0.000 000 109 992 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 109 992 8(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
0.000 000 109 992 8(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: 0.000 000 109 992 8.

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) 0.000 000 109 992 8 × 2 = 0 + 0.000 000 219 985 6;
  • 2) 0.000 000 219 985 6 × 2 = 0 + 0.000 000 439 971 2;
  • 3) 0.000 000 439 971 2 × 2 = 0 + 0.000 000 879 942 4;
  • 4) 0.000 000 879 942 4 × 2 = 0 + 0.000 001 759 884 8;
  • 5) 0.000 001 759 884 8 × 2 = 0 + 0.000 003 519 769 6;
  • 6) 0.000 003 519 769 6 × 2 = 0 + 0.000 007 039 539 2;
  • 7) 0.000 007 039 539 2 × 2 = 0 + 0.000 014 079 078 4;
  • 8) 0.000 014 079 078 4 × 2 = 0 + 0.000 028 158 156 8;
  • 9) 0.000 028 158 156 8 × 2 = 0 + 0.000 056 316 313 6;
  • 10) 0.000 056 316 313 6 × 2 = 0 + 0.000 112 632 627 2;
  • 11) 0.000 112 632 627 2 × 2 = 0 + 0.000 225 265 254 4;
  • 12) 0.000 225 265 254 4 × 2 = 0 + 0.000 450 530 508 8;
  • 13) 0.000 450 530 508 8 × 2 = 0 + 0.000 901 061 017 6;
  • 14) 0.000 901 061 017 6 × 2 = 0 + 0.001 802 122 035 2;
  • 15) 0.001 802 122 035 2 × 2 = 0 + 0.003 604 244 070 4;
  • 16) 0.003 604 244 070 4 × 2 = 0 + 0.007 208 488 140 8;
  • 17) 0.007 208 488 140 8 × 2 = 0 + 0.014 416 976 281 6;
  • 18) 0.014 416 976 281 6 × 2 = 0 + 0.028 833 952 563 2;
  • 19) 0.028 833 952 563 2 × 2 = 0 + 0.057 667 905 126 4;
  • 20) 0.057 667 905 126 4 × 2 = 0 + 0.115 335 810 252 8;
  • 21) 0.115 335 810 252 8 × 2 = 0 + 0.230 671 620 505 6;
  • 22) 0.230 671 620 505 6 × 2 = 0 + 0.461 343 241 011 2;
  • 23) 0.461 343 241 011 2 × 2 = 0 + 0.922 686 482 022 4;
  • 24) 0.922 686 482 022 4 × 2 = 1 + 0.845 372 964 044 8;
  • 25) 0.845 372 964 044 8 × 2 = 1 + 0.690 745 928 089 6;
  • 26) 0.690 745 928 089 6 × 2 = 1 + 0.381 491 856 179 2;
  • 27) 0.381 491 856 179 2 × 2 = 0 + 0.762 983 712 358 4;
  • 28) 0.762 983 712 358 4 × 2 = 1 + 0.525 967 424 716 8;
  • 29) 0.525 967 424 716 8 × 2 = 1 + 0.051 934 849 433 6;
  • 30) 0.051 934 849 433 6 × 2 = 0 + 0.103 869 698 867 2;
  • 31) 0.103 869 698 867 2 × 2 = 0 + 0.207 739 397 734 4;
  • 32) 0.207 739 397 734 4 × 2 = 0 + 0.415 478 795 468 8;
  • 33) 0.415 478 795 468 8 × 2 = 0 + 0.830 957 590 937 6;
  • 34) 0.830 957 590 937 6 × 2 = 1 + 0.661 915 181 875 2;
  • 35) 0.661 915 181 875 2 × 2 = 1 + 0.323 830 363 750 4;
  • 36) 0.323 830 363 750 4 × 2 = 0 + 0.647 660 727 500 8;
  • 37) 0.647 660 727 500 8 × 2 = 1 + 0.295 321 455 001 6;
  • 38) 0.295 321 455 001 6 × 2 = 0 + 0.590 642 910 003 2;
  • 39) 0.590 642 910 003 2 × 2 = 1 + 0.181 285 820 006 4;
  • 40) 0.181 285 820 006 4 × 2 = 0 + 0.362 571 640 012 8;
  • 41) 0.362 571 640 012 8 × 2 = 0 + 0.725 143 280 025 6;
  • 42) 0.725 143 280 025 6 × 2 = 1 + 0.450 286 560 051 2;
  • 43) 0.450 286 560 051 2 × 2 = 0 + 0.900 573 120 102 4;
  • 44) 0.900 573 120 102 4 × 2 = 1 + 0.801 146 240 204 8;
  • 45) 0.801 146 240 204 8 × 2 = 1 + 0.602 292 480 409 6;
  • 46) 0.602 292 480 409 6 × 2 = 1 + 0.204 584 960 819 2;
  • 47) 0.204 584 960 819 2 × 2 = 0 + 0.409 169 921 638 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:


0.000 000 109 992 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1010 0101 110(2)

5. Positive number before normalization:

0.000 000 109 992 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1010 0101 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:


0.000 000 109 992 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1010 0101 110(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 1010 0101 110(2) × 20 =

1.1101 1000 0110 1010 0101 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 1010 0101 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 0101 0010 1110 =

110 1100 0011 0101 0010 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 0101 0010 1110


Decimal number 0.000 000 109 992 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 0101 0010 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