-0.105 000 000 108 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.105 000 000 108(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
-0.105 000 000 108(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. Start with the positive version of the number:

|-0.105 000 000 108| = 0.105 000 000 108


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.105 000 000 108.

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) 0.105 000 000 108 × 2 = 0 + 0.210 000 000 216;
  • 2) 0.210 000 000 216 × 2 = 0 + 0.420 000 000 432;
  • 3) 0.420 000 000 432 × 2 = 0 + 0.840 000 000 864;
  • 4) 0.840 000 000 864 × 2 = 1 + 0.680 000 001 728;
  • 5) 0.680 000 001 728 × 2 = 1 + 0.360 000 003 456;
  • 6) 0.360 000 003 456 × 2 = 0 + 0.720 000 006 912;
  • 7) 0.720 000 006 912 × 2 = 1 + 0.440 000 013 824;
  • 8) 0.440 000 013 824 × 2 = 0 + 0.880 000 027 648;
  • 9) 0.880 000 027 648 × 2 = 1 + 0.760 000 055 296;
  • 10) 0.760 000 055 296 × 2 = 1 + 0.520 000 110 592;
  • 11) 0.520 000 110 592 × 2 = 1 + 0.040 000 221 184;
  • 12) 0.040 000 221 184 × 2 = 0 + 0.080 000 442 368;
  • 13) 0.080 000 442 368 × 2 = 0 + 0.160 000 884 736;
  • 14) 0.160 000 884 736 × 2 = 0 + 0.320 001 769 472;
  • 15) 0.320 001 769 472 × 2 = 0 + 0.640 003 538 944;
  • 16) 0.640 003 538 944 × 2 = 1 + 0.280 007 077 888;
  • 17) 0.280 007 077 888 × 2 = 0 + 0.560 014 155 776;
  • 18) 0.560 014 155 776 × 2 = 1 + 0.120 028 311 552;
  • 19) 0.120 028 311 552 × 2 = 0 + 0.240 056 623 104;
  • 20) 0.240 056 623 104 × 2 = 0 + 0.480 113 246 208;
  • 21) 0.480 113 246 208 × 2 = 0 + 0.960 226 492 416;
  • 22) 0.960 226 492 416 × 2 = 1 + 0.920 452 984 832;
  • 23) 0.920 452 984 832 × 2 = 1 + 0.840 905 969 664;
  • 24) 0.840 905 969 664 × 2 = 1 + 0.681 811 939 328;
  • 25) 0.681 811 939 328 × 2 = 1 + 0.363 623 878 656;
  • 26) 0.363 623 878 656 × 2 = 0 + 0.727 247 757 312;
  • 27) 0.727 247 757 312 × 2 = 1 + 0.454 495 514 624;
  • 28) 0.454 495 514 624 × 2 = 0 + 0.908 991 029 248;
  • 29) 0.908 991 029 248 × 2 = 1 + 0.817 982 058 496;
  • 30) 0.817 982 058 496 × 2 = 1 + 0.635 964 116 992;
  • 31) 0.635 964 116 992 × 2 = 1 + 0.271 928 233 984;
  • 32) 0.271 928 233 984 × 2 = 0 + 0.543 856 467 968;
  • 33) 0.543 856 467 968 × 2 = 1 + 0.087 712 935 936;
  • 34) 0.087 712 935 936 × 2 = 0 + 0.175 425 871 872;
  • 35) 0.175 425 871 872 × 2 = 0 + 0.350 851 743 744;
  • 36) 0.350 851 743 744 × 2 = 0 + 0.701 703 487 488;
  • 37) 0.701 703 487 488 × 2 = 1 + 0.403 406 974 976;
  • 38) 0.403 406 974 976 × 2 = 0 + 0.806 813 949 952;
  • 39) 0.806 813 949 952 × 2 = 1 + 0.613 627 899 904;
  • 40) 0.613 627 899 904 × 2 = 1 + 0.227 255 799 808;
  • 41) 0.227 255 799 808 × 2 = 0 + 0.454 511 599 616;
  • 42) 0.454 511 599 616 × 2 = 0 + 0.909 023 199 232;
  • 43) 0.909 023 199 232 × 2 = 1 + 0.818 046 398 464;
  • 44) 0.818 046 398 464 × 2 = 1 + 0.636 092 796 928;
  • 45) 0.636 092 796 928 × 2 = 1 + 0.272 185 593 856;
  • 46) 0.272 185 593 856 × 2 = 0 + 0.544 371 187 712;
  • 47) 0.544 371 187 712 × 2 = 1 + 0.088 742 375 424;
  • 48) 0.088 742 375 424 × 2 = 0 + 0.177 484 750 848;
  • 49) 0.177 484 750 848 × 2 = 0 + 0.354 969 501 696;
  • 50) 0.354 969 501 696 × 2 = 0 + 0.709 939 003 392;
  • 51) 0.709 939 003 392 × 2 = 1 + 0.419 878 006 784;
  • 52) 0.419 878 006 784 × 2 = 0 + 0.839 756 013 568;
  • 53) 0.839 756 013 568 × 2 = 1 + 0.679 512 027 136;
  • 54) 0.679 512 027 136 × 2 = 1 + 0.359 024 054 272;
  • 55) 0.359 024 054 272 × 2 = 0 + 0.718 048 108 544;
  • 56) 0.718 048 108 544 × 2 = 1 + 0.436 096 217 088;

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).


5. 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:


0.105 000 000 108(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101(2)

6. Positive number before normalization:

0.105 000 000 108(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.105 000 000 108(10) =

0.0001 1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101(2) =

0.0001 1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101(2) × 20 =

1.1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101(2) × 2-4


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):
1.1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-4 + 2(11-1) - 1 =

(-4 + 1 023)(10) =

1 019(10)


10. 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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1019(10) =

011 1111 1011(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101 =

1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1011

Mantissa (52 bits) =
1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101


Decimal number -0.105 000 000 108 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1011 - 1010 1110 0001 0100 0111 1010 1110 1000 1011 0011 1010 0010 1101

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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