0.000 000 037 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 037 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
0.000 000 037 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: 0.000 000 037 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) 0.000 000 037 1 × 2 = 0 + 0.000 000 074 2;
  • 2) 0.000 000 074 2 × 2 = 0 + 0.000 000 148 4;
  • 3) 0.000 000 148 4 × 2 = 0 + 0.000 000 296 8;
  • 4) 0.000 000 296 8 × 2 = 0 + 0.000 000 593 6;
  • 5) 0.000 000 593 6 × 2 = 0 + 0.000 001 187 2;
  • 6) 0.000 001 187 2 × 2 = 0 + 0.000 002 374 4;
  • 7) 0.000 002 374 4 × 2 = 0 + 0.000 004 748 8;
  • 8) 0.000 004 748 8 × 2 = 0 + 0.000 009 497 6;
  • 9) 0.000 009 497 6 × 2 = 0 + 0.000 018 995 2;
  • 10) 0.000 018 995 2 × 2 = 0 + 0.000 037 990 4;
  • 11) 0.000 037 990 4 × 2 = 0 + 0.000 075 980 8;
  • 12) 0.000 075 980 8 × 2 = 0 + 0.000 151 961 6;
  • 13) 0.000 151 961 6 × 2 = 0 + 0.000 303 923 2;
  • 14) 0.000 303 923 2 × 2 = 0 + 0.000 607 846 4;
  • 15) 0.000 607 846 4 × 2 = 0 + 0.001 215 692 8;
  • 16) 0.001 215 692 8 × 2 = 0 + 0.002 431 385 6;
  • 17) 0.002 431 385 6 × 2 = 0 + 0.004 862 771 2;
  • 18) 0.004 862 771 2 × 2 = 0 + 0.009 725 542 4;
  • 19) 0.009 725 542 4 × 2 = 0 + 0.019 451 084 8;
  • 20) 0.019 451 084 8 × 2 = 0 + 0.038 902 169 6;
  • 21) 0.038 902 169 6 × 2 = 0 + 0.077 804 339 2;
  • 22) 0.077 804 339 2 × 2 = 0 + 0.155 608 678 4;
  • 23) 0.155 608 678 4 × 2 = 0 + 0.311 217 356 8;
  • 24) 0.311 217 356 8 × 2 = 0 + 0.622 434 713 6;
  • 25) 0.622 434 713 6 × 2 = 1 + 0.244 869 427 2;
  • 26) 0.244 869 427 2 × 2 = 0 + 0.489 738 854 4;
  • 27) 0.489 738 854 4 × 2 = 0 + 0.979 477 708 8;
  • 28) 0.979 477 708 8 × 2 = 1 + 0.958 955 417 6;
  • 29) 0.958 955 417 6 × 2 = 1 + 0.917 910 835 2;
  • 30) 0.917 910 835 2 × 2 = 1 + 0.835 821 670 4;
  • 31) 0.835 821 670 4 × 2 = 1 + 0.671 643 340 8;
  • 32) 0.671 643 340 8 × 2 = 1 + 0.343 286 681 6;
  • 33) 0.343 286 681 6 × 2 = 0 + 0.686 573 363 2;
  • 34) 0.686 573 363 2 × 2 = 1 + 0.373 146 726 4;
  • 35) 0.373 146 726 4 × 2 = 0 + 0.746 293 452 8;
  • 36) 0.746 293 452 8 × 2 = 1 + 0.492 586 905 6;
  • 37) 0.492 586 905 6 × 2 = 0 + 0.985 173 811 2;
  • 38) 0.985 173 811 2 × 2 = 1 + 0.970 347 622 4;
  • 39) 0.970 347 622 4 × 2 = 1 + 0.940 695 244 8;
  • 40) 0.940 695 244 8 × 2 = 1 + 0.881 390 489 6;
  • 41) 0.881 390 489 6 × 2 = 1 + 0.762 780 979 2;
  • 42) 0.762 780 979 2 × 2 = 1 + 0.525 561 958 4;
  • 43) 0.525 561 958 4 × 2 = 1 + 0.051 123 916 8;
  • 44) 0.051 123 916 8 × 2 = 0 + 0.102 247 833 6;
  • 45) 0.102 247 833 6 × 2 = 0 + 0.204 495 667 2;
  • 46) 0.204 495 667 2 × 2 = 0 + 0.408 991 334 4;
  • 47) 0.408 991 334 4 × 2 = 0 + 0.817 982 668 8;
  • 48) 0.817 982 668 8 × 2 = 1 + 0.635 965 337 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:


0.000 000 037 1(10) =

0.0000 0000 0000 0000 0000 0000 1001 1111 0101 0111 1110 0001(2)

5. Positive number before normalization:

0.000 000 037 1(10) =

0.0000 0000 0000 0000 0000 0000 1001 1111 0101 0111 1110 0001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 037 1(10) =

0.0000 0000 0000 0000 0000 0000 1001 1111 0101 0111 1110 0001(2) =

0.0000 0000 0000 0000 0000 0000 1001 1111 0101 0111 1110 0001(2) × 20 =

1.0011 1110 1010 1111 1100 001(2) × 2-25


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): -25

Mantissa (not normalized):
1.0011 1110 1010 1111 1100 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(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;
  • 102 ÷ 2 = 51 + 0;
  • 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) =

102(10) =

0110 0110(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. 001 1111 0101 0111 1110 0001 =

001 1111 0101 0111 1110 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 0110

Mantissa (23 bits) =
001 1111 0101 0111 1110 0001


Decimal number 0.000 000 037 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0110 - 001 1111 0101 0111 1110 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