0.000 000 110 007 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 007 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
0.000 000 110 007 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: 0.000 000 110 007 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) 0.000 000 110 007 7 × 2 = 0 + 0.000 000 220 015 4;
  • 2) 0.000 000 220 015 4 × 2 = 0 + 0.000 000 440 030 8;
  • 3) 0.000 000 440 030 8 × 2 = 0 + 0.000 000 880 061 6;
  • 4) 0.000 000 880 061 6 × 2 = 0 + 0.000 001 760 123 2;
  • 5) 0.000 001 760 123 2 × 2 = 0 + 0.000 003 520 246 4;
  • 6) 0.000 003 520 246 4 × 2 = 0 + 0.000 007 040 492 8;
  • 7) 0.000 007 040 492 8 × 2 = 0 + 0.000 014 080 985 6;
  • 8) 0.000 014 080 985 6 × 2 = 0 + 0.000 028 161 971 2;
  • 9) 0.000 028 161 971 2 × 2 = 0 + 0.000 056 323 942 4;
  • 10) 0.000 056 323 942 4 × 2 = 0 + 0.000 112 647 884 8;
  • 11) 0.000 112 647 884 8 × 2 = 0 + 0.000 225 295 769 6;
  • 12) 0.000 225 295 769 6 × 2 = 0 + 0.000 450 591 539 2;
  • 13) 0.000 450 591 539 2 × 2 = 0 + 0.000 901 183 078 4;
  • 14) 0.000 901 183 078 4 × 2 = 0 + 0.001 802 366 156 8;
  • 15) 0.001 802 366 156 8 × 2 = 0 + 0.003 604 732 313 6;
  • 16) 0.003 604 732 313 6 × 2 = 0 + 0.007 209 464 627 2;
  • 17) 0.007 209 464 627 2 × 2 = 0 + 0.014 418 929 254 4;
  • 18) 0.014 418 929 254 4 × 2 = 0 + 0.028 837 858 508 8;
  • 19) 0.028 837 858 508 8 × 2 = 0 + 0.057 675 717 017 6;
  • 20) 0.057 675 717 017 6 × 2 = 0 + 0.115 351 434 035 2;
  • 21) 0.115 351 434 035 2 × 2 = 0 + 0.230 702 868 070 4;
  • 22) 0.230 702 868 070 4 × 2 = 0 + 0.461 405 736 140 8;
  • 23) 0.461 405 736 140 8 × 2 = 0 + 0.922 811 472 281 6;
  • 24) 0.922 811 472 281 6 × 2 = 1 + 0.845 622 944 563 2;
  • 25) 0.845 622 944 563 2 × 2 = 1 + 0.691 245 889 126 4;
  • 26) 0.691 245 889 126 4 × 2 = 1 + 0.382 491 778 252 8;
  • 27) 0.382 491 778 252 8 × 2 = 0 + 0.764 983 556 505 6;
  • 28) 0.764 983 556 505 6 × 2 = 1 + 0.529 967 113 011 2;
  • 29) 0.529 967 113 011 2 × 2 = 1 + 0.059 934 226 022 4;
  • 30) 0.059 934 226 022 4 × 2 = 0 + 0.119 868 452 044 8;
  • 31) 0.119 868 452 044 8 × 2 = 0 + 0.239 736 904 089 6;
  • 32) 0.239 736 904 089 6 × 2 = 0 + 0.479 473 808 179 2;
  • 33) 0.479 473 808 179 2 × 2 = 0 + 0.958 947 616 358 4;
  • 34) 0.958 947 616 358 4 × 2 = 1 + 0.917 895 232 716 8;
  • 35) 0.917 895 232 716 8 × 2 = 1 + 0.835 790 465 433 6;
  • 36) 0.835 790 465 433 6 × 2 = 1 + 0.671 580 930 867 2;
  • 37) 0.671 580 930 867 2 × 2 = 1 + 0.343 161 861 734 4;
  • 38) 0.343 161 861 734 4 × 2 = 0 + 0.686 323 723 468 8;
  • 39) 0.686 323 723 468 8 × 2 = 1 + 0.372 647 446 937 6;
  • 40) 0.372 647 446 937 6 × 2 = 0 + 0.745 294 893 875 2;
  • 41) 0.745 294 893 875 2 × 2 = 1 + 0.490 589 787 750 4;
  • 42) 0.490 589 787 750 4 × 2 = 0 + 0.981 179 575 500 8;
  • 43) 0.981 179 575 500 8 × 2 = 1 + 0.962 359 151 001 6;
  • 44) 0.962 359 151 001 6 × 2 = 1 + 0.924 718 302 003 2;
  • 45) 0.924 718 302 003 2 × 2 = 1 + 0.849 436 604 006 4;
  • 46) 0.849 436 604 006 4 × 2 = 1 + 0.698 873 208 012 8;
  • 47) 0.698 873 208 012 8 × 2 = 1 + 0.397 746 416 025 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:


0.000 000 110 007 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1010 1011 111(2)

5. Positive number before normalization:

0.000 000 110 007 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1010 1011 111(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:


0.000 000 110 007 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1010 1011 111(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1010 1011 111(2) × 20 =

1.1101 1000 0111 1010 1011 111(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 1010 1011 111


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 1101 0101 1111 =

110 1100 0011 1101 0101 1111


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 1101 0101 1111


Decimal number 0.000 000 110 007 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1101 0101 1111

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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