Decimal to 32 Bit IEEE 754 Binary: Convert Number 115.783 397 248 051 829 803 540 722 76 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number 115.783 397 248 051 829 803 540 722 76(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 115.
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;
  • 115 ÷ 2 = 57 + 1;
  • 57 ÷ 2 = 28 + 1;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

115(10) =

111 0011(2)


3. Convert to binary (base 2) the fractional part: 0.783 397 248 051 829 803 540 722 76.

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) 0.783 397 248 051 829 803 540 722 76 × 2 = 1 + 0.566 794 496 103 659 607 081 445 52;
  • 2) 0.566 794 496 103 659 607 081 445 52 × 2 = 1 + 0.133 588 992 207 319 214 162 891 04;
  • 3) 0.133 588 992 207 319 214 162 891 04 × 2 = 0 + 0.267 177 984 414 638 428 325 782 08;
  • 4) 0.267 177 984 414 638 428 325 782 08 × 2 = 0 + 0.534 355 968 829 276 856 651 564 16;
  • 5) 0.534 355 968 829 276 856 651 564 16 × 2 = 1 + 0.068 711 937 658 553 713 303 128 32;
  • 6) 0.068 711 937 658 553 713 303 128 32 × 2 = 0 + 0.137 423 875 317 107 426 606 256 64;
  • 7) 0.137 423 875 317 107 426 606 256 64 × 2 = 0 + 0.274 847 750 634 214 853 212 513 28;
  • 8) 0.274 847 750 634 214 853 212 513 28 × 2 = 0 + 0.549 695 501 268 429 706 425 026 56;
  • 9) 0.549 695 501 268 429 706 425 026 56 × 2 = 1 + 0.099 391 002 536 859 412 850 053 12;
  • 10) 0.099 391 002 536 859 412 850 053 12 × 2 = 0 + 0.198 782 005 073 718 825 700 106 24;
  • 11) 0.198 782 005 073 718 825 700 106 24 × 2 = 0 + 0.397 564 010 147 437 651 400 212 48;
  • 12) 0.397 564 010 147 437 651 400 212 48 × 2 = 0 + 0.795 128 020 294 875 302 800 424 96;
  • 13) 0.795 128 020 294 875 302 800 424 96 × 2 = 1 + 0.590 256 040 589 750 605 600 849 92;
  • 14) 0.590 256 040 589 750 605 600 849 92 × 2 = 1 + 0.180 512 081 179 501 211 201 699 84;
  • 15) 0.180 512 081 179 501 211 201 699 84 × 2 = 0 + 0.361 024 162 359 002 422 403 399 68;
  • 16) 0.361 024 162 359 002 422 403 399 68 × 2 = 0 + 0.722 048 324 718 004 844 806 799 36;
  • 17) 0.722 048 324 718 004 844 806 799 36 × 2 = 1 + 0.444 096 649 436 009 689 613 598 72;
  • 18) 0.444 096 649 436 009 689 613 598 72 × 2 = 0 + 0.888 193 298 872 019 379 227 197 44;
  • 19) 0.888 193 298 872 019 379 227 197 44 × 2 = 1 + 0.776 386 597 744 038 758 454 394 88;
  • 20) 0.776 386 597 744 038 758 454 394 88 × 2 = 1 + 0.552 773 195 488 077 516 908 789 76;
  • 21) 0.552 773 195 488 077 516 908 789 76 × 2 = 1 + 0.105 546 390 976 155 033 817 579 52;
  • 22) 0.105 546 390 976 155 033 817 579 52 × 2 = 0 + 0.211 092 781 952 310 067 635 159 04;
  • 23) 0.211 092 781 952 310 067 635 159 04 × 2 = 0 + 0.422 185 563 904 620 135 270 318 08;
  • 24) 0.422 185 563 904 620 135 270 318 08 × 2 = 0 + 0.844 371 127 809 240 270 540 636 16;

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:


0.783 397 248 051 829 803 540 722 76(10) =

0.1100 1000 1000 1100 1011 1000(2)

5. Positive number before normalization:

115.783 397 248 051 829 803 540 722 76(10) =

111 0011.1100 1000 1000 1100 1011 1000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 6 positions to the left, so that only one non zero digit remains to the left of it:


115.783 397 248 051 829 803 540 722 76(10) =

111 0011.1100 1000 1000 1100 1011 1000(2) =

111 0011.1100 1000 1000 1100 1011 1000(2) × 20 =

1.1100 1111 0010 0010 0011 0010 1110 00(2) × 26


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): 6

Mantissa (not normalized):
1.1100 1111 0010 0010 0011 0010 1110 00


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

6 + 2(8-1) - 1 =

(6 + 127)(10) =

133(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;
  • 133 ÷ 2 = 66 + 1;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

133(10) =

1000 0101(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 110 0111 1001 0001 0001 1001 011 1000 =

110 0111 1001 0001 0001 1001


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) =
1000 0101

Mantissa (23 bits) =
110 0111 1001 0001 0001 1001


The base ten decimal number 115.783 397 248 051 829 803 540 722 76 converted and written in 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0101 - 110 0111 1001 0001 0001 1001

» Decimal number in base ten 115.783 397 248 051 829 803 540 722 57 converted and written as 32 bit single precision IEEE 754 binary floating point representation standard

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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