0.000 000 060 54 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 060 54(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
0.000 000 060 54(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: 0.000 000 060 54.

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) 0.000 000 060 54 × 2 = 0 + 0.000 000 121 08;
  • 2) 0.000 000 121 08 × 2 = 0 + 0.000 000 242 16;
  • 3) 0.000 000 242 16 × 2 = 0 + 0.000 000 484 32;
  • 4) 0.000 000 484 32 × 2 = 0 + 0.000 000 968 64;
  • 5) 0.000 000 968 64 × 2 = 0 + 0.000 001 937 28;
  • 6) 0.000 001 937 28 × 2 = 0 + 0.000 003 874 56;
  • 7) 0.000 003 874 56 × 2 = 0 + 0.000 007 749 12;
  • 8) 0.000 007 749 12 × 2 = 0 + 0.000 015 498 24;
  • 9) 0.000 015 498 24 × 2 = 0 + 0.000 030 996 48;
  • 10) 0.000 030 996 48 × 2 = 0 + 0.000 061 992 96;
  • 11) 0.000 061 992 96 × 2 = 0 + 0.000 123 985 92;
  • 12) 0.000 123 985 92 × 2 = 0 + 0.000 247 971 84;
  • 13) 0.000 247 971 84 × 2 = 0 + 0.000 495 943 68;
  • 14) 0.000 495 943 68 × 2 = 0 + 0.000 991 887 36;
  • 15) 0.000 991 887 36 × 2 = 0 + 0.001 983 774 72;
  • 16) 0.001 983 774 72 × 2 = 0 + 0.003 967 549 44;
  • 17) 0.003 967 549 44 × 2 = 0 + 0.007 935 098 88;
  • 18) 0.007 935 098 88 × 2 = 0 + 0.015 870 197 76;
  • 19) 0.015 870 197 76 × 2 = 0 + 0.031 740 395 52;
  • 20) 0.031 740 395 52 × 2 = 0 + 0.063 480 791 04;
  • 21) 0.063 480 791 04 × 2 = 0 + 0.126 961 582 08;
  • 22) 0.126 961 582 08 × 2 = 0 + 0.253 923 164 16;
  • 23) 0.253 923 164 16 × 2 = 0 + 0.507 846 328 32;
  • 24) 0.507 846 328 32 × 2 = 1 + 0.015 692 656 64;
  • 25) 0.015 692 656 64 × 2 = 0 + 0.031 385 313 28;
  • 26) 0.031 385 313 28 × 2 = 0 + 0.062 770 626 56;
  • 27) 0.062 770 626 56 × 2 = 0 + 0.125 541 253 12;
  • 28) 0.125 541 253 12 × 2 = 0 + 0.251 082 506 24;
  • 29) 0.251 082 506 24 × 2 = 0 + 0.502 165 012 48;
  • 30) 0.502 165 012 48 × 2 = 1 + 0.004 330 024 96;
  • 31) 0.004 330 024 96 × 2 = 0 + 0.008 660 049 92;
  • 32) 0.008 660 049 92 × 2 = 0 + 0.017 320 099 84;
  • 33) 0.017 320 099 84 × 2 = 0 + 0.034 640 199 68;
  • 34) 0.034 640 199 68 × 2 = 0 + 0.069 280 399 36;
  • 35) 0.069 280 399 36 × 2 = 0 + 0.138 560 798 72;
  • 36) 0.138 560 798 72 × 2 = 0 + 0.277 121 597 44;
  • 37) 0.277 121 597 44 × 2 = 0 + 0.554 243 194 88;
  • 38) 0.554 243 194 88 × 2 = 1 + 0.108 486 389 76;
  • 39) 0.108 486 389 76 × 2 = 0 + 0.216 972 779 52;
  • 40) 0.216 972 779 52 × 2 = 0 + 0.433 945 559 04;
  • 41) 0.433 945 559 04 × 2 = 0 + 0.867 891 118 08;
  • 42) 0.867 891 118 08 × 2 = 1 + 0.735 782 236 16;
  • 43) 0.735 782 236 16 × 2 = 1 + 0.471 564 472 32;
  • 44) 0.471 564 472 32 × 2 = 0 + 0.943 128 944 64;
  • 45) 0.943 128 944 64 × 2 = 1 + 0.886 257 889 28;
  • 46) 0.886 257 889 28 × 2 = 1 + 0.772 515 778 56;
  • 47) 0.772 515 778 56 × 2 = 1 + 0.545 031 557 12;

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:


0.000 000 060 54(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0100 0110 111(2)

5. Positive number before normalization:

0.000 000 060 54(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0100 0110 111(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:


0.000 000 060 54(10) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0100 0110 111(2) =

0.0000 0000 0000 0000 0000 0001 0000 0100 0000 0100 0110 111(2) × 20 =

1.0000 0100 0000 0100 0110 111(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):
1.0000 0100 0000 0100 0110 111


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 0010 0000 0010 0011 0111 =

000 0010 0000 0010 0011 0111


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 0010 0000 0010 0011 0111


Decimal number 0.000 000 060 54 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 000 0010 0000 0010 0011 0111

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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