-0.000 000 000 311 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 311(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 311(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 311| = 0.000 000 000 311


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: 0.000 000 000 311.

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) 0.000 000 000 311 × 2 = 0 + 0.000 000 000 622;
  • 2) 0.000 000 000 622 × 2 = 0 + 0.000 000 001 244;
  • 3) 0.000 000 001 244 × 2 = 0 + 0.000 000 002 488;
  • 4) 0.000 000 002 488 × 2 = 0 + 0.000 000 004 976;
  • 5) 0.000 000 004 976 × 2 = 0 + 0.000 000 009 952;
  • 6) 0.000 000 009 952 × 2 = 0 + 0.000 000 019 904;
  • 7) 0.000 000 019 904 × 2 = 0 + 0.000 000 039 808;
  • 8) 0.000 000 039 808 × 2 = 0 + 0.000 000 079 616;
  • 9) 0.000 000 079 616 × 2 = 0 + 0.000 000 159 232;
  • 10) 0.000 000 159 232 × 2 = 0 + 0.000 000 318 464;
  • 11) 0.000 000 318 464 × 2 = 0 + 0.000 000 636 928;
  • 12) 0.000 000 636 928 × 2 = 0 + 0.000 001 273 856;
  • 13) 0.000 001 273 856 × 2 = 0 + 0.000 002 547 712;
  • 14) 0.000 002 547 712 × 2 = 0 + 0.000 005 095 424;
  • 15) 0.000 005 095 424 × 2 = 0 + 0.000 010 190 848;
  • 16) 0.000 010 190 848 × 2 = 0 + 0.000 020 381 696;
  • 17) 0.000 020 381 696 × 2 = 0 + 0.000 040 763 392;
  • 18) 0.000 040 763 392 × 2 = 0 + 0.000 081 526 784;
  • 19) 0.000 081 526 784 × 2 = 0 + 0.000 163 053 568;
  • 20) 0.000 163 053 568 × 2 = 0 + 0.000 326 107 136;
  • 21) 0.000 326 107 136 × 2 = 0 + 0.000 652 214 272;
  • 22) 0.000 652 214 272 × 2 = 0 + 0.001 304 428 544;
  • 23) 0.001 304 428 544 × 2 = 0 + 0.002 608 857 088;
  • 24) 0.002 608 857 088 × 2 = 0 + 0.005 217 714 176;
  • 25) 0.005 217 714 176 × 2 = 0 + 0.010 435 428 352;
  • 26) 0.010 435 428 352 × 2 = 0 + 0.020 870 856 704;
  • 27) 0.020 870 856 704 × 2 = 0 + 0.041 741 713 408;
  • 28) 0.041 741 713 408 × 2 = 0 + 0.083 483 426 816;
  • 29) 0.083 483 426 816 × 2 = 0 + 0.166 966 853 632;
  • 30) 0.166 966 853 632 × 2 = 0 + 0.333 933 707 264;
  • 31) 0.333 933 707 264 × 2 = 0 + 0.667 867 414 528;
  • 32) 0.667 867 414 528 × 2 = 1 + 0.335 734 829 056;
  • 33) 0.335 734 829 056 × 2 = 0 + 0.671 469 658 112;
  • 34) 0.671 469 658 112 × 2 = 1 + 0.342 939 316 224;
  • 35) 0.342 939 316 224 × 2 = 0 + 0.685 878 632 448;
  • 36) 0.685 878 632 448 × 2 = 1 + 0.371 757 264 896;
  • 37) 0.371 757 264 896 × 2 = 0 + 0.743 514 529 792;
  • 38) 0.743 514 529 792 × 2 = 1 + 0.487 029 059 584;
  • 39) 0.487 029 059 584 × 2 = 0 + 0.974 058 119 168;
  • 40) 0.974 058 119 168 × 2 = 1 + 0.948 116 238 336;
  • 41) 0.948 116 238 336 × 2 = 1 + 0.896 232 476 672;
  • 42) 0.896 232 476 672 × 2 = 1 + 0.792 464 953 344;
  • 43) 0.792 464 953 344 × 2 = 1 + 0.584 929 906 688;
  • 44) 0.584 929 906 688 × 2 = 1 + 0.169 859 813 376;
  • 45) 0.169 859 813 376 × 2 = 0 + 0.339 719 626 752;
  • 46) 0.339 719 626 752 × 2 = 0 + 0.679 439 253 504;
  • 47) 0.679 439 253 504 × 2 = 1 + 0.358 878 507 008;
  • 48) 0.358 878 507 008 × 2 = 0 + 0.717 757 014 016;
  • 49) 0.717 757 014 016 × 2 = 1 + 0.435 514 028 032;
  • 50) 0.435 514 028 032 × 2 = 0 + 0.871 028 056 064;
  • 51) 0.871 028 056 064 × 2 = 1 + 0.742 056 112 128;
  • 52) 0.742 056 112 128 × 2 = 1 + 0.484 112 224 256;
  • 53) 0.484 112 224 256 × 2 = 0 + 0.968 224 448 512;
  • 54) 0.968 224 448 512 × 2 = 1 + 0.936 448 897 024;
  • 55) 0.936 448 897 024 × 2 = 1 + 0.872 897 794 048;

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:


0.000 000 000 311(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0101 0101 1111 0010 1011 011(2)

6. Positive number before normalization:

0.000 000 000 311(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0101 0101 1111 0010 1011 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 311(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0101 0101 1111 0010 1011 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0101 0101 1111 0010 1011 011(2) × 20 =

1.0101 0101 1111 0010 1011 011(2) × 2-32


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -32

Mantissa (not normalized):
1.0101 0101 1111 0010 1011 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

95(10) =

0101 1111(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 010 1010 1111 1001 0101 1011 =

010 1010 1111 1001 0101 1011


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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

Exponent (8 bits) =
0101 1111

Mantissa (23 bits) =
010 1010 1111 1001 0101 1011


Decimal number -0.000 000 000 311 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0101 1111 - 010 1010 1111 1001 0101 1011

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

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

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111