0.000 000 110 016 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 016 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
0.000 000 110 016 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: 0.000 000 110 016 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) 0.000 000 110 016 8 × 2 = 0 + 0.000 000 220 033 6;
  • 2) 0.000 000 220 033 6 × 2 = 0 + 0.000 000 440 067 2;
  • 3) 0.000 000 440 067 2 × 2 = 0 + 0.000 000 880 134 4;
  • 4) 0.000 000 880 134 4 × 2 = 0 + 0.000 001 760 268 8;
  • 5) 0.000 001 760 268 8 × 2 = 0 + 0.000 003 520 537 6;
  • 6) 0.000 003 520 537 6 × 2 = 0 + 0.000 007 041 075 2;
  • 7) 0.000 007 041 075 2 × 2 = 0 + 0.000 014 082 150 4;
  • 8) 0.000 014 082 150 4 × 2 = 0 + 0.000 028 164 300 8;
  • 9) 0.000 028 164 300 8 × 2 = 0 + 0.000 056 328 601 6;
  • 10) 0.000 056 328 601 6 × 2 = 0 + 0.000 112 657 203 2;
  • 11) 0.000 112 657 203 2 × 2 = 0 + 0.000 225 314 406 4;
  • 12) 0.000 225 314 406 4 × 2 = 0 + 0.000 450 628 812 8;
  • 13) 0.000 450 628 812 8 × 2 = 0 + 0.000 901 257 625 6;
  • 14) 0.000 901 257 625 6 × 2 = 0 + 0.001 802 515 251 2;
  • 15) 0.001 802 515 251 2 × 2 = 0 + 0.003 605 030 502 4;
  • 16) 0.003 605 030 502 4 × 2 = 0 + 0.007 210 061 004 8;
  • 17) 0.007 210 061 004 8 × 2 = 0 + 0.014 420 122 009 6;
  • 18) 0.014 420 122 009 6 × 2 = 0 + 0.028 840 244 019 2;
  • 19) 0.028 840 244 019 2 × 2 = 0 + 0.057 680 488 038 4;
  • 20) 0.057 680 488 038 4 × 2 = 0 + 0.115 360 976 076 8;
  • 21) 0.115 360 976 076 8 × 2 = 0 + 0.230 721 952 153 6;
  • 22) 0.230 721 952 153 6 × 2 = 0 + 0.461 443 904 307 2;
  • 23) 0.461 443 904 307 2 × 2 = 0 + 0.922 887 808 614 4;
  • 24) 0.922 887 808 614 4 × 2 = 1 + 0.845 775 617 228 8;
  • 25) 0.845 775 617 228 8 × 2 = 1 + 0.691 551 234 457 6;
  • 26) 0.691 551 234 457 6 × 2 = 1 + 0.383 102 468 915 2;
  • 27) 0.383 102 468 915 2 × 2 = 0 + 0.766 204 937 830 4;
  • 28) 0.766 204 937 830 4 × 2 = 1 + 0.532 409 875 660 8;
  • 29) 0.532 409 875 660 8 × 2 = 1 + 0.064 819 751 321 6;
  • 30) 0.064 819 751 321 6 × 2 = 0 + 0.129 639 502 643 2;
  • 31) 0.129 639 502 643 2 × 2 = 0 + 0.259 279 005 286 4;
  • 32) 0.259 279 005 286 4 × 2 = 0 + 0.518 558 010 572 8;
  • 33) 0.518 558 010 572 8 × 2 = 1 + 0.037 116 021 145 6;
  • 34) 0.037 116 021 145 6 × 2 = 0 + 0.074 232 042 291 2;
  • 35) 0.074 232 042 291 2 × 2 = 0 + 0.148 464 084 582 4;
  • 36) 0.148 464 084 582 4 × 2 = 0 + 0.296 928 169 164 8;
  • 37) 0.296 928 169 164 8 × 2 = 0 + 0.593 856 338 329 6;
  • 38) 0.593 856 338 329 6 × 2 = 1 + 0.187 712 676 659 2;
  • 39) 0.187 712 676 659 2 × 2 = 0 + 0.375 425 353 318 4;
  • 40) 0.375 425 353 318 4 × 2 = 0 + 0.750 850 706 636 8;
  • 41) 0.750 850 706 636 8 × 2 = 1 + 0.501 701 413 273 6;
  • 42) 0.501 701 413 273 6 × 2 = 1 + 0.003 402 826 547 2;
  • 43) 0.003 402 826 547 2 × 2 = 0 + 0.006 805 653 094 4;
  • 44) 0.006 805 653 094 4 × 2 = 0 + 0.013 611 306 188 8;
  • 45) 0.013 611 306 188 8 × 2 = 0 + 0.027 222 612 377 6;
  • 46) 0.027 222 612 377 6 × 2 = 0 + 0.054 445 224 755 2;
  • 47) 0.054 445 224 755 2 × 2 = 0 + 0.108 890 449 510 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:


0.000 000 110 016 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0100 1100 000(2)

5. Positive number before normalization:

0.000 000 110 016 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0100 1100 000(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:


0.000 000 110 016 8(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0100 1100 000(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0100 1100 000(2) × 20 =

1.1101 1000 1000 0100 1100 000(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 1000 0100 1100 000


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 0100 0010 0110 0000 =

110 1100 0100 0010 0110 0000


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 0100 0010 0110 0000


Decimal number 0.000 000 110 016 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 0010 0110 0000

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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