0.000 000 110 005 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 005 1(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
0.000 000 110 005 1(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: 0.000 000 110 005 1.

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) 0.000 000 110 005 1 × 2 = 0 + 0.000 000 220 010 2;
  • 2) 0.000 000 220 010 2 × 2 = 0 + 0.000 000 440 020 4;
  • 3) 0.000 000 440 020 4 × 2 = 0 + 0.000 000 880 040 8;
  • 4) 0.000 000 880 040 8 × 2 = 0 + 0.000 001 760 081 6;
  • 5) 0.000 001 760 081 6 × 2 = 0 + 0.000 003 520 163 2;
  • 6) 0.000 003 520 163 2 × 2 = 0 + 0.000 007 040 326 4;
  • 7) 0.000 007 040 326 4 × 2 = 0 + 0.000 014 080 652 8;
  • 8) 0.000 014 080 652 8 × 2 = 0 + 0.000 028 161 305 6;
  • 9) 0.000 028 161 305 6 × 2 = 0 + 0.000 056 322 611 2;
  • 10) 0.000 056 322 611 2 × 2 = 0 + 0.000 112 645 222 4;
  • 11) 0.000 112 645 222 4 × 2 = 0 + 0.000 225 290 444 8;
  • 12) 0.000 225 290 444 8 × 2 = 0 + 0.000 450 580 889 6;
  • 13) 0.000 450 580 889 6 × 2 = 0 + 0.000 901 161 779 2;
  • 14) 0.000 901 161 779 2 × 2 = 0 + 0.001 802 323 558 4;
  • 15) 0.001 802 323 558 4 × 2 = 0 + 0.003 604 647 116 8;
  • 16) 0.003 604 647 116 8 × 2 = 0 + 0.007 209 294 233 6;
  • 17) 0.007 209 294 233 6 × 2 = 0 + 0.014 418 588 467 2;
  • 18) 0.014 418 588 467 2 × 2 = 0 + 0.028 837 176 934 4;
  • 19) 0.028 837 176 934 4 × 2 = 0 + 0.057 674 353 868 8;
  • 20) 0.057 674 353 868 8 × 2 = 0 + 0.115 348 707 737 6;
  • 21) 0.115 348 707 737 6 × 2 = 0 + 0.230 697 415 475 2;
  • 22) 0.230 697 415 475 2 × 2 = 0 + 0.461 394 830 950 4;
  • 23) 0.461 394 830 950 4 × 2 = 0 + 0.922 789 661 900 8;
  • 24) 0.922 789 661 900 8 × 2 = 1 + 0.845 579 323 801 6;
  • 25) 0.845 579 323 801 6 × 2 = 1 + 0.691 158 647 603 2;
  • 26) 0.691 158 647 603 2 × 2 = 1 + 0.382 317 295 206 4;
  • 27) 0.382 317 295 206 4 × 2 = 0 + 0.764 634 590 412 8;
  • 28) 0.764 634 590 412 8 × 2 = 1 + 0.529 269 180 825 6;
  • 29) 0.529 269 180 825 6 × 2 = 1 + 0.058 538 361 651 2;
  • 30) 0.058 538 361 651 2 × 2 = 0 + 0.117 076 723 302 4;
  • 31) 0.117 076 723 302 4 × 2 = 0 + 0.234 153 446 604 8;
  • 32) 0.234 153 446 604 8 × 2 = 0 + 0.468 306 893 209 6;
  • 33) 0.468 306 893 209 6 × 2 = 0 + 0.936 613 786 419 2;
  • 34) 0.936 613 786 419 2 × 2 = 1 + 0.873 227 572 838 4;
  • 35) 0.873 227 572 838 4 × 2 = 1 + 0.746 455 145 676 8;
  • 36) 0.746 455 145 676 8 × 2 = 1 + 0.492 910 291 353 6;
  • 37) 0.492 910 291 353 6 × 2 = 0 + 0.985 820 582 707 2;
  • 38) 0.985 820 582 707 2 × 2 = 1 + 0.971 641 165 414 4;
  • 39) 0.971 641 165 414 4 × 2 = 1 + 0.943 282 330 828 8;
  • 40) 0.943 282 330 828 8 × 2 = 1 + 0.886 564 661 657 6;
  • 41) 0.886 564 661 657 6 × 2 = 1 + 0.773 129 323 315 2;
  • 42) 0.773 129 323 315 2 × 2 = 1 + 0.546 258 646 630 4;
  • 43) 0.546 258 646 630 4 × 2 = 1 + 0.092 517 293 260 8;
  • 44) 0.092 517 293 260 8 × 2 = 0 + 0.185 034 586 521 6;
  • 45) 0.185 034 586 521 6 × 2 = 0 + 0.370 069 173 043 2;
  • 46) 0.370 069 173 043 2 × 2 = 0 + 0.740 138 346 086 4;
  • 47) 0.740 138 346 086 4 × 2 = 1 + 0.480 276 692 172 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:


0.000 000 110 005 1(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0111 1110 001(2)

5. Positive number before normalization:

0.000 000 110 005 1(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0111 1110 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:


0.000 000 110 005 1(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0111 1110 001(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 0111 1110 001(2) × 20 =

1.1101 1000 0111 0111 1110 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 0111 0111 1110 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 1011 1111 0001 =

110 1100 0011 1011 1111 0001


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 1011 1111 0001


Decimal number 0.000 000 110 005 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1011 1111 0001

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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