1.117 587 089 538 718 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.117 587 089 538 718(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
1.117 587 089 538 718(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: 0.117 587 089 538 718.

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) 0.117 587 089 538 718 × 2 = 0 + 0.235 174 179 077 436;
  • 2) 0.235 174 179 077 436 × 2 = 0 + 0.470 348 358 154 872;
  • 3) 0.470 348 358 154 872 × 2 = 0 + 0.940 696 716 309 744;
  • 4) 0.940 696 716 309 744 × 2 = 1 + 0.881 393 432 619 488;
  • 5) 0.881 393 432 619 488 × 2 = 1 + 0.762 786 865 238 976;
  • 6) 0.762 786 865 238 976 × 2 = 1 + 0.525 573 730 477 952;
  • 7) 0.525 573 730 477 952 × 2 = 1 + 0.051 147 460 955 904;
  • 8) 0.051 147 460 955 904 × 2 = 0 + 0.102 294 921 911 808;
  • 9) 0.102 294 921 911 808 × 2 = 0 + 0.204 589 843 823 616;
  • 10) 0.204 589 843 823 616 × 2 = 0 + 0.409 179 687 647 232;
  • 11) 0.409 179 687 647 232 × 2 = 0 + 0.818 359 375 294 464;
  • 12) 0.818 359 375 294 464 × 2 = 1 + 0.636 718 750 588 928;
  • 13) 0.636 718 750 588 928 × 2 = 1 + 0.273 437 501 177 856;
  • 14) 0.273 437 501 177 856 × 2 = 0 + 0.546 875 002 355 712;
  • 15) 0.546 875 002 355 712 × 2 = 1 + 0.093 750 004 711 424;
  • 16) 0.093 750 004 711 424 × 2 = 0 + 0.187 500 009 422 848;
  • 17) 0.187 500 009 422 848 × 2 = 0 + 0.375 000 018 845 696;
  • 18) 0.375 000 018 845 696 × 2 = 0 + 0.750 000 037 691 392;
  • 19) 0.750 000 037 691 392 × 2 = 1 + 0.500 000 075 382 784;
  • 20) 0.500 000 075 382 784 × 2 = 1 + 0.000 000 150 765 568;
  • 21) 0.000 000 150 765 568 × 2 = 0 + 0.000 000 301 531 136;
  • 22) 0.000 000 301 531 136 × 2 = 0 + 0.000 000 603 062 272;
  • 23) 0.000 000 603 062 272 × 2 = 0 + 0.000 001 206 124 544;
  • 24) 0.000 001 206 124 544 × 2 = 0 + 0.000 002 412 249 088;
  • 25) 0.000 002 412 249 088 × 2 = 0 + 0.000 004 824 498 176;
  • 26) 0.000 004 824 498 176 × 2 = 0 + 0.000 009 648 996 352;
  • 27) 0.000 009 648 996 352 × 2 = 0 + 0.000 019 297 992 704;
  • 28) 0.000 019 297 992 704 × 2 = 0 + 0.000 038 595 985 408;
  • 29) 0.000 038 595 985 408 × 2 = 0 + 0.000 077 191 970 816;
  • 30) 0.000 077 191 970 816 × 2 = 0 + 0.000 154 383 941 632;
  • 31) 0.000 154 383 941 632 × 2 = 0 + 0.000 308 767 883 264;
  • 32) 0.000 308 767 883 264 × 2 = 0 + 0.000 617 535 766 528;
  • 33) 0.000 617 535 766 528 × 2 = 0 + 0.001 235 071 533 056;
  • 34) 0.001 235 071 533 056 × 2 = 0 + 0.002 470 143 066 112;
  • 35) 0.002 470 143 066 112 × 2 = 0 + 0.004 940 286 132 224;
  • 36) 0.004 940 286 132 224 × 2 = 0 + 0.009 880 572 264 448;
  • 37) 0.009 880 572 264 448 × 2 = 0 + 0.019 761 144 528 896;
  • 38) 0.019 761 144 528 896 × 2 = 0 + 0.039 522 289 057 792;
  • 39) 0.039 522 289 057 792 × 2 = 0 + 0.079 044 578 115 584;
  • 40) 0.079 044 578 115 584 × 2 = 0 + 0.158 089 156 231 168;
  • 41) 0.158 089 156 231 168 × 2 = 0 + 0.316 178 312 462 336;
  • 42) 0.316 178 312 462 336 × 2 = 0 + 0.632 356 624 924 672;
  • 43) 0.632 356 624 924 672 × 2 = 1 + 0.264 713 249 849 344;
  • 44) 0.264 713 249 849 344 × 2 = 0 + 0.529 426 499 698 688;
  • 45) 0.529 426 499 698 688 × 2 = 1 + 0.058 852 999 397 376;
  • 46) 0.058 852 999 397 376 × 2 = 0 + 0.117 705 998 794 752;
  • 47) 0.117 705 998 794 752 × 2 = 0 + 0.235 411 997 589 504;
  • 48) 0.235 411 997 589 504 × 2 = 0 + 0.470 823 995 179 008;
  • 49) 0.470 823 995 179 008 × 2 = 0 + 0.941 647 990 358 016;
  • 50) 0.941 647 990 358 016 × 2 = 1 + 0.883 295 980 716 032;
  • 51) 0.883 295 980 716 032 × 2 = 1 + 0.766 591 961 432 064;
  • 52) 0.766 591 961 432 064 × 2 = 1 + 0.533 183 922 864 128;
  • 53) 0.533 183 922 864 128 × 2 = 1 + 0.066 367 845 728 256;

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:


0.117 587 089 538 718(10) =

0.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1000 0111 1(2)

5. Positive number before normalization:

1.117 587 089 538 718(10) =

1.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1000 0111 1(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:


1.117 587 089 538 718(10) =

1.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1000 0111 1(2) =

1.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1000 0111 1(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):
1.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1000 0111 1


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. 0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1000 0111 1 =

0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1000 0111


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) =
0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1000 0111


Decimal number 1.117 587 089 538 718 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1000 0111

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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