0.000 000 253 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 253(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
0.000 000 253(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: 0.000 000 253.

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) 0.000 000 253 × 2 = 0 + 0.000 000 506;
  • 2) 0.000 000 506 × 2 = 0 + 0.000 001 012;
  • 3) 0.000 001 012 × 2 = 0 + 0.000 002 024;
  • 4) 0.000 002 024 × 2 = 0 + 0.000 004 048;
  • 5) 0.000 004 048 × 2 = 0 + 0.000 008 096;
  • 6) 0.000 008 096 × 2 = 0 + 0.000 016 192;
  • 7) 0.000 016 192 × 2 = 0 + 0.000 032 384;
  • 8) 0.000 032 384 × 2 = 0 + 0.000 064 768;
  • 9) 0.000 064 768 × 2 = 0 + 0.000 129 536;
  • 10) 0.000 129 536 × 2 = 0 + 0.000 259 072;
  • 11) 0.000 259 072 × 2 = 0 + 0.000 518 144;
  • 12) 0.000 518 144 × 2 = 0 + 0.001 036 288;
  • 13) 0.001 036 288 × 2 = 0 + 0.002 072 576;
  • 14) 0.002 072 576 × 2 = 0 + 0.004 145 152;
  • 15) 0.004 145 152 × 2 = 0 + 0.008 290 304;
  • 16) 0.008 290 304 × 2 = 0 + 0.016 580 608;
  • 17) 0.016 580 608 × 2 = 0 + 0.033 161 216;
  • 18) 0.033 161 216 × 2 = 0 + 0.066 322 432;
  • 19) 0.066 322 432 × 2 = 0 + 0.132 644 864;
  • 20) 0.132 644 864 × 2 = 0 + 0.265 289 728;
  • 21) 0.265 289 728 × 2 = 0 + 0.530 579 456;
  • 22) 0.530 579 456 × 2 = 1 + 0.061 158 912;
  • 23) 0.061 158 912 × 2 = 0 + 0.122 317 824;
  • 24) 0.122 317 824 × 2 = 0 + 0.244 635 648;
  • 25) 0.244 635 648 × 2 = 0 + 0.489 271 296;
  • 26) 0.489 271 296 × 2 = 0 + 0.978 542 592;
  • 27) 0.978 542 592 × 2 = 1 + 0.957 085 184;
  • 28) 0.957 085 184 × 2 = 1 + 0.914 170 368;
  • 29) 0.914 170 368 × 2 = 1 + 0.828 340 736;
  • 30) 0.828 340 736 × 2 = 1 + 0.656 681 472;
  • 31) 0.656 681 472 × 2 = 1 + 0.313 362 944;
  • 32) 0.313 362 944 × 2 = 0 + 0.626 725 888;
  • 33) 0.626 725 888 × 2 = 1 + 0.253 451 776;
  • 34) 0.253 451 776 × 2 = 0 + 0.506 903 552;
  • 35) 0.506 903 552 × 2 = 1 + 0.013 807 104;
  • 36) 0.013 807 104 × 2 = 0 + 0.027 614 208;
  • 37) 0.027 614 208 × 2 = 0 + 0.055 228 416;
  • 38) 0.055 228 416 × 2 = 0 + 0.110 456 832;
  • 39) 0.110 456 832 × 2 = 0 + 0.220 913 664;
  • 40) 0.220 913 664 × 2 = 0 + 0.441 827 328;
  • 41) 0.441 827 328 × 2 = 0 + 0.883 654 656;
  • 42) 0.883 654 656 × 2 = 1 + 0.767 309 312;
  • 43) 0.767 309 312 × 2 = 1 + 0.534 618 624;
  • 44) 0.534 618 624 × 2 = 1 + 0.069 237 248;
  • 45) 0.069 237 248 × 2 = 0 + 0.138 474 496;

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:


0.000 000 253(10) =

0.0000 0000 0000 0000 0000 0100 0011 1110 1010 0000 0111 0(2)

5. Positive number before normalization:

0.000 000 253(10) =

0.0000 0000 0000 0000 0000 0100 0011 1110 1010 0000 0111 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 253(10) =

0.0000 0000 0000 0000 0000 0100 0011 1110 1010 0000 0111 0(2) =

0.0000 0000 0000 0000 0000 0100 0011 1110 1010 0000 0111 0(2) × 20 =

1.0000 1111 1010 1000 0001 110(2) × 2-22


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): -22

Mantissa (not normalized):
1.0000 1111 1010 1000 0001 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(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;
  • 105 ÷ 2 = 52 + 1;
  • 52 ÷ 2 = 26 + 0;
  • 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) =

105(10) =

0110 1001(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 0111 1101 0100 0000 1110 =

000 0111 1101 0100 0000 1110


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 1001

Mantissa (23 bits) =
000 0111 1101 0100 0000 1110


Decimal number 0.000 000 253 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 000 0111 1101 0100 0000 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