0.000 000 110 011 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 011 7(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
0.000 000 110 011 7(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: 0.000 000 110 011 7.

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) 0.000 000 110 011 7 × 2 = 0 + 0.000 000 220 023 4;
  • 2) 0.000 000 220 023 4 × 2 = 0 + 0.000 000 440 046 8;
  • 3) 0.000 000 440 046 8 × 2 = 0 + 0.000 000 880 093 6;
  • 4) 0.000 000 880 093 6 × 2 = 0 + 0.000 001 760 187 2;
  • 5) 0.000 001 760 187 2 × 2 = 0 + 0.000 003 520 374 4;
  • 6) 0.000 003 520 374 4 × 2 = 0 + 0.000 007 040 748 8;
  • 7) 0.000 007 040 748 8 × 2 = 0 + 0.000 014 081 497 6;
  • 8) 0.000 014 081 497 6 × 2 = 0 + 0.000 028 162 995 2;
  • 9) 0.000 028 162 995 2 × 2 = 0 + 0.000 056 325 990 4;
  • 10) 0.000 056 325 990 4 × 2 = 0 + 0.000 112 651 980 8;
  • 11) 0.000 112 651 980 8 × 2 = 0 + 0.000 225 303 961 6;
  • 12) 0.000 225 303 961 6 × 2 = 0 + 0.000 450 607 923 2;
  • 13) 0.000 450 607 923 2 × 2 = 0 + 0.000 901 215 846 4;
  • 14) 0.000 901 215 846 4 × 2 = 0 + 0.001 802 431 692 8;
  • 15) 0.001 802 431 692 8 × 2 = 0 + 0.003 604 863 385 6;
  • 16) 0.003 604 863 385 6 × 2 = 0 + 0.007 209 726 771 2;
  • 17) 0.007 209 726 771 2 × 2 = 0 + 0.014 419 453 542 4;
  • 18) 0.014 419 453 542 4 × 2 = 0 + 0.028 838 907 084 8;
  • 19) 0.028 838 907 084 8 × 2 = 0 + 0.057 677 814 169 6;
  • 20) 0.057 677 814 169 6 × 2 = 0 + 0.115 355 628 339 2;
  • 21) 0.115 355 628 339 2 × 2 = 0 + 0.230 711 256 678 4;
  • 22) 0.230 711 256 678 4 × 2 = 0 + 0.461 422 513 356 8;
  • 23) 0.461 422 513 356 8 × 2 = 0 + 0.922 845 026 713 6;
  • 24) 0.922 845 026 713 6 × 2 = 1 + 0.845 690 053 427 2;
  • 25) 0.845 690 053 427 2 × 2 = 1 + 0.691 380 106 854 4;
  • 26) 0.691 380 106 854 4 × 2 = 1 + 0.382 760 213 708 8;
  • 27) 0.382 760 213 708 8 × 2 = 0 + 0.765 520 427 417 6;
  • 28) 0.765 520 427 417 6 × 2 = 1 + 0.531 040 854 835 2;
  • 29) 0.531 040 854 835 2 × 2 = 1 + 0.062 081 709 670 4;
  • 30) 0.062 081 709 670 4 × 2 = 0 + 0.124 163 419 340 8;
  • 31) 0.124 163 419 340 8 × 2 = 0 + 0.248 326 838 681 6;
  • 32) 0.248 326 838 681 6 × 2 = 0 + 0.496 653 677 363 2;
  • 33) 0.496 653 677 363 2 × 2 = 0 + 0.993 307 354 726 4;
  • 34) 0.993 307 354 726 4 × 2 = 1 + 0.986 614 709 452 8;
  • 35) 0.986 614 709 452 8 × 2 = 1 + 0.973 229 418 905 6;
  • 36) 0.973 229 418 905 6 × 2 = 1 + 0.946 458 837 811 2;
  • 37) 0.946 458 837 811 2 × 2 = 1 + 0.892 917 675 622 4;
  • 38) 0.892 917 675 622 4 × 2 = 1 + 0.785 835 351 244 8;
  • 39) 0.785 835 351 244 8 × 2 = 1 + 0.571 670 702 489 6;
  • 40) 0.571 670 702 489 6 × 2 = 1 + 0.143 341 404 979 2;
  • 41) 0.143 341 404 979 2 × 2 = 0 + 0.286 682 809 958 4;
  • 42) 0.286 682 809 958 4 × 2 = 0 + 0.573 365 619 916 8;
  • 43) 0.573 365 619 916 8 × 2 = 1 + 0.146 731 239 833 6;
  • 44) 0.146 731 239 833 6 × 2 = 0 + 0.293 462 479 667 2;
  • 45) 0.293 462 479 667 2 × 2 = 0 + 0.586 924 959 334 4;
  • 46) 0.586 924 959 334 4 × 2 = 1 + 0.173 849 918 668 8;
  • 47) 0.173 849 918 668 8 × 2 = 0 + 0.347 699 837 337 6;

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:


0.000 000 110 011 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1111 0010 010(2)

5. Positive number before normalization:

0.000 000 110 011 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1111 0010 010(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:


0.000 000 110 011 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1111 0010 010(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1111 0010 010(2) × 20 =

1.1101 1000 0111 1111 0010 010(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 0111 1111 0010 010


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 0011 1111 1001 0010 =

110 1100 0011 1111 1001 0010


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 0011 1111 1001 0010


Decimal number 0.000 000 110 011 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1111 1001 0010

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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