-0.000 000 341 99 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 341 99(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
-0.000 000 341 99(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. Start with the positive version of the number:

|-0.000 000 341 99| = 0.000 000 341 99


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 341 99.

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) 0.000 000 341 99 × 2 = 0 + 0.000 000 683 98;
  • 2) 0.000 000 683 98 × 2 = 0 + 0.000 001 367 96;
  • 3) 0.000 001 367 96 × 2 = 0 + 0.000 002 735 92;
  • 4) 0.000 002 735 92 × 2 = 0 + 0.000 005 471 84;
  • 5) 0.000 005 471 84 × 2 = 0 + 0.000 010 943 68;
  • 6) 0.000 010 943 68 × 2 = 0 + 0.000 021 887 36;
  • 7) 0.000 021 887 36 × 2 = 0 + 0.000 043 774 72;
  • 8) 0.000 043 774 72 × 2 = 0 + 0.000 087 549 44;
  • 9) 0.000 087 549 44 × 2 = 0 + 0.000 175 098 88;
  • 10) 0.000 175 098 88 × 2 = 0 + 0.000 350 197 76;
  • 11) 0.000 350 197 76 × 2 = 0 + 0.000 700 395 52;
  • 12) 0.000 700 395 52 × 2 = 0 + 0.001 400 791 04;
  • 13) 0.001 400 791 04 × 2 = 0 + 0.002 801 582 08;
  • 14) 0.002 801 582 08 × 2 = 0 + 0.005 603 164 16;
  • 15) 0.005 603 164 16 × 2 = 0 + 0.011 206 328 32;
  • 16) 0.011 206 328 32 × 2 = 0 + 0.022 412 656 64;
  • 17) 0.022 412 656 64 × 2 = 0 + 0.044 825 313 28;
  • 18) 0.044 825 313 28 × 2 = 0 + 0.089 650 626 56;
  • 19) 0.089 650 626 56 × 2 = 0 + 0.179 301 253 12;
  • 20) 0.179 301 253 12 × 2 = 0 + 0.358 602 506 24;
  • 21) 0.358 602 506 24 × 2 = 0 + 0.717 205 012 48;
  • 22) 0.717 205 012 48 × 2 = 1 + 0.434 410 024 96;
  • 23) 0.434 410 024 96 × 2 = 0 + 0.868 820 049 92;
  • 24) 0.868 820 049 92 × 2 = 1 + 0.737 640 099 84;
  • 25) 0.737 640 099 84 × 2 = 1 + 0.475 280 199 68;
  • 26) 0.475 280 199 68 × 2 = 0 + 0.950 560 399 36;
  • 27) 0.950 560 399 36 × 2 = 1 + 0.901 120 798 72;
  • 28) 0.901 120 798 72 × 2 = 1 + 0.802 241 597 44;
  • 29) 0.802 241 597 44 × 2 = 1 + 0.604 483 194 88;
  • 30) 0.604 483 194 88 × 2 = 1 + 0.208 966 389 76;
  • 31) 0.208 966 389 76 × 2 = 0 + 0.417 932 779 52;
  • 32) 0.417 932 779 52 × 2 = 0 + 0.835 865 559 04;
  • 33) 0.835 865 559 04 × 2 = 1 + 0.671 731 118 08;
  • 34) 0.671 731 118 08 × 2 = 1 + 0.343 462 236 16;
  • 35) 0.343 462 236 16 × 2 = 0 + 0.686 924 472 32;
  • 36) 0.686 924 472 32 × 2 = 1 + 0.373 848 944 64;
  • 37) 0.373 848 944 64 × 2 = 0 + 0.747 697 889 28;
  • 38) 0.747 697 889 28 × 2 = 1 + 0.495 395 778 56;
  • 39) 0.495 395 778 56 × 2 = 0 + 0.990 791 557 12;
  • 40) 0.990 791 557 12 × 2 = 1 + 0.981 583 114 24;
  • 41) 0.981 583 114 24 × 2 = 1 + 0.963 166 228 48;
  • 42) 0.963 166 228 48 × 2 = 1 + 0.926 332 456 96;
  • 43) 0.926 332 456 96 × 2 = 1 + 0.852 664 913 92;
  • 44) 0.852 664 913 92 × 2 = 1 + 0.705 329 827 84;
  • 45) 0.705 329 827 84 × 2 = 1 + 0.410 659 655 68;

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).


5. 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:


0.000 000 341 99(10) =

0.0000 0000 0000 0000 0000 0101 1011 1100 1101 0101 1111 1(2)

6. Positive number before normalization:

0.000 000 341 99(10) =

0.0000 0000 0000 0000 0000 0101 1011 1100 1101 0101 1111 1(2)

7. 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:


0.000 000 341 99(10) =

0.0000 0000 0000 0000 0000 0101 1011 1100 1101 0101 1111 1(2) =

0.0000 0000 0000 0000 0000 0101 1011 1100 1101 0101 1111 1(2) × 20 =

1.0110 1111 0011 0101 0111 111(2) × 2-22


8. 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 1 (a negative number)

Exponent (unadjusted): -22

Mantissa (not normalized):
1.0110 1111 0011 0101 0111 111


9. 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)


10. 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;

11. 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)


12. 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 0111 1001 1010 1011 1111 =

011 0111 1001 1010 1011 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 1001

Mantissa (23 bits) =
011 0111 1001 1010 1011 1111


Decimal number -0.000 000 341 99 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1001 - 011 0111 1001 1010 1011 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