-0.135 000 186 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.135 000 186(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
-0.135 000 186(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:

|-0.135 000 186| = 0.135 000 186


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: 0.135 000 186.

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) 0.135 000 186 × 2 = 0 + 0.270 000 372;
  • 2) 0.270 000 372 × 2 = 0 + 0.540 000 744;
  • 3) 0.540 000 744 × 2 = 1 + 0.080 001 488;
  • 4) 0.080 001 488 × 2 = 0 + 0.160 002 976;
  • 5) 0.160 002 976 × 2 = 0 + 0.320 005 952;
  • 6) 0.320 005 952 × 2 = 0 + 0.640 011 904;
  • 7) 0.640 011 904 × 2 = 1 + 0.280 023 808;
  • 8) 0.280 023 808 × 2 = 0 + 0.560 047 616;
  • 9) 0.560 047 616 × 2 = 1 + 0.120 095 232;
  • 10) 0.120 095 232 × 2 = 0 + 0.240 190 464;
  • 11) 0.240 190 464 × 2 = 0 + 0.480 380 928;
  • 12) 0.480 380 928 × 2 = 0 + 0.960 761 856;
  • 13) 0.960 761 856 × 2 = 1 + 0.921 523 712;
  • 14) 0.921 523 712 × 2 = 1 + 0.843 047 424;
  • 15) 0.843 047 424 × 2 = 1 + 0.686 094 848;
  • 16) 0.686 094 848 × 2 = 1 + 0.372 189 696;
  • 17) 0.372 189 696 × 2 = 0 + 0.744 379 392;
  • 18) 0.744 379 392 × 2 = 1 + 0.488 758 784;
  • 19) 0.488 758 784 × 2 = 0 + 0.977 517 568;
  • 20) 0.977 517 568 × 2 = 1 + 0.955 035 136;
  • 21) 0.955 035 136 × 2 = 1 + 0.910 070 272;
  • 22) 0.910 070 272 × 2 = 1 + 0.820 140 544;
  • 23) 0.820 140 544 × 2 = 1 + 0.640 281 088;
  • 24) 0.640 281 088 × 2 = 1 + 0.280 562 176;
  • 25) 0.280 562 176 × 2 = 0 + 0.561 124 352;
  • 26) 0.561 124 352 × 2 = 1 + 0.122 248 704;
  • 27) 0.122 248 704 × 2 = 0 + 0.244 497 408;
  • 28) 0.244 497 408 × 2 = 0 + 0.488 994 816;
  • 29) 0.488 994 816 × 2 = 0 + 0.977 989 632;
  • 30) 0.977 989 632 × 2 = 1 + 0.955 979 264;
  • 31) 0.955 979 264 × 2 = 1 + 0.911 958 528;
  • 32) 0.911 958 528 × 2 = 1 + 0.823 917 056;
  • 33) 0.823 917 056 × 2 = 1 + 0.647 834 112;
  • 34) 0.647 834 112 × 2 = 1 + 0.295 668 224;
  • 35) 0.295 668 224 × 2 = 0 + 0.591 336 448;
  • 36) 0.591 336 448 × 2 = 1 + 0.182 672 896;
  • 37) 0.182 672 896 × 2 = 0 + 0.365 345 792;
  • 38) 0.365 345 792 × 2 = 0 + 0.730 691 584;
  • 39) 0.730 691 584 × 2 = 1 + 0.461 383 168;
  • 40) 0.461 383 168 × 2 = 0 + 0.922 766 336;
  • 41) 0.922 766 336 × 2 = 1 + 0.845 532 672;
  • 42) 0.845 532 672 × 2 = 1 + 0.691 065 344;
  • 43) 0.691 065 344 × 2 = 1 + 0.382 130 688;
  • 44) 0.382 130 688 × 2 = 0 + 0.764 261 376;
  • 45) 0.764 261 376 × 2 = 1 + 0.528 522 752;
  • 46) 0.528 522 752 × 2 = 1 + 0.057 045 504;
  • 47) 0.057 045 504 × 2 = 0 + 0.114 091 008;
  • 48) 0.114 091 008 × 2 = 0 + 0.228 182 016;
  • 49) 0.228 182 016 × 2 = 0 + 0.456 364 032;
  • 50) 0.456 364 032 × 2 = 0 + 0.912 728 064;
  • 51) 0.912 728 064 × 2 = 1 + 0.825 456 128;
  • 52) 0.825 456 128 × 2 = 1 + 0.650 912 256;
  • 53) 0.650 912 256 × 2 = 1 + 0.301 824 512;
  • 54) 0.301 824 512 × 2 = 0 + 0.603 649 024;
  • 55) 0.603 649 024 × 2 = 1 + 0.207 298 048;

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:


0.135 000 186(10) =

0.0010 0010 1000 1111 0101 1111 0100 0111 1101 0010 1110 1100 0011 101(2)

6. Positive number before normalization:

0.135 000 186(10) =

0.0010 0010 1000 1111 0101 1111 0100 0111 1101 0010 1110 1100 0011 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.135 000 186(10) =

0.0010 0010 1000 1111 0101 1111 0100 0111 1101 0010 1110 1100 0011 101(2) =

0.0010 0010 1000 1111 0101 1111 0100 0111 1101 0010 1110 1100 0011 101(2) × 20 =

1.0001 0100 0111 1010 1111 1010 0011 1110 1001 0111 0110 0001 1101(2) × 2-3


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): -3

Mantissa (not normalized):
1.0001 0100 0111 1010 1111 1010 0011 1110 1001 0111 0110 0001 1101


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 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;

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) =

1020(10) =

011 1111 1100(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. 0001 0100 0111 1010 1111 1010 0011 1110 1001 0111 0110 0001 1101 =

0001 0100 0111 1010 1111 1010 0011 1110 1001 0111 0110 0001 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 1100

Mantissa (52 bits) =
0001 0100 0111 1010 1111 1010 0011 1110 1001 0111 0110 0001 1101


Decimal number -0.135 000 186 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0001 0100 0111 1010 1111 1010 0011 1110 1001 0111 0110 0001 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