0.000 000 067 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 067(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
0.000 000 067(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: 0.000 000 067.

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) 0.000 000 067 × 2 = 0 + 0.000 000 134;
  • 2) 0.000 000 134 × 2 = 0 + 0.000 000 268;
  • 3) 0.000 000 268 × 2 = 0 + 0.000 000 536;
  • 4) 0.000 000 536 × 2 = 0 + 0.000 001 072;
  • 5) 0.000 001 072 × 2 = 0 + 0.000 002 144;
  • 6) 0.000 002 144 × 2 = 0 + 0.000 004 288;
  • 7) 0.000 004 288 × 2 = 0 + 0.000 008 576;
  • 8) 0.000 008 576 × 2 = 0 + 0.000 017 152;
  • 9) 0.000 017 152 × 2 = 0 + 0.000 034 304;
  • 10) 0.000 034 304 × 2 = 0 + 0.000 068 608;
  • 11) 0.000 068 608 × 2 = 0 + 0.000 137 216;
  • 12) 0.000 137 216 × 2 = 0 + 0.000 274 432;
  • 13) 0.000 274 432 × 2 = 0 + 0.000 548 864;
  • 14) 0.000 548 864 × 2 = 0 + 0.001 097 728;
  • 15) 0.001 097 728 × 2 = 0 + 0.002 195 456;
  • 16) 0.002 195 456 × 2 = 0 + 0.004 390 912;
  • 17) 0.004 390 912 × 2 = 0 + 0.008 781 824;
  • 18) 0.008 781 824 × 2 = 0 + 0.017 563 648;
  • 19) 0.017 563 648 × 2 = 0 + 0.035 127 296;
  • 20) 0.035 127 296 × 2 = 0 + 0.070 254 592;
  • 21) 0.070 254 592 × 2 = 0 + 0.140 509 184;
  • 22) 0.140 509 184 × 2 = 0 + 0.281 018 368;
  • 23) 0.281 018 368 × 2 = 0 + 0.562 036 736;
  • 24) 0.562 036 736 × 2 = 1 + 0.124 073 472;
  • 25) 0.124 073 472 × 2 = 0 + 0.248 146 944;
  • 26) 0.248 146 944 × 2 = 0 + 0.496 293 888;
  • 27) 0.496 293 888 × 2 = 0 + 0.992 587 776;
  • 28) 0.992 587 776 × 2 = 1 + 0.985 175 552;
  • 29) 0.985 175 552 × 2 = 1 + 0.970 351 104;
  • 30) 0.970 351 104 × 2 = 1 + 0.940 702 208;
  • 31) 0.940 702 208 × 2 = 1 + 0.881 404 416;
  • 32) 0.881 404 416 × 2 = 1 + 0.762 808 832;
  • 33) 0.762 808 832 × 2 = 1 + 0.525 617 664;
  • 34) 0.525 617 664 × 2 = 1 + 0.051 235 328;
  • 35) 0.051 235 328 × 2 = 0 + 0.102 470 656;
  • 36) 0.102 470 656 × 2 = 0 + 0.204 941 312;
  • 37) 0.204 941 312 × 2 = 0 + 0.409 882 624;
  • 38) 0.409 882 624 × 2 = 0 + 0.819 765 248;
  • 39) 0.819 765 248 × 2 = 1 + 0.639 530 496;
  • 40) 0.639 530 496 × 2 = 1 + 0.279 060 992;
  • 41) 0.279 060 992 × 2 = 0 + 0.558 121 984;
  • 42) 0.558 121 984 × 2 = 1 + 0.116 243 968;
  • 43) 0.116 243 968 × 2 = 0 + 0.232 487 936;
  • 44) 0.232 487 936 × 2 = 0 + 0.464 975 872;
  • 45) 0.464 975 872 × 2 = 0 + 0.929 951 744;
  • 46) 0.929 951 744 × 2 = 1 + 0.859 903 488;
  • 47) 0.859 903 488 × 2 = 1 + 0.719 806 976;

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:


0.000 000 067(10) =

0.0000 0000 0000 0000 0000 0001 0001 1111 1100 0011 0100 011(2)

5. Positive number before normalization:

0.000 000 067(10) =

0.0000 0000 0000 0000 0000 0001 0001 1111 1100 0011 0100 011(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:


0.000 000 067(10) =

0.0000 0000 0000 0000 0000 0001 0001 1111 1100 0011 0100 011(2) =

0.0000 0000 0000 0000 0000 0001 0001 1111 1100 0011 0100 011(2) × 20 =

1.0001 1111 1100 0011 0100 011(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):
1.0001 1111 1100 0011 0100 011


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. 000 1111 1110 0001 1010 0011 =

000 1111 1110 0001 1010 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) =
0110 0111

Mantissa (23 bits) =
000 1111 1110 0001 1010 0011


Decimal number 0.000 000 067 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 000 1111 1110 0001 1010 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