1.117 587 089 538 738 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.117 587 089 538 738(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
1.117 587 089 538 738(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: 0.117 587 089 538 738.

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) 0.117 587 089 538 738 × 2 = 0 + 0.235 174 179 077 476;
  • 2) 0.235 174 179 077 476 × 2 = 0 + 0.470 348 358 154 952;
  • 3) 0.470 348 358 154 952 × 2 = 0 + 0.940 696 716 309 904;
  • 4) 0.940 696 716 309 904 × 2 = 1 + 0.881 393 432 619 808;
  • 5) 0.881 393 432 619 808 × 2 = 1 + 0.762 786 865 239 616;
  • 6) 0.762 786 865 239 616 × 2 = 1 + 0.525 573 730 479 232;
  • 7) 0.525 573 730 479 232 × 2 = 1 + 0.051 147 460 958 464;
  • 8) 0.051 147 460 958 464 × 2 = 0 + 0.102 294 921 916 928;
  • 9) 0.102 294 921 916 928 × 2 = 0 + 0.204 589 843 833 856;
  • 10) 0.204 589 843 833 856 × 2 = 0 + 0.409 179 687 667 712;
  • 11) 0.409 179 687 667 712 × 2 = 0 + 0.818 359 375 335 424;
  • 12) 0.818 359 375 335 424 × 2 = 1 + 0.636 718 750 670 848;
  • 13) 0.636 718 750 670 848 × 2 = 1 + 0.273 437 501 341 696;
  • 14) 0.273 437 501 341 696 × 2 = 0 + 0.546 875 002 683 392;
  • 15) 0.546 875 002 683 392 × 2 = 1 + 0.093 750 005 366 784;
  • 16) 0.093 750 005 366 784 × 2 = 0 + 0.187 500 010 733 568;
  • 17) 0.187 500 010 733 568 × 2 = 0 + 0.375 000 021 467 136;
  • 18) 0.375 000 021 467 136 × 2 = 0 + 0.750 000 042 934 272;
  • 19) 0.750 000 042 934 272 × 2 = 1 + 0.500 000 085 868 544;
  • 20) 0.500 000 085 868 544 × 2 = 1 + 0.000 000 171 737 088;
  • 21) 0.000 000 171 737 088 × 2 = 0 + 0.000 000 343 474 176;
  • 22) 0.000 000 343 474 176 × 2 = 0 + 0.000 000 686 948 352;
  • 23) 0.000 000 686 948 352 × 2 = 0 + 0.000 001 373 896 704;
  • 24) 0.000 001 373 896 704 × 2 = 0 + 0.000 002 747 793 408;
  • 25) 0.000 002 747 793 408 × 2 = 0 + 0.000 005 495 586 816;
  • 26) 0.000 005 495 586 816 × 2 = 0 + 0.000 010 991 173 632;
  • 27) 0.000 010 991 173 632 × 2 = 0 + 0.000 021 982 347 264;
  • 28) 0.000 021 982 347 264 × 2 = 0 + 0.000 043 964 694 528;
  • 29) 0.000 043 964 694 528 × 2 = 0 + 0.000 087 929 389 056;
  • 30) 0.000 087 929 389 056 × 2 = 0 + 0.000 175 858 778 112;
  • 31) 0.000 175 858 778 112 × 2 = 0 + 0.000 351 717 556 224;
  • 32) 0.000 351 717 556 224 × 2 = 0 + 0.000 703 435 112 448;
  • 33) 0.000 703 435 112 448 × 2 = 0 + 0.001 406 870 224 896;
  • 34) 0.001 406 870 224 896 × 2 = 0 + 0.002 813 740 449 792;
  • 35) 0.002 813 740 449 792 × 2 = 0 + 0.005 627 480 899 584;
  • 36) 0.005 627 480 899 584 × 2 = 0 + 0.011 254 961 799 168;
  • 37) 0.011 254 961 799 168 × 2 = 0 + 0.022 509 923 598 336;
  • 38) 0.022 509 923 598 336 × 2 = 0 + 0.045 019 847 196 672;
  • 39) 0.045 019 847 196 672 × 2 = 0 + 0.090 039 694 393 344;
  • 40) 0.090 039 694 393 344 × 2 = 0 + 0.180 079 388 786 688;
  • 41) 0.180 079 388 786 688 × 2 = 0 + 0.360 158 777 573 376;
  • 42) 0.360 158 777 573 376 × 2 = 0 + 0.720 317 555 146 752;
  • 43) 0.720 317 555 146 752 × 2 = 1 + 0.440 635 110 293 504;
  • 44) 0.440 635 110 293 504 × 2 = 0 + 0.881 270 220 587 008;
  • 45) 0.881 270 220 587 008 × 2 = 1 + 0.762 540 441 174 016;
  • 46) 0.762 540 441 174 016 × 2 = 1 + 0.525 080 882 348 032;
  • 47) 0.525 080 882 348 032 × 2 = 1 + 0.050 161 764 696 064;
  • 48) 0.050 161 764 696 064 × 2 = 0 + 0.100 323 529 392 128;
  • 49) 0.100 323 529 392 128 × 2 = 0 + 0.200 647 058 784 256;
  • 50) 0.200 647 058 784 256 × 2 = 0 + 0.401 294 117 568 512;
  • 51) 0.401 294 117 568 512 × 2 = 0 + 0.802 588 235 137 024;
  • 52) 0.802 588 235 137 024 × 2 = 1 + 0.605 176 470 274 048;
  • 53) 0.605 176 470 274 048 × 2 = 1 + 0.210 352 940 548 096;

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:


0.117 587 089 538 738(10) =

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

5. Positive number before normalization:

1.117 587 089 538 738(10) =

1.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1110 0001 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:


1.117 587 089 538 738(10) =

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

1.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1110 0001 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 1110 0001 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 1110 0001 1 =

0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 1110 0001


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 1110 0001


Decimal number 1.117 587 089 538 738 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 1110 0001

» Convert decimal 1.117 587 089 538 686 to 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