • 45) 0.992 187 538 700 171 571 363 84 × 2 = 1 + 0.984 375 077 400 343 142 727 68;
  • 46) 0.984 375 077 400 343 142 727 68 × 2 = 1 + 0.968 750 154 800 686 285 455 36;
  • 47) 0.968 750 154 800 686 285 455 36 × 2 = 1 + 0.937 500 309 601 372 570 910 72;
  • 48) 0.937 500 309 601 372 570 910 72 × 2 = 1 + 0.875 000 619 202 745 141 821 44;
  • 49) 0.875 000 619 202 745 141 821 44 × 2 = 1 + 0.750 001 238 405 490 283 642 88;
  • 50) 0.750 001 238 405 490 283 642 88 × 2 = 1 + 0.500 002 476 810 980 567 285 76;
  • 51) 0.500 002 476 810 980 567 285 76 × 2 = 1 + 0.000 004 953 621 961 134 571 52;
  • 52) 0.000 004 953 621 961 134 571 52 × 2 = 0 + 0.000 009 907 243 922 269 143 04;
  • 53) 0.000 009 907 243 922 269 143 04 × 2 = 0 + 0.000 019 814 487 844 538 286 08;

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).


4. 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:


0.999 999 999 999 999 555 912 99(10) =

0.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0(2)

5. Positive number before normalization:

6.999 999 999 999 999 555 912 99(10) =

110.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


6.999 999 999 999 999 555 912 99(10) =

110.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0(2) =

110.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1110 0(2) × 20 =

1.1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 100(2) × 22


7. 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 0 (a positive number)

Exponent (unadjusted): 2

Mantissa (not normalized):
1.1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(10)


9. 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;
  • 1 025 ÷ 2 = 512 + 1;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1025(10) =

100 0000 0001(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 100 =

1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
100 0000 0001

Mantissa (52 bits) =
1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number 6.999 999 999 999 999 555 912 99 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

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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