0.000 000 109 995 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 109 995 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
0.000 000 109 995 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. 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: 0.000 000 109 995 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) 0.000 000 109 995 3 × 2 = 0 + 0.000 000 219 990 6;
  • 2) 0.000 000 219 990 6 × 2 = 0 + 0.000 000 439 981 2;
  • 3) 0.000 000 439 981 2 × 2 = 0 + 0.000 000 879 962 4;
  • 4) 0.000 000 879 962 4 × 2 = 0 + 0.000 001 759 924 8;
  • 5) 0.000 001 759 924 8 × 2 = 0 + 0.000 003 519 849 6;
  • 6) 0.000 003 519 849 6 × 2 = 0 + 0.000 007 039 699 2;
  • 7) 0.000 007 039 699 2 × 2 = 0 + 0.000 014 079 398 4;
  • 8) 0.000 014 079 398 4 × 2 = 0 + 0.000 028 158 796 8;
  • 9) 0.000 028 158 796 8 × 2 = 0 + 0.000 056 317 593 6;
  • 10) 0.000 056 317 593 6 × 2 = 0 + 0.000 112 635 187 2;
  • 11) 0.000 112 635 187 2 × 2 = 0 + 0.000 225 270 374 4;
  • 12) 0.000 225 270 374 4 × 2 = 0 + 0.000 450 540 748 8;
  • 13) 0.000 450 540 748 8 × 2 = 0 + 0.000 901 081 497 6;
  • 14) 0.000 901 081 497 6 × 2 = 0 + 0.001 802 162 995 2;
  • 15) 0.001 802 162 995 2 × 2 = 0 + 0.003 604 325 990 4;
  • 16) 0.003 604 325 990 4 × 2 = 0 + 0.007 208 651 980 8;
  • 17) 0.007 208 651 980 8 × 2 = 0 + 0.014 417 303 961 6;
  • 18) 0.014 417 303 961 6 × 2 = 0 + 0.028 834 607 923 2;
  • 19) 0.028 834 607 923 2 × 2 = 0 + 0.057 669 215 846 4;
  • 20) 0.057 669 215 846 4 × 2 = 0 + 0.115 338 431 692 8;
  • 21) 0.115 338 431 692 8 × 2 = 0 + 0.230 676 863 385 6;
  • 22) 0.230 676 863 385 6 × 2 = 0 + 0.461 353 726 771 2;
  • 23) 0.461 353 726 771 2 × 2 = 0 + 0.922 707 453 542 4;
  • 24) 0.922 707 453 542 4 × 2 = 1 + 0.845 414 907 084 8;
  • 25) 0.845 414 907 084 8 × 2 = 1 + 0.690 829 814 169 6;
  • 26) 0.690 829 814 169 6 × 2 = 1 + 0.381 659 628 339 2;
  • 27) 0.381 659 628 339 2 × 2 = 0 + 0.763 319 256 678 4;
  • 28) 0.763 319 256 678 4 × 2 = 1 + 0.526 638 513 356 8;
  • 29) 0.526 638 513 356 8 × 2 = 1 + 0.053 277 026 713 6;
  • 30) 0.053 277 026 713 6 × 2 = 0 + 0.106 554 053 427 2;
  • 31) 0.106 554 053 427 2 × 2 = 0 + 0.213 108 106 854 4;
  • 32) 0.213 108 106 854 4 × 2 = 0 + 0.426 216 213 708 8;
  • 33) 0.426 216 213 708 8 × 2 = 0 + 0.852 432 427 417 6;
  • 34) 0.852 432 427 417 6 × 2 = 1 + 0.704 864 854 835 2;
  • 35) 0.704 864 854 835 2 × 2 = 1 + 0.409 729 709 670 4;
  • 36) 0.409 729 709 670 4 × 2 = 0 + 0.819 459 419 340 8;
  • 37) 0.819 459 419 340 8 × 2 = 1 + 0.638 918 838 681 6;
  • 38) 0.638 918 838 681 6 × 2 = 1 + 0.277 837 677 363 2;
  • 39) 0.277 837 677 363 2 × 2 = 0 + 0.555 675 354 726 4;
  • 40) 0.555 675 354 726 4 × 2 = 1 + 0.111 350 709 452 8;
  • 41) 0.111 350 709 452 8 × 2 = 0 + 0.222 701 418 905 6;
  • 42) 0.222 701 418 905 6 × 2 = 0 + 0.445 402 837 811 2;
  • 43) 0.445 402 837 811 2 × 2 = 0 + 0.890 805 675 622 4;
  • 44) 0.890 805 675 622 4 × 2 = 1 + 0.781 611 351 244 8;
  • 45) 0.781 611 351 244 8 × 2 = 1 + 0.563 222 702 489 6;
  • 46) 0.563 222 702 489 6 × 2 = 1 + 0.126 445 404 979 2;
  • 47) 0.126 445 404 979 2 × 2 = 0 + 0.252 890 809 958 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).


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:


0.000 000 109 995 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1101 0001 110(2)

5. Positive number before normalization:

0.000 000 109 995 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1101 0001 110(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:


0.000 000 109 995 3(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1101 0001 110(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1101 0001 110(2) × 20 =

1.1101 1000 0110 1101 0001 110(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 0110 1101 0001 110


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 0110 1000 1110 =

110 1100 0011 0110 1000 1110


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 0110 1000 1110


Decimal number 0.000 000 109 995 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 0110 1000 1110

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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