-0.000 000 000 881 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 881(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
-0.000 000 000 881(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:

|-0.000 000 000 881| = 0.000 000 000 881


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: 0.000 000 000 881.

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) 0.000 000 000 881 × 2 = 0 + 0.000 000 001 762;
  • 2) 0.000 000 001 762 × 2 = 0 + 0.000 000 003 524;
  • 3) 0.000 000 003 524 × 2 = 0 + 0.000 000 007 048;
  • 4) 0.000 000 007 048 × 2 = 0 + 0.000 000 014 096;
  • 5) 0.000 000 014 096 × 2 = 0 + 0.000 000 028 192;
  • 6) 0.000 000 028 192 × 2 = 0 + 0.000 000 056 384;
  • 7) 0.000 000 056 384 × 2 = 0 + 0.000 000 112 768;
  • 8) 0.000 000 112 768 × 2 = 0 + 0.000 000 225 536;
  • 9) 0.000 000 225 536 × 2 = 0 + 0.000 000 451 072;
  • 10) 0.000 000 451 072 × 2 = 0 + 0.000 000 902 144;
  • 11) 0.000 000 902 144 × 2 = 0 + 0.000 001 804 288;
  • 12) 0.000 001 804 288 × 2 = 0 + 0.000 003 608 576;
  • 13) 0.000 003 608 576 × 2 = 0 + 0.000 007 217 152;
  • 14) 0.000 007 217 152 × 2 = 0 + 0.000 014 434 304;
  • 15) 0.000 014 434 304 × 2 = 0 + 0.000 028 868 608;
  • 16) 0.000 028 868 608 × 2 = 0 + 0.000 057 737 216;
  • 17) 0.000 057 737 216 × 2 = 0 + 0.000 115 474 432;
  • 18) 0.000 115 474 432 × 2 = 0 + 0.000 230 948 864;
  • 19) 0.000 230 948 864 × 2 = 0 + 0.000 461 897 728;
  • 20) 0.000 461 897 728 × 2 = 0 + 0.000 923 795 456;
  • 21) 0.000 923 795 456 × 2 = 0 + 0.001 847 590 912;
  • 22) 0.001 847 590 912 × 2 = 0 + 0.003 695 181 824;
  • 23) 0.003 695 181 824 × 2 = 0 + 0.007 390 363 648;
  • 24) 0.007 390 363 648 × 2 = 0 + 0.014 780 727 296;
  • 25) 0.014 780 727 296 × 2 = 0 + 0.029 561 454 592;
  • 26) 0.029 561 454 592 × 2 = 0 + 0.059 122 909 184;
  • 27) 0.059 122 909 184 × 2 = 0 + 0.118 245 818 368;
  • 28) 0.118 245 818 368 × 2 = 0 + 0.236 491 636 736;
  • 29) 0.236 491 636 736 × 2 = 0 + 0.472 983 273 472;
  • 30) 0.472 983 273 472 × 2 = 0 + 0.945 966 546 944;
  • 31) 0.945 966 546 944 × 2 = 1 + 0.891 933 093 888;
  • 32) 0.891 933 093 888 × 2 = 1 + 0.783 866 187 776;
  • 33) 0.783 866 187 776 × 2 = 1 + 0.567 732 375 552;
  • 34) 0.567 732 375 552 × 2 = 1 + 0.135 464 751 104;
  • 35) 0.135 464 751 104 × 2 = 0 + 0.270 929 502 208;
  • 36) 0.270 929 502 208 × 2 = 0 + 0.541 859 004 416;
  • 37) 0.541 859 004 416 × 2 = 1 + 0.083 718 008 832;
  • 38) 0.083 718 008 832 × 2 = 0 + 0.167 436 017 664;
  • 39) 0.167 436 017 664 × 2 = 0 + 0.334 872 035 328;
  • 40) 0.334 872 035 328 × 2 = 0 + 0.669 744 070 656;
  • 41) 0.669 744 070 656 × 2 = 1 + 0.339 488 141 312;
  • 42) 0.339 488 141 312 × 2 = 0 + 0.678 976 282 624;
  • 43) 0.678 976 282 624 × 2 = 1 + 0.357 952 565 248;
  • 44) 0.357 952 565 248 × 2 = 0 + 0.715 905 130 496;
  • 45) 0.715 905 130 496 × 2 = 1 + 0.431 810 260 992;
  • 46) 0.431 810 260 992 × 2 = 0 + 0.863 620 521 984;
  • 47) 0.863 620 521 984 × 2 = 1 + 0.727 241 043 968;
  • 48) 0.727 241 043 968 × 2 = 1 + 0.454 482 087 936;
  • 49) 0.454 482 087 936 × 2 = 0 + 0.908 964 175 872;
  • 50) 0.908 964 175 872 × 2 = 1 + 0.817 928 351 744;
  • 51) 0.817 928 351 744 × 2 = 1 + 0.635 856 703 488;
  • 52) 0.635 856 703 488 × 2 = 1 + 0.271 713 406 976;
  • 53) 0.271 713 406 976 × 2 = 0 + 0.543 426 813 952;
  • 54) 0.543 426 813 952 × 2 = 1 + 0.086 853 627 904;

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:


0.000 000 000 881(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 1000 1010 1011 0111 01(2)

6. Positive number before normalization:

0.000 000 000 881(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 1000 1010 1011 0111 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 881(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 1000 1010 1011 0111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1100 1000 1010 1011 0111 01(2) × 20 =

1.1110 0100 0101 0101 1011 101(2) × 2-31


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): -31

Mantissa (not normalized):
1.1110 0100 0101 0101 1011 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

96(10) =

0110 0000(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. 111 0010 0010 1010 1101 1101 =

111 0010 0010 1010 1101 1101


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) =
0110 0000

Mantissa (23 bits) =
111 0010 0010 1010 1101 1101


Decimal number -0.000 000 000 881 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 111 0010 0010 1010 1101 1101

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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