-0.000 000 000 725 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 725(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
-0.000 000 000 725(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:

|-0.000 000 000 725| = 0.000 000 000 725


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: 0.000 000 000 725.

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) 0.000 000 000 725 × 2 = 0 + 0.000 000 001 45;
  • 2) 0.000 000 001 45 × 2 = 0 + 0.000 000 002 9;
  • 3) 0.000 000 002 9 × 2 = 0 + 0.000 000 005 8;
  • 4) 0.000 000 005 8 × 2 = 0 + 0.000 000 011 6;
  • 5) 0.000 000 011 6 × 2 = 0 + 0.000 000 023 2;
  • 6) 0.000 000 023 2 × 2 = 0 + 0.000 000 046 4;
  • 7) 0.000 000 046 4 × 2 = 0 + 0.000 000 092 8;
  • 8) 0.000 000 092 8 × 2 = 0 + 0.000 000 185 6;
  • 9) 0.000 000 185 6 × 2 = 0 + 0.000 000 371 2;
  • 10) 0.000 000 371 2 × 2 = 0 + 0.000 000 742 4;
  • 11) 0.000 000 742 4 × 2 = 0 + 0.000 001 484 8;
  • 12) 0.000 001 484 8 × 2 = 0 + 0.000 002 969 6;
  • 13) 0.000 002 969 6 × 2 = 0 + 0.000 005 939 2;
  • 14) 0.000 005 939 2 × 2 = 0 + 0.000 011 878 4;
  • 15) 0.000 011 878 4 × 2 = 0 + 0.000 023 756 8;
  • 16) 0.000 023 756 8 × 2 = 0 + 0.000 047 513 6;
  • 17) 0.000 047 513 6 × 2 = 0 + 0.000 095 027 2;
  • 18) 0.000 095 027 2 × 2 = 0 + 0.000 190 054 4;
  • 19) 0.000 190 054 4 × 2 = 0 + 0.000 380 108 8;
  • 20) 0.000 380 108 8 × 2 = 0 + 0.000 760 217 6;
  • 21) 0.000 760 217 6 × 2 = 0 + 0.001 520 435 2;
  • 22) 0.001 520 435 2 × 2 = 0 + 0.003 040 870 4;
  • 23) 0.003 040 870 4 × 2 = 0 + 0.006 081 740 8;
  • 24) 0.006 081 740 8 × 2 = 0 + 0.012 163 481 6;
  • 25) 0.012 163 481 6 × 2 = 0 + 0.024 326 963 2;
  • 26) 0.024 326 963 2 × 2 = 0 + 0.048 653 926 4;
  • 27) 0.048 653 926 4 × 2 = 0 + 0.097 307 852 8;
  • 28) 0.097 307 852 8 × 2 = 0 + 0.194 615 705 6;
  • 29) 0.194 615 705 6 × 2 = 0 + 0.389 231 411 2;
  • 30) 0.389 231 411 2 × 2 = 0 + 0.778 462 822 4;
  • 31) 0.778 462 822 4 × 2 = 1 + 0.556 925 644 8;
  • 32) 0.556 925 644 8 × 2 = 1 + 0.113 851 289 6;
  • 33) 0.113 851 289 6 × 2 = 0 + 0.227 702 579 2;
  • 34) 0.227 702 579 2 × 2 = 0 + 0.455 405 158 4;
  • 35) 0.455 405 158 4 × 2 = 0 + 0.910 810 316 8;
  • 36) 0.910 810 316 8 × 2 = 1 + 0.821 620 633 6;
  • 37) 0.821 620 633 6 × 2 = 1 + 0.643 241 267 2;
  • 38) 0.643 241 267 2 × 2 = 1 + 0.286 482 534 4;
  • 39) 0.286 482 534 4 × 2 = 0 + 0.572 965 068 8;
  • 40) 0.572 965 068 8 × 2 = 1 + 0.145 930 137 6;
  • 41) 0.145 930 137 6 × 2 = 0 + 0.291 860 275 2;
  • 42) 0.291 860 275 2 × 2 = 0 + 0.583 720 550 4;
  • 43) 0.583 720 550 4 × 2 = 1 + 0.167 441 100 8;
  • 44) 0.167 441 100 8 × 2 = 0 + 0.334 882 201 6;
  • 45) 0.334 882 201 6 × 2 = 0 + 0.669 764 403 2;
  • 46) 0.669 764 403 2 × 2 = 1 + 0.339 528 806 4;
  • 47) 0.339 528 806 4 × 2 = 0 + 0.679 057 612 8;
  • 48) 0.679 057 612 8 × 2 = 1 + 0.358 115 225 6;
  • 49) 0.358 115 225 6 × 2 = 0 + 0.716 230 451 2;
  • 50) 0.716 230 451 2 × 2 = 1 + 0.432 460 902 4;
  • 51) 0.432 460 902 4 × 2 = 0 + 0.864 921 804 8;
  • 52) 0.864 921 804 8 × 2 = 1 + 0.729 843 609 6;
  • 53) 0.729 843 609 6 × 2 = 1 + 0.459 687 219 2;
  • 54) 0.459 687 219 2 × 2 = 0 + 0.919 374 438 4;

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:


0.000 000 000 725(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1101 0010 0101 0101 10(2)

6. Positive number before normalization:

0.000 000 000 725(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1101 0010 0101 0101 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:


0.000 000 000 725(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1101 0010 0101 0101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0001 1101 0010 0101 0101 10(2) × 20 =

1.1000 1110 1001 0010 1010 110(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):
1.1000 1110 1001 0010 1010 110


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 0111 0100 1001 0101 0110 =

100 0111 0100 1001 0101 0110


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 0111 0100 1001 0101 0110


Decimal number -0.000 000 000 725 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 0111 0100 1001 0101 0110

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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