0.000 000 109 985 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 109 985 6(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
0.000 000 109 985 6(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: 0.000 000 109 985 6.

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) 0.000 000 109 985 6 × 2 = 0 + 0.000 000 219 971 2;
  • 2) 0.000 000 219 971 2 × 2 = 0 + 0.000 000 439 942 4;
  • 3) 0.000 000 439 942 4 × 2 = 0 + 0.000 000 879 884 8;
  • 4) 0.000 000 879 884 8 × 2 = 0 + 0.000 001 759 769 6;
  • 5) 0.000 001 759 769 6 × 2 = 0 + 0.000 003 519 539 2;
  • 6) 0.000 003 519 539 2 × 2 = 0 + 0.000 007 039 078 4;
  • 7) 0.000 007 039 078 4 × 2 = 0 + 0.000 014 078 156 8;
  • 8) 0.000 014 078 156 8 × 2 = 0 + 0.000 028 156 313 6;
  • 9) 0.000 028 156 313 6 × 2 = 0 + 0.000 056 312 627 2;
  • 10) 0.000 056 312 627 2 × 2 = 0 + 0.000 112 625 254 4;
  • 11) 0.000 112 625 254 4 × 2 = 0 + 0.000 225 250 508 8;
  • 12) 0.000 225 250 508 8 × 2 = 0 + 0.000 450 501 017 6;
  • 13) 0.000 450 501 017 6 × 2 = 0 + 0.000 901 002 035 2;
  • 14) 0.000 901 002 035 2 × 2 = 0 + 0.001 802 004 070 4;
  • 15) 0.001 802 004 070 4 × 2 = 0 + 0.003 604 008 140 8;
  • 16) 0.003 604 008 140 8 × 2 = 0 + 0.007 208 016 281 6;
  • 17) 0.007 208 016 281 6 × 2 = 0 + 0.014 416 032 563 2;
  • 18) 0.014 416 032 563 2 × 2 = 0 + 0.028 832 065 126 4;
  • 19) 0.028 832 065 126 4 × 2 = 0 + 0.057 664 130 252 8;
  • 20) 0.057 664 130 252 8 × 2 = 0 + 0.115 328 260 505 6;
  • 21) 0.115 328 260 505 6 × 2 = 0 + 0.230 656 521 011 2;
  • 22) 0.230 656 521 011 2 × 2 = 0 + 0.461 313 042 022 4;
  • 23) 0.461 313 042 022 4 × 2 = 0 + 0.922 626 084 044 8;
  • 24) 0.922 626 084 044 8 × 2 = 1 + 0.845 252 168 089 6;
  • 25) 0.845 252 168 089 6 × 2 = 1 + 0.690 504 336 179 2;
  • 26) 0.690 504 336 179 2 × 2 = 1 + 0.381 008 672 358 4;
  • 27) 0.381 008 672 358 4 × 2 = 0 + 0.762 017 344 716 8;
  • 28) 0.762 017 344 716 8 × 2 = 1 + 0.524 034 689 433 6;
  • 29) 0.524 034 689 433 6 × 2 = 1 + 0.048 069 378 867 2;
  • 30) 0.048 069 378 867 2 × 2 = 0 + 0.096 138 757 734 4;
  • 31) 0.096 138 757 734 4 × 2 = 0 + 0.192 277 515 468 8;
  • 32) 0.192 277 515 468 8 × 2 = 0 + 0.384 555 030 937 6;
  • 33) 0.384 555 030 937 6 × 2 = 0 + 0.769 110 061 875 2;
  • 34) 0.769 110 061 875 2 × 2 = 1 + 0.538 220 123 750 4;
  • 35) 0.538 220 123 750 4 × 2 = 1 + 0.076 440 247 500 8;
  • 36) 0.076 440 247 500 8 × 2 = 0 + 0.152 880 495 001 6;
  • 37) 0.152 880 495 001 6 × 2 = 0 + 0.305 760 990 003 2;
  • 38) 0.305 760 990 003 2 × 2 = 0 + 0.611 521 980 006 4;
  • 39) 0.611 521 980 006 4 × 2 = 1 + 0.223 043 960 012 8;
  • 40) 0.223 043 960 012 8 × 2 = 0 + 0.446 087 920 025 6;
  • 41) 0.446 087 920 025 6 × 2 = 0 + 0.892 175 840 051 2;
  • 42) 0.892 175 840 051 2 × 2 = 1 + 0.784 351 680 102 4;
  • 43) 0.784 351 680 102 4 × 2 = 1 + 0.568 703 360 204 8;
  • 44) 0.568 703 360 204 8 × 2 = 1 + 0.137 406 720 409 6;
  • 45) 0.137 406 720 409 6 × 2 = 0 + 0.274 813 440 819 2;
  • 46) 0.274 813 440 819 2 × 2 = 0 + 0.549 626 881 638 4;
  • 47) 0.549 626 881 638 4 × 2 = 1 + 0.099 253 763 276 8;

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:


0.000 000 109 985 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0010 0111 001(2)

5. Positive number before normalization:

0.000 000 109 985 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0010 0111 001(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:


0.000 000 109 985 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0010 0111 001(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0010 0111 001(2) × 20 =

1.1101 1000 0110 0010 0111 001(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 0010 0111 001


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 0001 0011 1001 =

110 1100 0011 0001 0011 1001


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 0001 0011 1001


Decimal number 0.000 000 109 985 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 0001 0011 1001

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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