32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 038 894 545 816 760 25 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 038 894 545 816 760 25(10) converted and written in 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: 0.000 038 894 545 816 760 25.

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) 0.000 038 894 545 816 760 25 × 2 = 0 + 0.000 077 789 091 633 520 5;
  • 2) 0.000 077 789 091 633 520 5 × 2 = 0 + 0.000 155 578 183 267 041;
  • 3) 0.000 155 578 183 267 041 × 2 = 0 + 0.000 311 156 366 534 082;
  • 4) 0.000 311 156 366 534 082 × 2 = 0 + 0.000 622 312 733 068 164;
  • 5) 0.000 622 312 733 068 164 × 2 = 0 + 0.001 244 625 466 136 328;
  • 6) 0.001 244 625 466 136 328 × 2 = 0 + 0.002 489 250 932 272 656;
  • 7) 0.002 489 250 932 272 656 × 2 = 0 + 0.004 978 501 864 545 312;
  • 8) 0.004 978 501 864 545 312 × 2 = 0 + 0.009 957 003 729 090 624;
  • 9) 0.009 957 003 729 090 624 × 2 = 0 + 0.019 914 007 458 181 248;
  • 10) 0.019 914 007 458 181 248 × 2 = 0 + 0.039 828 014 916 362 496;
  • 11) 0.039 828 014 916 362 496 × 2 = 0 + 0.079 656 029 832 724 992;
  • 12) 0.079 656 029 832 724 992 × 2 = 0 + 0.159 312 059 665 449 984;
  • 13) 0.159 312 059 665 449 984 × 2 = 0 + 0.318 624 119 330 899 968;
  • 14) 0.318 624 119 330 899 968 × 2 = 0 + 0.637 248 238 661 799 936;
  • 15) 0.637 248 238 661 799 936 × 2 = 1 + 0.274 496 477 323 599 872;
  • 16) 0.274 496 477 323 599 872 × 2 = 0 + 0.548 992 954 647 199 744;
  • 17) 0.548 992 954 647 199 744 × 2 = 1 + 0.097 985 909 294 399 488;
  • 18) 0.097 985 909 294 399 488 × 2 = 0 + 0.195 971 818 588 798 976;
  • 19) 0.195 971 818 588 798 976 × 2 = 0 + 0.391 943 637 177 597 952;
  • 20) 0.391 943 637 177 597 952 × 2 = 0 + 0.783 887 274 355 195 904;
  • 21) 0.783 887 274 355 195 904 × 2 = 1 + 0.567 774 548 710 391 808;
  • 22) 0.567 774 548 710 391 808 × 2 = 1 + 0.135 549 097 420 783 616;
  • 23) 0.135 549 097 420 783 616 × 2 = 0 + 0.271 098 194 841 567 232;
  • 24) 0.271 098 194 841 567 232 × 2 = 0 + 0.542 196 389 683 134 464;
  • 25) 0.542 196 389 683 134 464 × 2 = 1 + 0.084 392 779 366 268 928;
  • 26) 0.084 392 779 366 268 928 × 2 = 0 + 0.168 785 558 732 537 856;
  • 27) 0.168 785 558 732 537 856 × 2 = 0 + 0.337 571 117 465 075 712;
  • 28) 0.337 571 117 465 075 712 × 2 = 0 + 0.675 142 234 930 151 424;
  • 29) 0.675 142 234 930 151 424 × 2 = 1 + 0.350 284 469 860 302 848;
  • 30) 0.350 284 469 860 302 848 × 2 = 0 + 0.700 568 939 720 605 696;
  • 31) 0.700 568 939 720 605 696 × 2 = 1 + 0.401 137 879 441 211 392;
  • 32) 0.401 137 879 441 211 392 × 2 = 0 + 0.802 275 758 882 422 784;
  • 33) 0.802 275 758 882 422 784 × 2 = 1 + 0.604 551 517 764 845 568;
  • 34) 0.604 551 517 764 845 568 × 2 = 1 + 0.209 103 035 529 691 136;
  • 35) 0.209 103 035 529 691 136 × 2 = 0 + 0.418 206 071 059 382 272;
  • 36) 0.418 206 071 059 382 272 × 2 = 0 + 0.836 412 142 118 764 544;
  • 37) 0.836 412 142 118 764 544 × 2 = 1 + 0.672 824 284 237 529 088;
  • 38) 0.672 824 284 237 529 088 × 2 = 1 + 0.345 648 568 475 058 176;

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...)


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:


0.000 038 894 545 816 760 25(10) =

0.0000 0000 0000 0010 1000 1100 1000 1010 1100 11(2)


5. Positive number before normalization:

0.000 038 894 545 816 760 25(10) =

0.0000 0000 0000 0010 1000 1100 1000 1010 1100 11(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 038 894 545 816 760 25(10) =

0.0000 0000 0000 0010 1000 1100 1000 1010 1100 11(2) =

0.0000 0000 0000 0010 1000 1100 1000 1010 1100 11(2) × 20 =

1.0100 0110 0100 0101 0110 011(2) × 2-15


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): -15

Mantissa (not normalized):
1.0100 0110 0100 0101 0110 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-15 + 2(8-1) - 1 =

(-15 + 127)(10) =

112(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;
  • 112 ÷ 2 = 56 + 0;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

112(10) =

0111 0000(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. 010 0011 0010 0010 1011 0011 =

010 0011 0010 0010 1011 0011


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) =
0111 0000

Mantissa (23 bits) =
010 0011 0010 0010 1011 0011


The base ten decimal number 0.000 038 894 545 816 760 25 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 0000 - 010 0011 0010 0010 1011 0011

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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