115.783 397 248 051 829 803 540 229 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 115.783 397 248 051 829 803 540 229(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
115.783 397 248 051 829 803 540 229(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: 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: 0.783 397 248 051 829 803 540 229.

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) 0.783 397 248 051 829 803 540 229 × 2 = 1 + 0.566 794 496 103 659 607 080 458;
  • 2) 0.566 794 496 103 659 607 080 458 × 2 = 1 + 0.133 588 992 207 319 214 160 916;
  • 3) 0.133 588 992 207 319 214 160 916 × 2 = 0 + 0.267 177 984 414 638 428 321 832;
  • 4) 0.267 177 984 414 638 428 321 832 × 2 = 0 + 0.534 355 968 829 276 856 643 664;
  • 5) 0.534 355 968 829 276 856 643 664 × 2 = 1 + 0.068 711 937 658 553 713 287 328;
  • 6) 0.068 711 937 658 553 713 287 328 × 2 = 0 + 0.137 423 875 317 107 426 574 656;
  • 7) 0.137 423 875 317 107 426 574 656 × 2 = 0 + 0.274 847 750 634 214 853 149 312;
  • 8) 0.274 847 750 634 214 853 149 312 × 2 = 0 + 0.549 695 501 268 429 706 298 624;
  • 9) 0.549 695 501 268 429 706 298 624 × 2 = 1 + 0.099 391 002 536 859 412 597 248;
  • 10) 0.099 391 002 536 859 412 597 248 × 2 = 0 + 0.198 782 005 073 718 825 194 496;
  • 11) 0.198 782 005 073 718 825 194 496 × 2 = 0 + 0.397 564 010 147 437 650 388 992;
  • 12) 0.397 564 010 147 437 650 388 992 × 2 = 0 + 0.795 128 020 294 875 300 777 984;
  • 13) 0.795 128 020 294 875 300 777 984 × 2 = 1 + 0.590 256 040 589 750 601 555 968;
  • 14) 0.590 256 040 589 750 601 555 968 × 2 = 1 + 0.180 512 081 179 501 203 111 936;
  • 15) 0.180 512 081 179 501 203 111 936 × 2 = 0 + 0.361 024 162 359 002 406 223 872;
  • 16) 0.361 024 162 359 002 406 223 872 × 2 = 0 + 0.722 048 324 718 004 812 447 744;
  • 17) 0.722 048 324 718 004 812 447 744 × 2 = 1 + 0.444 096 649 436 009 624 895 488;
  • 18) 0.444 096 649 436 009 624 895 488 × 2 = 0 + 0.888 193 298 872 019 249 790 976;
  • 19) 0.888 193 298 872 019 249 790 976 × 2 = 1 + 0.776 386 597 744 038 499 581 952;
  • 20) 0.776 386 597 744 038 499 581 952 × 2 = 1 + 0.552 773 195 488 076 999 163 904;
  • 21) 0.552 773 195 488 076 999 163 904 × 2 = 1 + 0.105 546 390 976 153 998 327 808;
  • 22) 0.105 546 390 976 153 998 327 808 × 2 = 0 + 0.211 092 781 952 307 996 655 616;
  • 23) 0.211 092 781 952 307 996 655 616 × 2 = 0 + 0.422 185 563 904 615 993 311 232;
  • 24) 0.422 185 563 904 615 993 311 232 × 2 = 0 + 0.844 371 127 809 231 986 622 464;

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:


0.783 397 248 051 829 803 540 229(10) =

0.1100 1000 1000 1100 1011 1000(2)

5. Positive number before normalization:

115.783 397 248 051 829 803 540 229(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:


115.783 397 248 051 829 803 540 229(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


Decimal number 115.783 397 248 051 829 803 540 229 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0101 - 110 0111 1001 0001 0001 1001

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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