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

Convert decimal -0.000 000 037(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(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. Start with the positive version of the number:

|-0.000 000 037| = 0.000 000 037


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 037.

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 × 2 = 0 + 0.000 000 074;
  • 2) 0.000 000 074 × 2 = 0 + 0.000 000 148;
  • 3) 0.000 000 148 × 2 = 0 + 0.000 000 296;
  • 4) 0.000 000 296 × 2 = 0 + 0.000 000 592;
  • 5) 0.000 000 592 × 2 = 0 + 0.000 001 184;
  • 6) 0.000 001 184 × 2 = 0 + 0.000 002 368;
  • 7) 0.000 002 368 × 2 = 0 + 0.000 004 736;
  • 8) 0.000 004 736 × 2 = 0 + 0.000 009 472;
  • 9) 0.000 009 472 × 2 = 0 + 0.000 018 944;
  • 10) 0.000 018 944 × 2 = 0 + 0.000 037 888;
  • 11) 0.000 037 888 × 2 = 0 + 0.000 075 776;
  • 12) 0.000 075 776 × 2 = 0 + 0.000 151 552;
  • 13) 0.000 151 552 × 2 = 0 + 0.000 303 104;
  • 14) 0.000 303 104 × 2 = 0 + 0.000 606 208;
  • 15) 0.000 606 208 × 2 = 0 + 0.001 212 416;
  • 16) 0.001 212 416 × 2 = 0 + 0.002 424 832;
  • 17) 0.002 424 832 × 2 = 0 + 0.004 849 664;
  • 18) 0.004 849 664 × 2 = 0 + 0.009 699 328;
  • 19) 0.009 699 328 × 2 = 0 + 0.019 398 656;
  • 20) 0.019 398 656 × 2 = 0 + 0.038 797 312;
  • 21) 0.038 797 312 × 2 = 0 + 0.077 594 624;
  • 22) 0.077 594 624 × 2 = 0 + 0.155 189 248;
  • 23) 0.155 189 248 × 2 = 0 + 0.310 378 496;
  • 24) 0.310 378 496 × 2 = 0 + 0.620 756 992;
  • 25) 0.620 756 992 × 2 = 1 + 0.241 513 984;
  • 26) 0.241 513 984 × 2 = 0 + 0.483 027 968;
  • 27) 0.483 027 968 × 2 = 0 + 0.966 055 936;
  • 28) 0.966 055 936 × 2 = 1 + 0.932 111 872;
  • 29) 0.932 111 872 × 2 = 1 + 0.864 223 744;
  • 30) 0.864 223 744 × 2 = 1 + 0.728 447 488;
  • 31) 0.728 447 488 × 2 = 1 + 0.456 894 976;
  • 32) 0.456 894 976 × 2 = 0 + 0.913 789 952;
  • 33) 0.913 789 952 × 2 = 1 + 0.827 579 904;
  • 34) 0.827 579 904 × 2 = 1 + 0.655 159 808;
  • 35) 0.655 159 808 × 2 = 1 + 0.310 319 616;
  • 36) 0.310 319 616 × 2 = 0 + 0.620 639 232;
  • 37) 0.620 639 232 × 2 = 1 + 0.241 278 464;
  • 38) 0.241 278 464 × 2 = 0 + 0.482 556 928;
  • 39) 0.482 556 928 × 2 = 0 + 0.965 113 856;
  • 40) 0.965 113 856 × 2 = 1 + 0.930 227 712;
  • 41) 0.930 227 712 × 2 = 1 + 0.860 455 424;
  • 42) 0.860 455 424 × 2 = 1 + 0.720 910 848;
  • 43) 0.720 910 848 × 2 = 1 + 0.441 821 696;
  • 44) 0.441 821 696 × 2 = 0 + 0.883 643 392;
  • 45) 0.883 643 392 × 2 = 1 + 0.767 286 784;
  • 46) 0.767 286 784 × 2 = 1 + 0.534 573 568;
  • 47) 0.534 573 568 × 2 = 1 + 0.069 147 136;
  • 48) 0.069 147 136 × 2 = 0 + 0.138 294 272;

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).


5. 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(10) =

0.0000 0000 0000 0000 0000 0000 1001 1110 1110 1001 1110 1110(2)

6. Positive number before normalization:

0.000 000 037(10) =

0.0000 0000 0000 0000 0000 0000 1001 1110 1110 1001 1110 1110(2)

7. 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(10) =

0.0000 0000 0000 0000 0000 0000 1001 1110 1110 1001 1110 1110(2) =

0.0000 0000 0000 0000 0000 0000 1001 1110 1110 1001 1110 1110(2) × 20 =

1.0011 1101 1101 0011 1101 110(2) × 2-25


8. 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 1 (a negative number)

Exponent (unadjusted): -25

Mantissa (not normalized):
1.0011 1101 1101 0011 1101 110


9. 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)


10. 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;

11. 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)


12. 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 1110 1110 1001 1110 1110 =

001 1110 1110 1001 1110 1110


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0110

Mantissa (23 bits) =
001 1110 1110 1001 1110 1110


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

1 - 0110 0110 - 001 1110 1110 1001 1110 1110

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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