1.876 543 456 931 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.876 543 456 931(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
1.876 543 456 931(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.876 543 456 931.

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) 0.876 543 456 931 × 2 = 1 + 0.753 086 913 862;
  • 2) 0.753 086 913 862 × 2 = 1 + 0.506 173 827 724;
  • 3) 0.506 173 827 724 × 2 = 1 + 0.012 347 655 448;
  • 4) 0.012 347 655 448 × 2 = 0 + 0.024 695 310 896;
  • 5) 0.024 695 310 896 × 2 = 0 + 0.049 390 621 792;
  • 6) 0.049 390 621 792 × 2 = 0 + 0.098 781 243 584;
  • 7) 0.098 781 243 584 × 2 = 0 + 0.197 562 487 168;
  • 8) 0.197 562 487 168 × 2 = 0 + 0.395 124 974 336;
  • 9) 0.395 124 974 336 × 2 = 0 + 0.790 249 948 672;
  • 10) 0.790 249 948 672 × 2 = 1 + 0.580 499 897 344;
  • 11) 0.580 499 897 344 × 2 = 1 + 0.160 999 794 688;
  • 12) 0.160 999 794 688 × 2 = 0 + 0.321 999 589 376;
  • 13) 0.321 999 589 376 × 2 = 0 + 0.643 999 178 752;
  • 14) 0.643 999 178 752 × 2 = 1 + 0.287 998 357 504;
  • 15) 0.287 998 357 504 × 2 = 0 + 0.575 996 715 008;
  • 16) 0.575 996 715 008 × 2 = 1 + 0.151 993 430 016;
  • 17) 0.151 993 430 016 × 2 = 0 + 0.303 986 860 032;
  • 18) 0.303 986 860 032 × 2 = 0 + 0.607 973 720 064;
  • 19) 0.607 973 720 064 × 2 = 1 + 0.215 947 440 128;
  • 20) 0.215 947 440 128 × 2 = 0 + 0.431 894 880 256;
  • 21) 0.431 894 880 256 × 2 = 0 + 0.863 789 760 512;
  • 22) 0.863 789 760 512 × 2 = 1 + 0.727 579 521 024;
  • 23) 0.727 579 521 024 × 2 = 1 + 0.455 159 042 048;
  • 24) 0.455 159 042 048 × 2 = 0 + 0.910 318 084 096;
  • 25) 0.910 318 084 096 × 2 = 1 + 0.820 636 168 192;
  • 26) 0.820 636 168 192 × 2 = 1 + 0.641 272 336 384;
  • 27) 0.641 272 336 384 × 2 = 1 + 0.282 544 672 768;
  • 28) 0.282 544 672 768 × 2 = 0 + 0.565 089 345 536;
  • 29) 0.565 089 345 536 × 2 = 1 + 0.130 178 691 072;
  • 30) 0.130 178 691 072 × 2 = 0 + 0.260 357 382 144;
  • 31) 0.260 357 382 144 × 2 = 0 + 0.520 714 764 288;
  • 32) 0.520 714 764 288 × 2 = 1 + 0.041 429 528 576;
  • 33) 0.041 429 528 576 × 2 = 0 + 0.082 859 057 152;
  • 34) 0.082 859 057 152 × 2 = 0 + 0.165 718 114 304;
  • 35) 0.165 718 114 304 × 2 = 0 + 0.331 436 228 608;
  • 36) 0.331 436 228 608 × 2 = 0 + 0.662 872 457 216;
  • 37) 0.662 872 457 216 × 2 = 1 + 0.325 744 914 432;
  • 38) 0.325 744 914 432 × 2 = 0 + 0.651 489 828 864;
  • 39) 0.651 489 828 864 × 2 = 1 + 0.302 979 657 728;
  • 40) 0.302 979 657 728 × 2 = 0 + 0.605 959 315 456;
  • 41) 0.605 959 315 456 × 2 = 1 + 0.211 918 630 912;
  • 42) 0.211 918 630 912 × 2 = 0 + 0.423 837 261 824;
  • 43) 0.423 837 261 824 × 2 = 0 + 0.847 674 523 648;
  • 44) 0.847 674 523 648 × 2 = 1 + 0.695 349 047 296;
  • 45) 0.695 349 047 296 × 2 = 1 + 0.390 698 094 592;
  • 46) 0.390 698 094 592 × 2 = 0 + 0.781 396 189 184;
  • 47) 0.781 396 189 184 × 2 = 1 + 0.562 792 378 368;
  • 48) 0.562 792 378 368 × 2 = 1 + 0.125 584 756 736;
  • 49) 0.125 584 756 736 × 2 = 0 + 0.251 169 513 472;
  • 50) 0.251 169 513 472 × 2 = 0 + 0.502 339 026 944;
  • 51) 0.502 339 026 944 × 2 = 1 + 0.004 678 053 888;
  • 52) 0.004 678 053 888 × 2 = 0 + 0.009 356 107 776;
  • 53) 0.009 356 107 776 × 2 = 0 + 0.018 712 215 552;

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:


0.876 543 456 931(10) =

0.1110 0000 0110 0101 0010 0110 1110 1001 0000 1010 1001 1011 0010 0(2)

5. Positive number before normalization:

1.876 543 456 931(10) =

1.1110 0000 0110 0101 0010 0110 1110 1001 0000 1010 1001 1011 0010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.876 543 456 931(10) =

1.1110 0000 0110 0101 0010 0110 1110 1001 0000 1010 1001 1011 0010 0(2) =

1.1110 0000 0110 0101 0010 0110 1110 1001 0000 1010 1001 1011 0010 0(2) × 20


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): 0

Mantissa (not normalized):
1.1110 0000 0110 0101 0010 0110 1110 1001 0000 1010 1001 1011 0010 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 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;

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) =

1023(10) =

011 1111 1111(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. 1110 0000 0110 0101 0010 0110 1110 1001 0000 1010 1001 1011 0010 0 =

1110 0000 0110 0101 0010 0110 1110 1001 0000 1010 1001 1011 0010


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) =
011 1111 1111

Mantissa (52 bits) =
1110 0000 0110 0101 0010 0110 1110 1001 0000 1010 1001 1011 0010


Decimal number 1.876 543 456 931 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1110 0000 0110 0101 0010 0110 1110 1001 0000 1010 1001 1011 0010

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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