-0.000 000 000 737 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 737 5(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
-0.000 000 000 737 5(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:

|-0.000 000 000 737 5| = 0.000 000 000 737 5


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: 0.000 000 000 737 5.

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) 0.000 000 000 737 5 × 2 = 0 + 0.000 000 001 475;
  • 2) 0.000 000 001 475 × 2 = 0 + 0.000 000 002 95;
  • 3) 0.000 000 002 95 × 2 = 0 + 0.000 000 005 9;
  • 4) 0.000 000 005 9 × 2 = 0 + 0.000 000 011 8;
  • 5) 0.000 000 011 8 × 2 = 0 + 0.000 000 023 6;
  • 6) 0.000 000 023 6 × 2 = 0 + 0.000 000 047 2;
  • 7) 0.000 000 047 2 × 2 = 0 + 0.000 000 094 4;
  • 8) 0.000 000 094 4 × 2 = 0 + 0.000 000 188 8;
  • 9) 0.000 000 188 8 × 2 = 0 + 0.000 000 377 6;
  • 10) 0.000 000 377 6 × 2 = 0 + 0.000 000 755 2;
  • 11) 0.000 000 755 2 × 2 = 0 + 0.000 001 510 4;
  • 12) 0.000 001 510 4 × 2 = 0 + 0.000 003 020 8;
  • 13) 0.000 003 020 8 × 2 = 0 + 0.000 006 041 6;
  • 14) 0.000 006 041 6 × 2 = 0 + 0.000 012 083 2;
  • 15) 0.000 012 083 2 × 2 = 0 + 0.000 024 166 4;
  • 16) 0.000 024 166 4 × 2 = 0 + 0.000 048 332 8;
  • 17) 0.000 048 332 8 × 2 = 0 + 0.000 096 665 6;
  • 18) 0.000 096 665 6 × 2 = 0 + 0.000 193 331 2;
  • 19) 0.000 193 331 2 × 2 = 0 + 0.000 386 662 4;
  • 20) 0.000 386 662 4 × 2 = 0 + 0.000 773 324 8;
  • 21) 0.000 773 324 8 × 2 = 0 + 0.001 546 649 6;
  • 22) 0.001 546 649 6 × 2 = 0 + 0.003 093 299 2;
  • 23) 0.003 093 299 2 × 2 = 0 + 0.006 186 598 4;
  • 24) 0.006 186 598 4 × 2 = 0 + 0.012 373 196 8;
  • 25) 0.012 373 196 8 × 2 = 0 + 0.024 746 393 6;
  • 26) 0.024 746 393 6 × 2 = 0 + 0.049 492 787 2;
  • 27) 0.049 492 787 2 × 2 = 0 + 0.098 985 574 4;
  • 28) 0.098 985 574 4 × 2 = 0 + 0.197 971 148 8;
  • 29) 0.197 971 148 8 × 2 = 0 + 0.395 942 297 6;
  • 30) 0.395 942 297 6 × 2 = 0 + 0.791 884 595 2;
  • 31) 0.791 884 595 2 × 2 = 1 + 0.583 769 190 4;
  • 32) 0.583 769 190 4 × 2 = 1 + 0.167 538 380 8;
  • 33) 0.167 538 380 8 × 2 = 0 + 0.335 076 761 6;
  • 34) 0.335 076 761 6 × 2 = 0 + 0.670 153 523 2;
  • 35) 0.670 153 523 2 × 2 = 1 + 0.340 307 046 4;
  • 36) 0.340 307 046 4 × 2 = 0 + 0.680 614 092 8;
  • 37) 0.680 614 092 8 × 2 = 1 + 0.361 228 185 6;
  • 38) 0.361 228 185 6 × 2 = 0 + 0.722 456 371 2;
  • 39) 0.722 456 371 2 × 2 = 1 + 0.444 912 742 4;
  • 40) 0.444 912 742 4 × 2 = 0 + 0.889 825 484 8;
  • 41) 0.889 825 484 8 × 2 = 1 + 0.779 650 969 6;
  • 42) 0.779 650 969 6 × 2 = 1 + 0.559 301 939 2;
  • 43) 0.559 301 939 2 × 2 = 1 + 0.118 603 878 4;
  • 44) 0.118 603 878 4 × 2 = 0 + 0.237 207 756 8;
  • 45) 0.237 207 756 8 × 2 = 0 + 0.474 415 513 6;
  • 46) 0.474 415 513 6 × 2 = 0 + 0.948 831 027 2;
  • 47) 0.948 831 027 2 × 2 = 1 + 0.897 662 054 4;
  • 48) 0.897 662 054 4 × 2 = 1 + 0.795 324 108 8;
  • 49) 0.795 324 108 8 × 2 = 1 + 0.590 648 217 6;
  • 50) 0.590 648 217 6 × 2 = 1 + 0.181 296 435 2;
  • 51) 0.181 296 435 2 × 2 = 0 + 0.362 592 870 4;
  • 52) 0.362 592 870 4 × 2 = 0 + 0.725 185 740 8;
  • 53) 0.725 185 740 8 × 2 = 1 + 0.450 371 481 6;
  • 54) 0.450 371 481 6 × 2 = 0 + 0.900 742 963 2;

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:


0.000 000 000 737 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1010 1110 0011 1100 10(2)

6. Positive number before normalization:

0.000 000 000 737 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1010 1110 0011 1100 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 737 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1010 1110 0011 1100 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1010 1110 0011 1100 10(2) × 20 =

1.1001 0101 0111 0001 1110 010(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0101 0111 0001 1110 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

96(10) =

0110 0000(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. 100 1010 1011 1000 1111 0010 =

100 1010 1011 1000 1111 0010


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 0000

Mantissa (23 bits) =
100 1010 1011 1000 1111 0010


Decimal number -0.000 000 000 737 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1010 1011 1000 1111 0010

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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