-0.000 000 111 111 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 111 111 3(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
-0.000 000 111 111 3(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:

|-0.000 000 111 111 3| = 0.000 000 111 111 3


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: 0.000 000 111 111 3.

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) 0.000 000 111 111 3 × 2 = 0 + 0.000 000 222 222 6;
  • 2) 0.000 000 222 222 6 × 2 = 0 + 0.000 000 444 445 2;
  • 3) 0.000 000 444 445 2 × 2 = 0 + 0.000 000 888 890 4;
  • 4) 0.000 000 888 890 4 × 2 = 0 + 0.000 001 777 780 8;
  • 5) 0.000 001 777 780 8 × 2 = 0 + 0.000 003 555 561 6;
  • 6) 0.000 003 555 561 6 × 2 = 0 + 0.000 007 111 123 2;
  • 7) 0.000 007 111 123 2 × 2 = 0 + 0.000 014 222 246 4;
  • 8) 0.000 014 222 246 4 × 2 = 0 + 0.000 028 444 492 8;
  • 9) 0.000 028 444 492 8 × 2 = 0 + 0.000 056 888 985 6;
  • 10) 0.000 056 888 985 6 × 2 = 0 + 0.000 113 777 971 2;
  • 11) 0.000 113 777 971 2 × 2 = 0 + 0.000 227 555 942 4;
  • 12) 0.000 227 555 942 4 × 2 = 0 + 0.000 455 111 884 8;
  • 13) 0.000 455 111 884 8 × 2 = 0 + 0.000 910 223 769 6;
  • 14) 0.000 910 223 769 6 × 2 = 0 + 0.001 820 447 539 2;
  • 15) 0.001 820 447 539 2 × 2 = 0 + 0.003 640 895 078 4;
  • 16) 0.003 640 895 078 4 × 2 = 0 + 0.007 281 790 156 8;
  • 17) 0.007 281 790 156 8 × 2 = 0 + 0.014 563 580 313 6;
  • 18) 0.014 563 580 313 6 × 2 = 0 + 0.029 127 160 627 2;
  • 19) 0.029 127 160 627 2 × 2 = 0 + 0.058 254 321 254 4;
  • 20) 0.058 254 321 254 4 × 2 = 0 + 0.116 508 642 508 8;
  • 21) 0.116 508 642 508 8 × 2 = 0 + 0.233 017 285 017 6;
  • 22) 0.233 017 285 017 6 × 2 = 0 + 0.466 034 570 035 2;
  • 23) 0.466 034 570 035 2 × 2 = 0 + 0.932 069 140 070 4;
  • 24) 0.932 069 140 070 4 × 2 = 1 + 0.864 138 280 140 8;
  • 25) 0.864 138 280 140 8 × 2 = 1 + 0.728 276 560 281 6;
  • 26) 0.728 276 560 281 6 × 2 = 1 + 0.456 553 120 563 2;
  • 27) 0.456 553 120 563 2 × 2 = 0 + 0.913 106 241 126 4;
  • 28) 0.913 106 241 126 4 × 2 = 1 + 0.826 212 482 252 8;
  • 29) 0.826 212 482 252 8 × 2 = 1 + 0.652 424 964 505 6;
  • 30) 0.652 424 964 505 6 × 2 = 1 + 0.304 849 929 011 2;
  • 31) 0.304 849 929 011 2 × 2 = 0 + 0.609 699 858 022 4;
  • 32) 0.609 699 858 022 4 × 2 = 1 + 0.219 399 716 044 8;
  • 33) 0.219 399 716 044 8 × 2 = 0 + 0.438 799 432 089 6;
  • 34) 0.438 799 432 089 6 × 2 = 0 + 0.877 598 864 179 2;
  • 35) 0.877 598 864 179 2 × 2 = 1 + 0.755 197 728 358 4;
  • 36) 0.755 197 728 358 4 × 2 = 1 + 0.510 395 456 716 8;
  • 37) 0.510 395 456 716 8 × 2 = 1 + 0.020 790 913 433 6;
  • 38) 0.020 790 913 433 6 × 2 = 0 + 0.041 581 826 867 2;
  • 39) 0.041 581 826 867 2 × 2 = 0 + 0.083 163 653 734 4;
  • 40) 0.083 163 653 734 4 × 2 = 0 + 0.166 327 307 468 8;
  • 41) 0.166 327 307 468 8 × 2 = 0 + 0.332 654 614 937 6;
  • 42) 0.332 654 614 937 6 × 2 = 0 + 0.665 309 229 875 2;
  • 43) 0.665 309 229 875 2 × 2 = 1 + 0.330 618 459 750 4;
  • 44) 0.330 618 459 750 4 × 2 = 0 + 0.661 236 919 500 8;
  • 45) 0.661 236 919 500 8 × 2 = 1 + 0.322 473 839 001 6;
  • 46) 0.322 473 839 001 6 × 2 = 0 + 0.644 947 678 003 2;
  • 47) 0.644 947 678 003 2 × 2 = 1 + 0.289 895 356 006 4;

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:


0.000 000 111 111 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 1000 0010 101(2)

6. Positive number before normalization:

0.000 000 111 111 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 1000 0010 101(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 111 111 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 1000 0010 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1101 0011 1000 0010 101(2) × 20 =

1.1101 1101 0011 1000 0010 101(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1101 0011 1000 0010 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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. 110 1110 1001 1100 0001 0101 =

110 1110 1001 1100 0001 0101


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 0111

Mantissa (23 bits) =
110 1110 1001 1100 0001 0101


Decimal number -0.000 000 111 111 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 110 1110 1001 1100 0001 0101

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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