-0.000 000 000 888 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 888(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
-0.000 000 000 888(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:

|-0.000 000 000 888| = 0.000 000 000 888


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: 0.000 000 000 888.

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) 0.000 000 000 888 × 2 = 0 + 0.000 000 001 776;
  • 2) 0.000 000 001 776 × 2 = 0 + 0.000 000 003 552;
  • 3) 0.000 000 003 552 × 2 = 0 + 0.000 000 007 104;
  • 4) 0.000 000 007 104 × 2 = 0 + 0.000 000 014 208;
  • 5) 0.000 000 014 208 × 2 = 0 + 0.000 000 028 416;
  • 6) 0.000 000 028 416 × 2 = 0 + 0.000 000 056 832;
  • 7) 0.000 000 056 832 × 2 = 0 + 0.000 000 113 664;
  • 8) 0.000 000 113 664 × 2 = 0 + 0.000 000 227 328;
  • 9) 0.000 000 227 328 × 2 = 0 + 0.000 000 454 656;
  • 10) 0.000 000 454 656 × 2 = 0 + 0.000 000 909 312;
  • 11) 0.000 000 909 312 × 2 = 0 + 0.000 001 818 624;
  • 12) 0.000 001 818 624 × 2 = 0 + 0.000 003 637 248;
  • 13) 0.000 003 637 248 × 2 = 0 + 0.000 007 274 496;
  • 14) 0.000 007 274 496 × 2 = 0 + 0.000 014 548 992;
  • 15) 0.000 014 548 992 × 2 = 0 + 0.000 029 097 984;
  • 16) 0.000 029 097 984 × 2 = 0 + 0.000 058 195 968;
  • 17) 0.000 058 195 968 × 2 = 0 + 0.000 116 391 936;
  • 18) 0.000 116 391 936 × 2 = 0 + 0.000 232 783 872;
  • 19) 0.000 232 783 872 × 2 = 0 + 0.000 465 567 744;
  • 20) 0.000 465 567 744 × 2 = 0 + 0.000 931 135 488;
  • 21) 0.000 931 135 488 × 2 = 0 + 0.001 862 270 976;
  • 22) 0.001 862 270 976 × 2 = 0 + 0.003 724 541 952;
  • 23) 0.003 724 541 952 × 2 = 0 + 0.007 449 083 904;
  • 24) 0.007 449 083 904 × 2 = 0 + 0.014 898 167 808;
  • 25) 0.014 898 167 808 × 2 = 0 + 0.029 796 335 616;
  • 26) 0.029 796 335 616 × 2 = 0 + 0.059 592 671 232;
  • 27) 0.059 592 671 232 × 2 = 0 + 0.119 185 342 464;
  • 28) 0.119 185 342 464 × 2 = 0 + 0.238 370 684 928;
  • 29) 0.238 370 684 928 × 2 = 0 + 0.476 741 369 856;
  • 30) 0.476 741 369 856 × 2 = 0 + 0.953 482 739 712;
  • 31) 0.953 482 739 712 × 2 = 1 + 0.906 965 479 424;
  • 32) 0.906 965 479 424 × 2 = 1 + 0.813 930 958 848;
  • 33) 0.813 930 958 848 × 2 = 1 + 0.627 861 917 696;
  • 34) 0.627 861 917 696 × 2 = 1 + 0.255 723 835 392;
  • 35) 0.255 723 835 392 × 2 = 0 + 0.511 447 670 784;
  • 36) 0.511 447 670 784 × 2 = 1 + 0.022 895 341 568;
  • 37) 0.022 895 341 568 × 2 = 0 + 0.045 790 683 136;
  • 38) 0.045 790 683 136 × 2 = 0 + 0.091 581 366 272;
  • 39) 0.091 581 366 272 × 2 = 0 + 0.183 162 732 544;
  • 40) 0.183 162 732 544 × 2 = 0 + 0.366 325 465 088;
  • 41) 0.366 325 465 088 × 2 = 0 + 0.732 650 930 176;
  • 42) 0.732 650 930 176 × 2 = 1 + 0.465 301 860 352;
  • 43) 0.465 301 860 352 × 2 = 0 + 0.930 603 720 704;
  • 44) 0.930 603 720 704 × 2 = 1 + 0.861 207 441 408;
  • 45) 0.861 207 441 408 × 2 = 1 + 0.722 414 882 816;
  • 46) 0.722 414 882 816 × 2 = 1 + 0.444 829 765 632;
  • 47) 0.444 829 765 632 × 2 = 0 + 0.889 659 531 264;
  • 48) 0.889 659 531 264 × 2 = 1 + 0.779 319 062 528;
  • 49) 0.779 319 062 528 × 2 = 1 + 0.558 638 125 056;
  • 50) 0.558 638 125 056 × 2 = 1 + 0.117 276 250 112;
  • 51) 0.117 276 250 112 × 2 = 0 + 0.234 552 500 224;
  • 52) 0.234 552 500 224 × 2 = 0 + 0.469 105 000 448;
  • 53) 0.469 105 000 448 × 2 = 0 + 0.938 210 000 896;
  • 54) 0.938 210 000 896 × 2 = 1 + 0.876 420 001 792;

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:


0.000 000 000 888(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1101 0000 0101 1101 1100 01(2)

6. Positive number before normalization:

0.000 000 000 888(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1101 0000 0101 1101 1100 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:


0.000 000 000 888(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1101 0000 0101 1101 1100 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1101 0000 0101 1101 1100 01(2) × 20 =

1.1110 1000 0010 1110 1110 001(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 1000 0010 1110 1110 001


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 0100 0001 0111 0111 0001 =

111 0100 0001 0111 0111 0001


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 0100 0001 0111 0111 0001


Decimal number -0.000 000 000 888 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 111 0100 0001 0111 0111 0001

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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