0.000 000 344 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 344(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
0.000 000 344(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: 0.000 000 344.

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) 0.000 000 344 × 2 = 0 + 0.000 000 688;
  • 2) 0.000 000 688 × 2 = 0 + 0.000 001 376;
  • 3) 0.000 001 376 × 2 = 0 + 0.000 002 752;
  • 4) 0.000 002 752 × 2 = 0 + 0.000 005 504;
  • 5) 0.000 005 504 × 2 = 0 + 0.000 011 008;
  • 6) 0.000 011 008 × 2 = 0 + 0.000 022 016;
  • 7) 0.000 022 016 × 2 = 0 + 0.000 044 032;
  • 8) 0.000 044 032 × 2 = 0 + 0.000 088 064;
  • 9) 0.000 088 064 × 2 = 0 + 0.000 176 128;
  • 10) 0.000 176 128 × 2 = 0 + 0.000 352 256;
  • 11) 0.000 352 256 × 2 = 0 + 0.000 704 512;
  • 12) 0.000 704 512 × 2 = 0 + 0.001 409 024;
  • 13) 0.001 409 024 × 2 = 0 + 0.002 818 048;
  • 14) 0.002 818 048 × 2 = 0 + 0.005 636 096;
  • 15) 0.005 636 096 × 2 = 0 + 0.011 272 192;
  • 16) 0.011 272 192 × 2 = 0 + 0.022 544 384;
  • 17) 0.022 544 384 × 2 = 0 + 0.045 088 768;
  • 18) 0.045 088 768 × 2 = 0 + 0.090 177 536;
  • 19) 0.090 177 536 × 2 = 0 + 0.180 355 072;
  • 20) 0.180 355 072 × 2 = 0 + 0.360 710 144;
  • 21) 0.360 710 144 × 2 = 0 + 0.721 420 288;
  • 22) 0.721 420 288 × 2 = 1 + 0.442 840 576;
  • 23) 0.442 840 576 × 2 = 0 + 0.885 681 152;
  • 24) 0.885 681 152 × 2 = 1 + 0.771 362 304;
  • 25) 0.771 362 304 × 2 = 1 + 0.542 724 608;
  • 26) 0.542 724 608 × 2 = 1 + 0.085 449 216;
  • 27) 0.085 449 216 × 2 = 0 + 0.170 898 432;
  • 28) 0.170 898 432 × 2 = 0 + 0.341 796 864;
  • 29) 0.341 796 864 × 2 = 0 + 0.683 593 728;
  • 30) 0.683 593 728 × 2 = 1 + 0.367 187 456;
  • 31) 0.367 187 456 × 2 = 0 + 0.734 374 912;
  • 32) 0.734 374 912 × 2 = 1 + 0.468 749 824;
  • 33) 0.468 749 824 × 2 = 0 + 0.937 499 648;
  • 34) 0.937 499 648 × 2 = 1 + 0.874 999 296;
  • 35) 0.874 999 296 × 2 = 1 + 0.749 998 592;
  • 36) 0.749 998 592 × 2 = 1 + 0.499 997 184;
  • 37) 0.499 997 184 × 2 = 0 + 0.999 994 368;
  • 38) 0.999 994 368 × 2 = 1 + 0.999 988 736;
  • 39) 0.999 988 736 × 2 = 1 + 0.999 977 472;
  • 40) 0.999 977 472 × 2 = 1 + 0.999 954 944;
  • 41) 0.999 954 944 × 2 = 1 + 0.999 909 888;
  • 42) 0.999 909 888 × 2 = 1 + 0.999 819 776;
  • 43) 0.999 819 776 × 2 = 1 + 0.999 639 552;
  • 44) 0.999 639 552 × 2 = 1 + 0.999 279 104;
  • 45) 0.999 279 104 × 2 = 1 + 0.998 558 208;

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:


0.000 000 344(10) =

0.0000 0000 0000 0000 0000 0101 1100 0101 0111 0111 1111 1(2)

5. Positive number before normalization:

0.000 000 344(10) =

0.0000 0000 0000 0000 0000 0101 1100 0101 0111 0111 1111 1(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:


0.000 000 344(10) =

0.0000 0000 0000 0000 0000 0101 1100 0101 0111 0111 1111 1(2) =

0.0000 0000 0000 0000 0000 0101 1100 0101 0111 0111 1111 1(2) × 20 =

1.0111 0001 0101 1101 1111 111(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):
1.0111 0001 0101 1101 1111 111


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. 011 1000 1010 1110 1111 1111 =

011 1000 1010 1110 1111 1111


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) =
011 1000 1010 1110 1111 1111


Decimal number 0.000 000 344 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1001 - 011 1000 1010 1110 1111 1111

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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