0.000 000 110 019 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 019(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
0.000 000 110 019(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: 0.000 000 110 019.

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) 0.000 000 110 019 × 2 = 0 + 0.000 000 220 038;
  • 2) 0.000 000 220 038 × 2 = 0 + 0.000 000 440 076;
  • 3) 0.000 000 440 076 × 2 = 0 + 0.000 000 880 152;
  • 4) 0.000 000 880 152 × 2 = 0 + 0.000 001 760 304;
  • 5) 0.000 001 760 304 × 2 = 0 + 0.000 003 520 608;
  • 6) 0.000 003 520 608 × 2 = 0 + 0.000 007 041 216;
  • 7) 0.000 007 041 216 × 2 = 0 + 0.000 014 082 432;
  • 8) 0.000 014 082 432 × 2 = 0 + 0.000 028 164 864;
  • 9) 0.000 028 164 864 × 2 = 0 + 0.000 056 329 728;
  • 10) 0.000 056 329 728 × 2 = 0 + 0.000 112 659 456;
  • 11) 0.000 112 659 456 × 2 = 0 + 0.000 225 318 912;
  • 12) 0.000 225 318 912 × 2 = 0 + 0.000 450 637 824;
  • 13) 0.000 450 637 824 × 2 = 0 + 0.000 901 275 648;
  • 14) 0.000 901 275 648 × 2 = 0 + 0.001 802 551 296;
  • 15) 0.001 802 551 296 × 2 = 0 + 0.003 605 102 592;
  • 16) 0.003 605 102 592 × 2 = 0 + 0.007 210 205 184;
  • 17) 0.007 210 205 184 × 2 = 0 + 0.014 420 410 368;
  • 18) 0.014 420 410 368 × 2 = 0 + 0.028 840 820 736;
  • 19) 0.028 840 820 736 × 2 = 0 + 0.057 681 641 472;
  • 20) 0.057 681 641 472 × 2 = 0 + 0.115 363 282 944;
  • 21) 0.115 363 282 944 × 2 = 0 + 0.230 726 565 888;
  • 22) 0.230 726 565 888 × 2 = 0 + 0.461 453 131 776;
  • 23) 0.461 453 131 776 × 2 = 0 + 0.922 906 263 552;
  • 24) 0.922 906 263 552 × 2 = 1 + 0.845 812 527 104;
  • 25) 0.845 812 527 104 × 2 = 1 + 0.691 625 054 208;
  • 26) 0.691 625 054 208 × 2 = 1 + 0.383 250 108 416;
  • 27) 0.383 250 108 416 × 2 = 0 + 0.766 500 216 832;
  • 28) 0.766 500 216 832 × 2 = 1 + 0.533 000 433 664;
  • 29) 0.533 000 433 664 × 2 = 1 + 0.066 000 867 328;
  • 30) 0.066 000 867 328 × 2 = 0 + 0.132 001 734 656;
  • 31) 0.132 001 734 656 × 2 = 0 + 0.264 003 469 312;
  • 32) 0.264 003 469 312 × 2 = 0 + 0.528 006 938 624;
  • 33) 0.528 006 938 624 × 2 = 1 + 0.056 013 877 248;
  • 34) 0.056 013 877 248 × 2 = 0 + 0.112 027 754 496;
  • 35) 0.112 027 754 496 × 2 = 0 + 0.224 055 508 992;
  • 36) 0.224 055 508 992 × 2 = 0 + 0.448 111 017 984;
  • 37) 0.448 111 017 984 × 2 = 0 + 0.896 222 035 968;
  • 38) 0.896 222 035 968 × 2 = 1 + 0.792 444 071 936;
  • 39) 0.792 444 071 936 × 2 = 1 + 0.584 888 143 872;
  • 40) 0.584 888 143 872 × 2 = 1 + 0.169 776 287 744;
  • 41) 0.169 776 287 744 × 2 = 0 + 0.339 552 575 488;
  • 42) 0.339 552 575 488 × 2 = 0 + 0.679 105 150 976;
  • 43) 0.679 105 150 976 × 2 = 1 + 0.358 210 301 952;
  • 44) 0.358 210 301 952 × 2 = 0 + 0.716 420 603 904;
  • 45) 0.716 420 603 904 × 2 = 1 + 0.432 841 207 808;
  • 46) 0.432 841 207 808 × 2 = 0 + 0.865 682 415 616;
  • 47) 0.865 682 415 616 × 2 = 1 + 0.731 364 831 232;

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:


0.000 000 110 019(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0111 0010 101(2)

5. Positive number before normalization:

0.000 000 110 019(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0111 0010 101(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:


0.000 000 110 019(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0111 0010 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0111 0010 101(2) × 20 =

1.1101 1000 1000 0111 0010 101(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 0111 0010 101


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 0011 1001 0101 =

110 1100 0100 0011 1001 0101


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 0011 1001 0101


Decimal number 0.000 000 110 019 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 0011 1001 0101

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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