1.117 587 089 538 689 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.117 587 089 538 689(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
1.117 587 089 538 689(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: 0.117 587 089 538 689.

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) 0.117 587 089 538 689 × 2 = 0 + 0.235 174 179 077 378;
  • 2) 0.235 174 179 077 378 × 2 = 0 + 0.470 348 358 154 756;
  • 3) 0.470 348 358 154 756 × 2 = 0 + 0.940 696 716 309 512;
  • 4) 0.940 696 716 309 512 × 2 = 1 + 0.881 393 432 619 024;
  • 5) 0.881 393 432 619 024 × 2 = 1 + 0.762 786 865 238 048;
  • 6) 0.762 786 865 238 048 × 2 = 1 + 0.525 573 730 476 096;
  • 7) 0.525 573 730 476 096 × 2 = 1 + 0.051 147 460 952 192;
  • 8) 0.051 147 460 952 192 × 2 = 0 + 0.102 294 921 904 384;
  • 9) 0.102 294 921 904 384 × 2 = 0 + 0.204 589 843 808 768;
  • 10) 0.204 589 843 808 768 × 2 = 0 + 0.409 179 687 617 536;
  • 11) 0.409 179 687 617 536 × 2 = 0 + 0.818 359 375 235 072;
  • 12) 0.818 359 375 235 072 × 2 = 1 + 0.636 718 750 470 144;
  • 13) 0.636 718 750 470 144 × 2 = 1 + 0.273 437 500 940 288;
  • 14) 0.273 437 500 940 288 × 2 = 0 + 0.546 875 001 880 576;
  • 15) 0.546 875 001 880 576 × 2 = 1 + 0.093 750 003 761 152;
  • 16) 0.093 750 003 761 152 × 2 = 0 + 0.187 500 007 522 304;
  • 17) 0.187 500 007 522 304 × 2 = 0 + 0.375 000 015 044 608;
  • 18) 0.375 000 015 044 608 × 2 = 0 + 0.750 000 030 089 216;
  • 19) 0.750 000 030 089 216 × 2 = 1 + 0.500 000 060 178 432;
  • 20) 0.500 000 060 178 432 × 2 = 1 + 0.000 000 120 356 864;
  • 21) 0.000 000 120 356 864 × 2 = 0 + 0.000 000 240 713 728;
  • 22) 0.000 000 240 713 728 × 2 = 0 + 0.000 000 481 427 456;
  • 23) 0.000 000 481 427 456 × 2 = 0 + 0.000 000 962 854 912;
  • 24) 0.000 000 962 854 912 × 2 = 0 + 0.000 001 925 709 824;
  • 25) 0.000 001 925 709 824 × 2 = 0 + 0.000 003 851 419 648;
  • 26) 0.000 003 851 419 648 × 2 = 0 + 0.000 007 702 839 296;
  • 27) 0.000 007 702 839 296 × 2 = 0 + 0.000 015 405 678 592;
  • 28) 0.000 015 405 678 592 × 2 = 0 + 0.000 030 811 357 184;
  • 29) 0.000 030 811 357 184 × 2 = 0 + 0.000 061 622 714 368;
  • 30) 0.000 061 622 714 368 × 2 = 0 + 0.000 123 245 428 736;
  • 31) 0.000 123 245 428 736 × 2 = 0 + 0.000 246 490 857 472;
  • 32) 0.000 246 490 857 472 × 2 = 0 + 0.000 492 981 714 944;
  • 33) 0.000 492 981 714 944 × 2 = 0 + 0.000 985 963 429 888;
  • 34) 0.000 985 963 429 888 × 2 = 0 + 0.001 971 926 859 776;
  • 35) 0.001 971 926 859 776 × 2 = 0 + 0.003 943 853 719 552;
  • 36) 0.003 943 853 719 552 × 2 = 0 + 0.007 887 707 439 104;
  • 37) 0.007 887 707 439 104 × 2 = 0 + 0.015 775 414 878 208;
  • 38) 0.015 775 414 878 208 × 2 = 0 + 0.031 550 829 756 416;
  • 39) 0.031 550 829 756 416 × 2 = 0 + 0.063 101 659 512 832;
  • 40) 0.063 101 659 512 832 × 2 = 0 + 0.126 203 319 025 664;
  • 41) 0.126 203 319 025 664 × 2 = 0 + 0.252 406 638 051 328;
  • 42) 0.252 406 638 051 328 × 2 = 0 + 0.504 813 276 102 656;
  • 43) 0.504 813 276 102 656 × 2 = 1 + 0.009 626 552 205 312;
  • 44) 0.009 626 552 205 312 × 2 = 0 + 0.019 253 104 410 624;
  • 45) 0.019 253 104 410 624 × 2 = 0 + 0.038 506 208 821 248;
  • 46) 0.038 506 208 821 248 × 2 = 0 + 0.077 012 417 642 496;
  • 47) 0.077 012 417 642 496 × 2 = 0 + 0.154 024 835 284 992;
  • 48) 0.154 024 835 284 992 × 2 = 0 + 0.308 049 670 569 984;
  • 49) 0.308 049 670 569 984 × 2 = 0 + 0.616 099 341 139 968;
  • 50) 0.616 099 341 139 968 × 2 = 1 + 0.232 198 682 279 936;
  • 51) 0.232 198 682 279 936 × 2 = 0 + 0.464 397 364 559 872;
  • 52) 0.464 397 364 559 872 × 2 = 0 + 0.928 794 729 119 744;
  • 53) 0.928 794 729 119 744 × 2 = 1 + 0.857 589 458 239 488;

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:


0.117 587 089 538 689(10) =

0.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 0000 0100 1(2)

5. Positive number before normalization:

1.117 587 089 538 689(10) =

1.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 0000 0100 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:


1.117 587 089 538 689(10) =

1.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 0000 0100 1(2) =

1.0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 0000 0100 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 0000 0100 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 0000 0100 1 =

0001 1110 0001 1010 0011 0000 0000 0000 0000 0000 0010 0000 0100


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 0000 0100


Decimal number 1.117 587 089 538 689 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 0000 0100

» Convert decimal 1.117 587 089 538 738 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