2.560 879 599 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.560 879 599 8(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
2.560 879 599 8(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: 0.560 879 599 8.

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) 0.560 879 599 8 × 2 = 1 + 0.121 759 199 6;
  • 2) 0.121 759 199 6 × 2 = 0 + 0.243 518 399 2;
  • 3) 0.243 518 399 2 × 2 = 0 + 0.487 036 798 4;
  • 4) 0.487 036 798 4 × 2 = 0 + 0.974 073 596 8;
  • 5) 0.974 073 596 8 × 2 = 1 + 0.948 147 193 6;
  • 6) 0.948 147 193 6 × 2 = 1 + 0.896 294 387 2;
  • 7) 0.896 294 387 2 × 2 = 1 + 0.792 588 774 4;
  • 8) 0.792 588 774 4 × 2 = 1 + 0.585 177 548 8;
  • 9) 0.585 177 548 8 × 2 = 1 + 0.170 355 097 6;
  • 10) 0.170 355 097 6 × 2 = 0 + 0.340 710 195 2;
  • 11) 0.340 710 195 2 × 2 = 0 + 0.681 420 390 4;
  • 12) 0.681 420 390 4 × 2 = 1 + 0.362 840 780 8;
  • 13) 0.362 840 780 8 × 2 = 0 + 0.725 681 561 6;
  • 14) 0.725 681 561 6 × 2 = 1 + 0.451 363 123 2;
  • 15) 0.451 363 123 2 × 2 = 0 + 0.902 726 246 4;
  • 16) 0.902 726 246 4 × 2 = 1 + 0.805 452 492 8;
  • 17) 0.805 452 492 8 × 2 = 1 + 0.610 904 985 6;
  • 18) 0.610 904 985 6 × 2 = 1 + 0.221 809 971 2;
  • 19) 0.221 809 971 2 × 2 = 0 + 0.443 619 942 4;
  • 20) 0.443 619 942 4 × 2 = 0 + 0.887 239 884 8;
  • 21) 0.887 239 884 8 × 2 = 1 + 0.774 479 769 6;
  • 22) 0.774 479 769 6 × 2 = 1 + 0.548 959 539 2;
  • 23) 0.548 959 539 2 × 2 = 1 + 0.097 919 078 4;
  • 24) 0.097 919 078 4 × 2 = 0 + 0.195 838 156 8;
  • 25) 0.195 838 156 8 × 2 = 0 + 0.391 676 313 6;
  • 26) 0.391 676 313 6 × 2 = 0 + 0.783 352 627 2;
  • 27) 0.783 352 627 2 × 2 = 1 + 0.566 705 254 4;
  • 28) 0.566 705 254 4 × 2 = 1 + 0.133 410 508 8;
  • 29) 0.133 410 508 8 × 2 = 0 + 0.266 821 017 6;
  • 30) 0.266 821 017 6 × 2 = 0 + 0.533 642 035 2;
  • 31) 0.533 642 035 2 × 2 = 1 + 0.067 284 070 4;
  • 32) 0.067 284 070 4 × 2 = 0 + 0.134 568 140 8;
  • 33) 0.134 568 140 8 × 2 = 0 + 0.269 136 281 6;
  • 34) 0.269 136 281 6 × 2 = 0 + 0.538 272 563 2;
  • 35) 0.538 272 563 2 × 2 = 1 + 0.076 545 126 4;
  • 36) 0.076 545 126 4 × 2 = 0 + 0.153 090 252 8;
  • 37) 0.153 090 252 8 × 2 = 0 + 0.306 180 505 6;
  • 38) 0.306 180 505 6 × 2 = 0 + 0.612 361 011 2;
  • 39) 0.612 361 011 2 × 2 = 1 + 0.224 722 022 4;
  • 40) 0.224 722 022 4 × 2 = 0 + 0.449 444 044 8;
  • 41) 0.449 444 044 8 × 2 = 0 + 0.898 888 089 6;
  • 42) 0.898 888 089 6 × 2 = 1 + 0.797 776 179 2;
  • 43) 0.797 776 179 2 × 2 = 1 + 0.595 552 358 4;
  • 44) 0.595 552 358 4 × 2 = 1 + 0.191 104 716 8;
  • 45) 0.191 104 716 8 × 2 = 0 + 0.382 209 433 6;
  • 46) 0.382 209 433 6 × 2 = 0 + 0.764 418 867 2;
  • 47) 0.764 418 867 2 × 2 = 1 + 0.528 837 734 4;
  • 48) 0.528 837 734 4 × 2 = 1 + 0.057 675 468 8;
  • 49) 0.057 675 468 8 × 2 = 0 + 0.115 350 937 6;
  • 50) 0.115 350 937 6 × 2 = 0 + 0.230 701 875 2;
  • 51) 0.230 701 875 2 × 2 = 0 + 0.461 403 750 4;
  • 52) 0.461 403 750 4 × 2 = 0 + 0.922 807 500 8;
  • 53) 0.922 807 500 8 × 2 = 1 + 0.845 615 001 6;

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:


0.560 879 599 8(10) =

0.1000 1111 1001 0101 1100 1110 0011 0010 0010 0010 0111 0011 0000 1(2)

5. Positive number before normalization:

2.560 879 599 8(10) =

10.1000 1111 1001 0101 1100 1110 0011 0010 0010 0010 0111 0011 0000 1(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:


2.560 879 599 8(10) =

10.1000 1111 1001 0101 1100 1110 0011 0010 0010 0010 0111 0011 0000 1(2) =

10.1000 1111 1001 0101 1100 1110 0011 0010 0010 0010 0111 0011 0000 1(2) × 20 =

1.0100 0111 1100 1010 1110 0111 0001 1001 0001 0001 0011 1001 1000 01(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):
1.0100 0111 1100 1010 1110 0111 0001 1001 0001 0001 0011 1001 1000 01


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. 0100 0111 1100 1010 1110 0111 0001 1001 0001 0001 0011 1001 1000 01 =

0100 0111 1100 1010 1110 0111 0001 1001 0001 0001 0011 1001 1000


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) =
0100 0111 1100 1010 1110 0111 0001 1001 0001 0001 0011 1001 1000


Decimal number 2.560 879 599 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0100 0111 1100 1010 1110 0111 0001 1001 0001 0001 0011 1001 1000

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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