Decimal to 64 Bit IEEE 754 Binary: Convert Number 6.095 719 648 259 7 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 6.095 719 648 259 7(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation standard (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: 0.095 719 648 259 7.

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) 0.095 719 648 259 7 × 2 = 0 + 0.191 439 296 519 4;
  • 2) 0.191 439 296 519 4 × 2 = 0 + 0.382 878 593 038 8;
  • 3) 0.382 878 593 038 8 × 2 = 0 + 0.765 757 186 077 6;
  • 4) 0.765 757 186 077 6 × 2 = 1 + 0.531 514 372 155 2;
  • 5) 0.531 514 372 155 2 × 2 = 1 + 0.063 028 744 310 4;
  • 6) 0.063 028 744 310 4 × 2 = 0 + 0.126 057 488 620 8;
  • 7) 0.126 057 488 620 8 × 2 = 0 + 0.252 114 977 241 6;
  • 8) 0.252 114 977 241 6 × 2 = 0 + 0.504 229 954 483 2;
  • 9) 0.504 229 954 483 2 × 2 = 1 + 0.008 459 908 966 4;
  • 10) 0.008 459 908 966 4 × 2 = 0 + 0.016 919 817 932 8;
  • 11) 0.016 919 817 932 8 × 2 = 0 + 0.033 839 635 865 6;
  • 12) 0.033 839 635 865 6 × 2 = 0 + 0.067 679 271 731 2;
  • 13) 0.067 679 271 731 2 × 2 = 0 + 0.135 358 543 462 4;
  • 14) 0.135 358 543 462 4 × 2 = 0 + 0.270 717 086 924 8;
  • 15) 0.270 717 086 924 8 × 2 = 0 + 0.541 434 173 849 6;
  • 16) 0.541 434 173 849 6 × 2 = 1 + 0.082 868 347 699 2;
  • 17) 0.082 868 347 699 2 × 2 = 0 + 0.165 736 695 398 4;
  • 18) 0.165 736 695 398 4 × 2 = 0 + 0.331 473 390 796 8;
  • 19) 0.331 473 390 796 8 × 2 = 0 + 0.662 946 781 593 6;
  • 20) 0.662 946 781 593 6 × 2 = 1 + 0.325 893 563 187 2;
  • 21) 0.325 893 563 187 2 × 2 = 0 + 0.651 787 126 374 4;
  • 22) 0.651 787 126 374 4 × 2 = 1 + 0.303 574 252 748 8;
  • 23) 0.303 574 252 748 8 × 2 = 0 + 0.607 148 505 497 6;
  • 24) 0.607 148 505 497 6 × 2 = 1 + 0.214 297 010 995 2;
  • 25) 0.214 297 010 995 2 × 2 = 0 + 0.428 594 021 990 4;
  • 26) 0.428 594 021 990 4 × 2 = 0 + 0.857 188 043 980 8;
  • 27) 0.857 188 043 980 8 × 2 = 1 + 0.714 376 087 961 6;
  • 28) 0.714 376 087 961 6 × 2 = 1 + 0.428 752 175 923 2;
  • 29) 0.428 752 175 923 2 × 2 = 0 + 0.857 504 351 846 4;
  • 30) 0.857 504 351 846 4 × 2 = 1 + 0.715 008 703 692 8;
  • 31) 0.715 008 703 692 8 × 2 = 1 + 0.430 017 407 385 6;
  • 32) 0.430 017 407 385 6 × 2 = 0 + 0.860 034 814 771 2;
  • 33) 0.860 034 814 771 2 × 2 = 1 + 0.720 069 629 542 4;
  • 34) 0.720 069 629 542 4 × 2 = 1 + 0.440 139 259 084 8;
  • 35) 0.440 139 259 084 8 × 2 = 0 + 0.880 278 518 169 6;
  • 36) 0.880 278 518 169 6 × 2 = 1 + 0.760 557 036 339 2;
  • 37) 0.760 557 036 339 2 × 2 = 1 + 0.521 114 072 678 4;
  • 38) 0.521 114 072 678 4 × 2 = 1 + 0.042 228 145 356 8;
  • 39) 0.042 228 145 356 8 × 2 = 0 + 0.084 456 290 713 6;
  • 40) 0.084 456 290 713 6 × 2 = 0 + 0.168 912 581 427 2;
  • 41) 0.168 912 581 427 2 × 2 = 0 + 0.337 825 162 854 4;
  • 42) 0.337 825 162 854 4 × 2 = 0 + 0.675 650 325 708 8;
  • 43) 0.675 650 325 708 8 × 2 = 1 + 0.351 300 651 417 6;
  • 44) 0.351 300 651 417 6 × 2 = 0 + 0.702 601 302 835 2;
  • 45) 0.702 601 302 835 2 × 2 = 1 + 0.405 202 605 670 4;
  • 46) 0.405 202 605 670 4 × 2 = 0 + 0.810 405 211 340 8;
  • 47) 0.810 405 211 340 8 × 2 = 1 + 0.620 810 422 681 6;
  • 48) 0.620 810 422 681 6 × 2 = 1 + 0.241 620 845 363 2;
  • 49) 0.241 620 845 363 2 × 2 = 0 + 0.483 241 690 726 4;
  • 50) 0.483 241 690 726 4 × 2 = 0 + 0.966 483 381 452 8;
  • 51) 0.966 483 381 452 8 × 2 = 1 + 0.932 966 762 905 6;
  • 52) 0.932 966 762 905 6 × 2 = 1 + 0.865 933 525 811 2;
  • 53) 0.865 933 525 811 2 × 2 = 1 + 0.731 867 051 622 4;

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:


0.095 719 648 259 7(10) =

0.0001 1000 1000 0001 0001 0101 0011 0110 1101 1100 0010 1011 0011 1(2)

5. Positive number before normalization:

6.095 719 648 259 7(10) =

110.0001 1000 1000 0001 0001 0101 0011 0110 1101 1100 0010 1011 0011 1(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:


6.095 719 648 259 7(10) =

110.0001 1000 1000 0001 0001 0101 0011 0110 1101 1100 0010 1011 0011 1(2) =

110.0001 1000 1000 0001 0001 0101 0011 0110 1101 1100 0010 1011 0011 1(2) × 20 =

1.1000 0110 0010 0000 0100 0101 0100 1101 1011 0111 0000 1010 1100 111(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):
1.1000 0110 0010 0000 0100 0101 0100 1101 1011 0111 0000 1010 1100 111


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. 1000 0110 0010 0000 0100 0101 0100 1101 1011 0111 0000 1010 1100 111 =

1000 0110 0010 0000 0100 0101 0100 1101 1011 0111 0000 1010 1100


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) =
1000 0110 0010 0000 0100 0101 0100 1101 1011 0111 0000 1010 1100


The base ten decimal number 6.095 719 648 259 7 converted and written in 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1000 0110 0010 0000 0100 0101 0100 1101 1011 0111 0000 1010 1100

» Decimal number in base ten 6.095 719 648 252 converted and written as 64 bit double precision IEEE 754 binary floating point representation standard

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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