0.000 000 110 020 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 020 5(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
0.000 000 110 020 5(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 110 020 5.

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) 0.000 000 110 020 5 × 2 = 0 + 0.000 000 220 041;
  • 2) 0.000 000 220 041 × 2 = 0 + 0.000 000 440 082;
  • 3) 0.000 000 440 082 × 2 = 0 + 0.000 000 880 164;
  • 4) 0.000 000 880 164 × 2 = 0 + 0.000 001 760 328;
  • 5) 0.000 001 760 328 × 2 = 0 + 0.000 003 520 656;
  • 6) 0.000 003 520 656 × 2 = 0 + 0.000 007 041 312;
  • 7) 0.000 007 041 312 × 2 = 0 + 0.000 014 082 624;
  • 8) 0.000 014 082 624 × 2 = 0 + 0.000 028 165 248;
  • 9) 0.000 028 165 248 × 2 = 0 + 0.000 056 330 496;
  • 10) 0.000 056 330 496 × 2 = 0 + 0.000 112 660 992;
  • 11) 0.000 112 660 992 × 2 = 0 + 0.000 225 321 984;
  • 12) 0.000 225 321 984 × 2 = 0 + 0.000 450 643 968;
  • 13) 0.000 450 643 968 × 2 = 0 + 0.000 901 287 936;
  • 14) 0.000 901 287 936 × 2 = 0 + 0.001 802 575 872;
  • 15) 0.001 802 575 872 × 2 = 0 + 0.003 605 151 744;
  • 16) 0.003 605 151 744 × 2 = 0 + 0.007 210 303 488;
  • 17) 0.007 210 303 488 × 2 = 0 + 0.014 420 606 976;
  • 18) 0.014 420 606 976 × 2 = 0 + 0.028 841 213 952;
  • 19) 0.028 841 213 952 × 2 = 0 + 0.057 682 427 904;
  • 20) 0.057 682 427 904 × 2 = 0 + 0.115 364 855 808;
  • 21) 0.115 364 855 808 × 2 = 0 + 0.230 729 711 616;
  • 22) 0.230 729 711 616 × 2 = 0 + 0.461 459 423 232;
  • 23) 0.461 459 423 232 × 2 = 0 + 0.922 918 846 464;
  • 24) 0.922 918 846 464 × 2 = 1 + 0.845 837 692 928;
  • 25) 0.845 837 692 928 × 2 = 1 + 0.691 675 385 856;
  • 26) 0.691 675 385 856 × 2 = 1 + 0.383 350 771 712;
  • 27) 0.383 350 771 712 × 2 = 0 + 0.766 701 543 424;
  • 28) 0.766 701 543 424 × 2 = 1 + 0.533 403 086 848;
  • 29) 0.533 403 086 848 × 2 = 1 + 0.066 806 173 696;
  • 30) 0.066 806 173 696 × 2 = 0 + 0.133 612 347 392;
  • 31) 0.133 612 347 392 × 2 = 0 + 0.267 224 694 784;
  • 32) 0.267 224 694 784 × 2 = 0 + 0.534 449 389 568;
  • 33) 0.534 449 389 568 × 2 = 1 + 0.068 898 779 136;
  • 34) 0.068 898 779 136 × 2 = 0 + 0.137 797 558 272;
  • 35) 0.137 797 558 272 × 2 = 0 + 0.275 595 116 544;
  • 36) 0.275 595 116 544 × 2 = 0 + 0.551 190 233 088;
  • 37) 0.551 190 233 088 × 2 = 1 + 0.102 380 466 176;
  • 38) 0.102 380 466 176 × 2 = 0 + 0.204 760 932 352;
  • 39) 0.204 760 932 352 × 2 = 0 + 0.409 521 864 704;
  • 40) 0.409 521 864 704 × 2 = 0 + 0.819 043 729 408;
  • 41) 0.819 043 729 408 × 2 = 1 + 0.638 087 458 816;
  • 42) 0.638 087 458 816 × 2 = 1 + 0.276 174 917 632;
  • 43) 0.276 174 917 632 × 2 = 0 + 0.552 349 835 264;
  • 44) 0.552 349 835 264 × 2 = 1 + 0.104 699 670 528;
  • 45) 0.104 699 670 528 × 2 = 0 + 0.209 399 341 056;
  • 46) 0.209 399 341 056 × 2 = 0 + 0.418 798 682 112;
  • 47) 0.418 798 682 112 × 2 = 0 + 0.837 597 364 224;

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).


4. 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:


0.000 000 110 020 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 1000 1101 000(2)

5. Positive number before normalization:

0.000 000 110 020 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 1000 1101 000(2)

6. 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:


0.000 000 110 020 5(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 1000 1101 000(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 1000 1101 000(2) × 20 =

1.1101 1000 1000 1000 1101 000(2) × 2-24


7. 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 0 (a positive number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1000 1000 1000 1101 000


8. 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)


9. 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;

10. 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)


11. 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 1100 0100 0100 0110 1000 =

110 1100 0100 0100 0110 1000


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

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0111

Mantissa (23 bits) =
110 1100 0100 0100 0110 1000


Decimal number 0.000 000 110 020 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 0100 0110 1000

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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