0.000 000 174 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 174(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
0.000 000 174(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: 0.000 000 174.

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) 0.000 000 174 × 2 = 0 + 0.000 000 348;
  • 2) 0.000 000 348 × 2 = 0 + 0.000 000 696;
  • 3) 0.000 000 696 × 2 = 0 + 0.000 001 392;
  • 4) 0.000 001 392 × 2 = 0 + 0.000 002 784;
  • 5) 0.000 002 784 × 2 = 0 + 0.000 005 568;
  • 6) 0.000 005 568 × 2 = 0 + 0.000 011 136;
  • 7) 0.000 011 136 × 2 = 0 + 0.000 022 272;
  • 8) 0.000 022 272 × 2 = 0 + 0.000 044 544;
  • 9) 0.000 044 544 × 2 = 0 + 0.000 089 088;
  • 10) 0.000 089 088 × 2 = 0 + 0.000 178 176;
  • 11) 0.000 178 176 × 2 = 0 + 0.000 356 352;
  • 12) 0.000 356 352 × 2 = 0 + 0.000 712 704;
  • 13) 0.000 712 704 × 2 = 0 + 0.001 425 408;
  • 14) 0.001 425 408 × 2 = 0 + 0.002 850 816;
  • 15) 0.002 850 816 × 2 = 0 + 0.005 701 632;
  • 16) 0.005 701 632 × 2 = 0 + 0.011 403 264;
  • 17) 0.011 403 264 × 2 = 0 + 0.022 806 528;
  • 18) 0.022 806 528 × 2 = 0 + 0.045 613 056;
  • 19) 0.045 613 056 × 2 = 0 + 0.091 226 112;
  • 20) 0.091 226 112 × 2 = 0 + 0.182 452 224;
  • 21) 0.182 452 224 × 2 = 0 + 0.364 904 448;
  • 22) 0.364 904 448 × 2 = 0 + 0.729 808 896;
  • 23) 0.729 808 896 × 2 = 1 + 0.459 617 792;
  • 24) 0.459 617 792 × 2 = 0 + 0.919 235 584;
  • 25) 0.919 235 584 × 2 = 1 + 0.838 471 168;
  • 26) 0.838 471 168 × 2 = 1 + 0.676 942 336;
  • 27) 0.676 942 336 × 2 = 1 + 0.353 884 672;
  • 28) 0.353 884 672 × 2 = 0 + 0.707 769 344;
  • 29) 0.707 769 344 × 2 = 1 + 0.415 538 688;
  • 30) 0.415 538 688 × 2 = 0 + 0.831 077 376;
  • 31) 0.831 077 376 × 2 = 1 + 0.662 154 752;
  • 32) 0.662 154 752 × 2 = 1 + 0.324 309 504;
  • 33) 0.324 309 504 × 2 = 0 + 0.648 619 008;
  • 34) 0.648 619 008 × 2 = 1 + 0.297 238 016;
  • 35) 0.297 238 016 × 2 = 0 + 0.594 476 032;
  • 36) 0.594 476 032 × 2 = 1 + 0.188 952 064;
  • 37) 0.188 952 064 × 2 = 0 + 0.377 904 128;
  • 38) 0.377 904 128 × 2 = 0 + 0.755 808 256;
  • 39) 0.755 808 256 × 2 = 1 + 0.511 616 512;
  • 40) 0.511 616 512 × 2 = 1 + 0.023 233 024;
  • 41) 0.023 233 024 × 2 = 0 + 0.046 466 048;
  • 42) 0.046 466 048 × 2 = 0 + 0.092 932 096;
  • 43) 0.092 932 096 × 2 = 0 + 0.185 864 192;
  • 44) 0.185 864 192 × 2 = 0 + 0.371 728 384;
  • 45) 0.371 728 384 × 2 = 0 + 0.743 456 768;
  • 46) 0.743 456 768 × 2 = 1 + 0.486 913 536;

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:


0.000 000 174(10) =

0.0000 0000 0000 0000 0000 0010 1110 1011 0101 0011 0000 01(2)

5. Positive number before normalization:

0.000 000 174(10) =

0.0000 0000 0000 0000 0000 0010 1110 1011 0101 0011 0000 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 23 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 174(10) =

0.0000 0000 0000 0000 0000 0010 1110 1011 0101 0011 0000 01(2) =

0.0000 0000 0000 0000 0000 0010 1110 1011 0101 0011 0000 01(2) × 20 =

1.0111 0101 1010 1001 1000 001(2) × 2-23


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): -23

Mantissa (not normalized):
1.0111 0101 1010 1001 1000 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-23 + 2(8-1) - 1 =

(-23 + 127)(10) =

104(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;
  • 104 ÷ 2 = 52 + 0;
  • 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) =

104(10) =

0110 1000(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. 011 1010 1101 0100 1100 0001 =

011 1010 1101 0100 1100 0001


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 1000

Mantissa (23 bits) =
011 1010 1101 0100 1100 0001


Decimal number 0.000 000 174 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1000 - 011 1010 1101 0100 1100 0001

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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