0.000 000 534 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 534(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
0.000 000 534(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: 0.000 000 534.

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) 0.000 000 534 × 2 = 0 + 0.000 001 068;
  • 2) 0.000 001 068 × 2 = 0 + 0.000 002 136;
  • 3) 0.000 002 136 × 2 = 0 + 0.000 004 272;
  • 4) 0.000 004 272 × 2 = 0 + 0.000 008 544;
  • 5) 0.000 008 544 × 2 = 0 + 0.000 017 088;
  • 6) 0.000 017 088 × 2 = 0 + 0.000 034 176;
  • 7) 0.000 034 176 × 2 = 0 + 0.000 068 352;
  • 8) 0.000 068 352 × 2 = 0 + 0.000 136 704;
  • 9) 0.000 136 704 × 2 = 0 + 0.000 273 408;
  • 10) 0.000 273 408 × 2 = 0 + 0.000 546 816;
  • 11) 0.000 546 816 × 2 = 0 + 0.001 093 632;
  • 12) 0.001 093 632 × 2 = 0 + 0.002 187 264;
  • 13) 0.002 187 264 × 2 = 0 + 0.004 374 528;
  • 14) 0.004 374 528 × 2 = 0 + 0.008 749 056;
  • 15) 0.008 749 056 × 2 = 0 + 0.017 498 112;
  • 16) 0.017 498 112 × 2 = 0 + 0.034 996 224;
  • 17) 0.034 996 224 × 2 = 0 + 0.069 992 448;
  • 18) 0.069 992 448 × 2 = 0 + 0.139 984 896;
  • 19) 0.139 984 896 × 2 = 0 + 0.279 969 792;
  • 20) 0.279 969 792 × 2 = 0 + 0.559 939 584;
  • 21) 0.559 939 584 × 2 = 1 + 0.119 879 168;
  • 22) 0.119 879 168 × 2 = 0 + 0.239 758 336;
  • 23) 0.239 758 336 × 2 = 0 + 0.479 516 672;
  • 24) 0.479 516 672 × 2 = 0 + 0.959 033 344;
  • 25) 0.959 033 344 × 2 = 1 + 0.918 066 688;
  • 26) 0.918 066 688 × 2 = 1 + 0.836 133 376;
  • 27) 0.836 133 376 × 2 = 1 + 0.672 266 752;
  • 28) 0.672 266 752 × 2 = 1 + 0.344 533 504;
  • 29) 0.344 533 504 × 2 = 0 + 0.689 067 008;
  • 30) 0.689 067 008 × 2 = 1 + 0.378 134 016;
  • 31) 0.378 134 016 × 2 = 0 + 0.756 268 032;
  • 32) 0.756 268 032 × 2 = 1 + 0.512 536 064;
  • 33) 0.512 536 064 × 2 = 1 + 0.025 072 128;
  • 34) 0.025 072 128 × 2 = 0 + 0.050 144 256;
  • 35) 0.050 144 256 × 2 = 0 + 0.100 288 512;
  • 36) 0.100 288 512 × 2 = 0 + 0.200 577 024;
  • 37) 0.200 577 024 × 2 = 0 + 0.401 154 048;
  • 38) 0.401 154 048 × 2 = 0 + 0.802 308 096;
  • 39) 0.802 308 096 × 2 = 1 + 0.604 616 192;
  • 40) 0.604 616 192 × 2 = 1 + 0.209 232 384;
  • 41) 0.209 232 384 × 2 = 0 + 0.418 464 768;
  • 42) 0.418 464 768 × 2 = 0 + 0.836 929 536;
  • 43) 0.836 929 536 × 2 = 1 + 0.673 859 072;
  • 44) 0.673 859 072 × 2 = 1 + 0.347 718 144;

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:


0.000 000 534(10) =

0.0000 0000 0000 0000 0000 1000 1111 0101 1000 0011 0011(2)

5. Positive number before normalization:

0.000 000 534(10) =

0.0000 0000 0000 0000 0000 1000 1111 0101 1000 0011 0011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 534(10) =

0.0000 0000 0000 0000 0000 1000 1111 0101 1000 0011 0011(2) =

0.0000 0000 0000 0000 0000 1000 1111 0101 1000 0011 0011(2) × 20 =

1.0001 1110 1011 0000 0110 011(2) × 2-21


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): -21

Mantissa (not normalized):
1.0001 1110 1011 0000 0110 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

106(10) =

0110 1010(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 0101 1000 0011 0011 =

000 1111 0101 1000 0011 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 1010

Mantissa (23 bits) =
000 1111 0101 1000 0011 0011


Decimal number 0.000 000 534 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 000 1111 0101 1000 0011 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