2.029 999 999 907 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.029 999 999 907(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
2.029 999 999 907(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: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.029 999 999 907.

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) 0.029 999 999 907 × 2 = 0 + 0.059 999 999 814;
  • 2) 0.059 999 999 814 × 2 = 0 + 0.119 999 999 628;
  • 3) 0.119 999 999 628 × 2 = 0 + 0.239 999 999 256;
  • 4) 0.239 999 999 256 × 2 = 0 + 0.479 999 998 512;
  • 5) 0.479 999 998 512 × 2 = 0 + 0.959 999 997 024;
  • 6) 0.959 999 997 024 × 2 = 1 + 0.919 999 994 048;
  • 7) 0.919 999 994 048 × 2 = 1 + 0.839 999 988 096;
  • 8) 0.839 999 988 096 × 2 = 1 + 0.679 999 976 192;
  • 9) 0.679 999 976 192 × 2 = 1 + 0.359 999 952 384;
  • 10) 0.359 999 952 384 × 2 = 0 + 0.719 999 904 768;
  • 11) 0.719 999 904 768 × 2 = 1 + 0.439 999 809 536;
  • 12) 0.439 999 809 536 × 2 = 0 + 0.879 999 619 072;
  • 13) 0.879 999 619 072 × 2 = 1 + 0.759 999 238 144;
  • 14) 0.759 999 238 144 × 2 = 1 + 0.519 998 476 288;
  • 15) 0.519 998 476 288 × 2 = 1 + 0.039 996 952 576;
  • 16) 0.039 996 952 576 × 2 = 0 + 0.079 993 905 152;
  • 17) 0.079 993 905 152 × 2 = 0 + 0.159 987 810 304;
  • 18) 0.159 987 810 304 × 2 = 0 + 0.319 975 620 608;
  • 19) 0.319 975 620 608 × 2 = 0 + 0.639 951 241 216;
  • 20) 0.639 951 241 216 × 2 = 1 + 0.279 902 482 432;
  • 21) 0.279 902 482 432 × 2 = 0 + 0.559 804 964 864;
  • 22) 0.559 804 964 864 × 2 = 1 + 0.119 609 929 728;
  • 23) 0.119 609 929 728 × 2 = 0 + 0.239 219 859 456;
  • 24) 0.239 219 859 456 × 2 = 0 + 0.478 439 718 912;
  • 25) 0.478 439 718 912 × 2 = 0 + 0.956 879 437 824;
  • 26) 0.956 879 437 824 × 2 = 1 + 0.913 758 875 648;
  • 27) 0.913 758 875 648 × 2 = 1 + 0.827 517 751 296;
  • 28) 0.827 517 751 296 × 2 = 1 + 0.655 035 502 592;
  • 29) 0.655 035 502 592 × 2 = 1 + 0.310 071 005 184;
  • 30) 0.310 071 005 184 × 2 = 0 + 0.620 142 010 368;
  • 31) 0.620 142 010 368 × 2 = 1 + 0.240 284 020 736;
  • 32) 0.240 284 020 736 × 2 = 0 + 0.480 568 041 472;
  • 33) 0.480 568 041 472 × 2 = 0 + 0.961 136 082 944;
  • 34) 0.961 136 082 944 × 2 = 1 + 0.922 272 165 888;
  • 35) 0.922 272 165 888 × 2 = 1 + 0.844 544 331 776;
  • 36) 0.844 544 331 776 × 2 = 1 + 0.689 088 663 552;
  • 37) 0.689 088 663 552 × 2 = 1 + 0.378 177 327 104;
  • 38) 0.378 177 327 104 × 2 = 0 + 0.756 354 654 208;
  • 39) 0.756 354 654 208 × 2 = 1 + 0.512 709 308 416;
  • 40) 0.512 709 308 416 × 2 = 1 + 0.025 418 616 832;
  • 41) 0.025 418 616 832 × 2 = 0 + 0.050 837 233 664;
  • 42) 0.050 837 233 664 × 2 = 0 + 0.101 674 467 328;
  • 43) 0.101 674 467 328 × 2 = 0 + 0.203 348 934 656;
  • 44) 0.203 348 934 656 × 2 = 0 + 0.406 697 869 312;
  • 45) 0.406 697 869 312 × 2 = 0 + 0.813 395 738 624;
  • 46) 0.813 395 738 624 × 2 = 1 + 0.626 791 477 248;
  • 47) 0.626 791 477 248 × 2 = 1 + 0.253 582 954 496;
  • 48) 0.253 582 954 496 × 2 = 0 + 0.507 165 908 992;
  • 49) 0.507 165 908 992 × 2 = 1 + 0.014 331 817 984;
  • 50) 0.014 331 817 984 × 2 = 0 + 0.028 663 635 968;
  • 51) 0.028 663 635 968 × 2 = 0 + 0.057 327 271 936;
  • 52) 0.057 327 271 936 × 2 = 0 + 0.114 654 543 872;
  • 53) 0.114 654 543 872 × 2 = 0 + 0.229 309 087 744;

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:


0.029 999 999 907(10) =

0.0000 0111 1010 1110 0001 0100 0111 1010 0111 1011 0000 0110 1000 0(2)

5. Positive number before normalization:

2.029 999 999 907(10) =

10.0000 0111 1010 1110 0001 0100 0111 1010 0111 1011 0000 0110 1000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


2.029 999 999 907(10) =

10.0000 0111 1010 1110 0001 0100 0111 1010 0111 1011 0000 0110 1000 0(2) =

10.0000 0111 1010 1110 0001 0100 0111 1010 0111 1011 0000 0110 1000 0(2) × 20 =

1.0000 0011 1101 0111 0000 1010 0011 1101 0011 1101 1000 0011 0100 00(2) × 21


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): 1

Mantissa (not normalized):
1.0000 0011 1101 0111 0000 1010 0011 1101 0011 1101 1000 0011 0100 00


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(11-1) - 1 =

(1 + 1 023)(10) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 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) =

1024(10) =

100 0000 0000(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. 0000 0011 1101 0111 0000 1010 0011 1101 0011 1101 1000 0011 0100 00 =

0000 0011 1101 0111 0000 1010 0011 1101 0011 1101 1000 0011 0100


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 0000

Mantissa (52 bits) =
0000 0011 1101 0111 0000 1010 0011 1101 0011 1101 1000 0011 0100


Decimal number 2.029 999 999 907 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0000 0011 1101 0111 0000 1010 0011 1101 0011 1101 1000 0011 0100

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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