32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 101 110 014 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 101 110 014(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. 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;

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.


0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 101 110 014.

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) 0.000 101 110 014 × 2 = 0 + 0.000 202 220 028;
  • 2) 0.000 202 220 028 × 2 = 0 + 0.000 404 440 056;
  • 3) 0.000 404 440 056 × 2 = 0 + 0.000 808 880 112;
  • 4) 0.000 808 880 112 × 2 = 0 + 0.001 617 760 224;
  • 5) 0.001 617 760 224 × 2 = 0 + 0.003 235 520 448;
  • 6) 0.003 235 520 448 × 2 = 0 + 0.006 471 040 896;
  • 7) 0.006 471 040 896 × 2 = 0 + 0.012 942 081 792;
  • 8) 0.012 942 081 792 × 2 = 0 + 0.025 884 163 584;
  • 9) 0.025 884 163 584 × 2 = 0 + 0.051 768 327 168;
  • 10) 0.051 768 327 168 × 2 = 0 + 0.103 536 654 336;
  • 11) 0.103 536 654 336 × 2 = 0 + 0.207 073 308 672;
  • 12) 0.207 073 308 672 × 2 = 0 + 0.414 146 617 344;
  • 13) 0.414 146 617 344 × 2 = 0 + 0.828 293 234 688;
  • 14) 0.828 293 234 688 × 2 = 1 + 0.656 586 469 376;
  • 15) 0.656 586 469 376 × 2 = 1 + 0.313 172 938 752;
  • 16) 0.313 172 938 752 × 2 = 0 + 0.626 345 877 504;
  • 17) 0.626 345 877 504 × 2 = 1 + 0.252 691 755 008;
  • 18) 0.252 691 755 008 × 2 = 0 + 0.505 383 510 016;
  • 19) 0.505 383 510 016 × 2 = 1 + 0.010 767 020 032;
  • 20) 0.010 767 020 032 × 2 = 0 + 0.021 534 040 064;
  • 21) 0.021 534 040 064 × 2 = 0 + 0.043 068 080 128;
  • 22) 0.043 068 080 128 × 2 = 0 + 0.086 136 160 256;
  • 23) 0.086 136 160 256 × 2 = 0 + 0.172 272 320 512;
  • 24) 0.172 272 320 512 × 2 = 0 + 0.344 544 641 024;
  • 25) 0.344 544 641 024 × 2 = 0 + 0.689 089 282 048;
  • 26) 0.689 089 282 048 × 2 = 1 + 0.378 178 564 096;
  • 27) 0.378 178 564 096 × 2 = 0 + 0.756 357 128 192;
  • 28) 0.756 357 128 192 × 2 = 1 + 0.512 714 256 384;
  • 29) 0.512 714 256 384 × 2 = 1 + 0.025 428 512 768;
  • 30) 0.025 428 512 768 × 2 = 0 + 0.050 857 025 536;
  • 31) 0.050 857 025 536 × 2 = 0 + 0.101 714 051 072;
  • 32) 0.101 714 051 072 × 2 = 0 + 0.203 428 102 144;
  • 33) 0.203 428 102 144 × 2 = 0 + 0.406 856 204 288;
  • 34) 0.406 856 204 288 × 2 = 0 + 0.813 712 408 576;
  • 35) 0.813 712 408 576 × 2 = 1 + 0.627 424 817 152;
  • 36) 0.627 424 817 152 × 2 = 1 + 0.254 849 634 304;
  • 37) 0.254 849 634 304 × 2 = 0 + 0.509 699 268 608;

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...)


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:


0.000 101 110 014(10) =

0.0000 0000 0000 0110 1010 0000 0101 1000 0011 0(2)


5. Positive number before normalization:

0.000 101 110 014(10) =

0.0000 0000 0000 0110 1010 0000 0101 1000 0011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 101 110 014(10) =

0.0000 0000 0000 0110 1010 0000 0101 1000 0011 0(2) =

0.0000 0000 0000 0110 1010 0000 0101 1000 0011 0(2) × 20 =

1.1010 1000 0001 0110 0000 110(2) × 2-14


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): -14

Mantissa (not normalized):
1.1010 1000 0001 0110 0000 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-14 + 2(8-1) - 1 =

(-14 + 127)(10) =

113(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;
  • 113 ÷ 2 = 56 + 1;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

113(10) =

0111 0001(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 101 0100 0000 1011 0000 0110 =

101 0100 0000 1011 0000 0110


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) =
0111 0001

Mantissa (23 bits) =
101 0100 0000 1011 0000 0110


The base ten decimal number 0.000 101 110 014 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 0001 - 101 0100 0000 1011 0000 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