32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 562 949 953 421 211 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 562 949 953 421 211(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. 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;
  • 562 949 953 421 211 ÷ 2 = 281 474 976 710 605 + 1;
  • 281 474 976 710 605 ÷ 2 = 140 737 488 355 302 + 1;
  • 140 737 488 355 302 ÷ 2 = 70 368 744 177 651 + 0;
  • 70 368 744 177 651 ÷ 2 = 35 184 372 088 825 + 1;
  • 35 184 372 088 825 ÷ 2 = 17 592 186 044 412 + 1;
  • 17 592 186 044 412 ÷ 2 = 8 796 093 022 206 + 0;
  • 8 796 093 022 206 ÷ 2 = 4 398 046 511 103 + 0;
  • 4 398 046 511 103 ÷ 2 = 2 199 023 255 551 + 1;
  • 2 199 023 255 551 ÷ 2 = 1 099 511 627 775 + 1;
  • 1 099 511 627 775 ÷ 2 = 549 755 813 887 + 1;
  • 549 755 813 887 ÷ 2 = 274 877 906 943 + 1;
  • 274 877 906 943 ÷ 2 = 137 438 953 471 + 1;
  • 137 438 953 471 ÷ 2 = 68 719 476 735 + 1;
  • 68 719 476 735 ÷ 2 = 34 359 738 367 + 1;
  • 34 359 738 367 ÷ 2 = 17 179 869 183 + 1;
  • 17 179 869 183 ÷ 2 = 8 589 934 591 + 1;
  • 8 589 934 591 ÷ 2 = 4 294 967 295 + 1;
  • 4 294 967 295 ÷ 2 = 2 147 483 647 + 1;
  • 2 147 483 647 ÷ 2 = 1 073 741 823 + 1;
  • 1 073 741 823 ÷ 2 = 536 870 911 + 1;
  • 536 870 911 ÷ 2 = 268 435 455 + 1;
  • 268 435 455 ÷ 2 = 134 217 727 + 1;
  • 134 217 727 ÷ 2 = 67 108 863 + 1;
  • 67 108 863 ÷ 2 = 33 554 431 + 1;
  • 33 554 431 ÷ 2 = 16 777 215 + 1;
  • 16 777 215 ÷ 2 = 8 388 607 + 1;
  • 8 388 607 ÷ 2 = 4 194 303 + 1;
  • 4 194 303 ÷ 2 = 2 097 151 + 1;
  • 2 097 151 ÷ 2 = 1 048 575 + 1;
  • 1 048 575 ÷ 2 = 524 287 + 1;
  • 524 287 ÷ 2 = 262 143 + 1;
  • 262 143 ÷ 2 = 131 071 + 1;
  • 131 071 ÷ 2 = 65 535 + 1;
  • 65 535 ÷ 2 = 32 767 + 1;
  • 32 767 ÷ 2 = 16 383 + 1;
  • 16 383 ÷ 2 = 8 191 + 1;
  • 8 191 ÷ 2 = 4 095 + 1;
  • 4 095 ÷ 2 = 2 047 + 1;
  • 2 047 ÷ 2 = 1 023 + 1;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the positive number.

Take all the remainders starting from the bottom of the list constructed above.


562 949 953 421 211(10) =

1 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1001 1011(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 48 positions to the left, so that only one non zero digit remains to the left of it:


562 949 953 421 211(10) =

1 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1001 1011(2) =

1 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1001 1011(2) × 20 =

1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1001 1011(2) × 248


4. 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): 48

Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1001 1011


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

48 + 2(8-1) - 1 =

(48 + 127)(10) =

175(10)


6. 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;
  • 175 ÷ 2 = 87 + 1;
  • 87 ÷ 2 = 43 + 1;
  • 43 ÷ 2 = 21 + 1;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

7. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

175(10) =

1010 1111(2)


8. 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. 111 1111 1111 1111 1111 1111 1 1111 1111 1111 1111 1001 1011 =

111 1111 1111 1111 1111 1111


9. 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) =
1010 1111

Mantissa (23 bits) =
111 1111 1111 1111 1111 1111


The base ten decimal number 562 949 953 421 211 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1010 1111 - 111 1111 1111 1111 1111 1111

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number 562 949 953 421 210 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number 562 949 953 421 212 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

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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