32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 000 000 814 907 254 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 000 000 814 907 254(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: 0.000 000 000 814 907 254.

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) 0.000 000 000 814 907 254 × 2 = 0 + 0.000 000 001 629 814 508;
  • 2) 0.000 000 001 629 814 508 × 2 = 0 + 0.000 000 003 259 629 016;
  • 3) 0.000 000 003 259 629 016 × 2 = 0 + 0.000 000 006 519 258 032;
  • 4) 0.000 000 006 519 258 032 × 2 = 0 + 0.000 000 013 038 516 064;
  • 5) 0.000 000 013 038 516 064 × 2 = 0 + 0.000 000 026 077 032 128;
  • 6) 0.000 000 026 077 032 128 × 2 = 0 + 0.000 000 052 154 064 256;
  • 7) 0.000 000 052 154 064 256 × 2 = 0 + 0.000 000 104 308 128 512;
  • 8) 0.000 000 104 308 128 512 × 2 = 0 + 0.000 000 208 616 257 024;
  • 9) 0.000 000 208 616 257 024 × 2 = 0 + 0.000 000 417 232 514 048;
  • 10) 0.000 000 417 232 514 048 × 2 = 0 + 0.000 000 834 465 028 096;
  • 11) 0.000 000 834 465 028 096 × 2 = 0 + 0.000 001 668 930 056 192;
  • 12) 0.000 001 668 930 056 192 × 2 = 0 + 0.000 003 337 860 112 384;
  • 13) 0.000 003 337 860 112 384 × 2 = 0 + 0.000 006 675 720 224 768;
  • 14) 0.000 006 675 720 224 768 × 2 = 0 + 0.000 013 351 440 449 536;
  • 15) 0.000 013 351 440 449 536 × 2 = 0 + 0.000 026 702 880 899 072;
  • 16) 0.000 026 702 880 899 072 × 2 = 0 + 0.000 053 405 761 798 144;
  • 17) 0.000 053 405 761 798 144 × 2 = 0 + 0.000 106 811 523 596 288;
  • 18) 0.000 106 811 523 596 288 × 2 = 0 + 0.000 213 623 047 192 576;
  • 19) 0.000 213 623 047 192 576 × 2 = 0 + 0.000 427 246 094 385 152;
  • 20) 0.000 427 246 094 385 152 × 2 = 0 + 0.000 854 492 188 770 304;
  • 21) 0.000 854 492 188 770 304 × 2 = 0 + 0.001 708 984 377 540 608;
  • 22) 0.001 708 984 377 540 608 × 2 = 0 + 0.003 417 968 755 081 216;
  • 23) 0.003 417 968 755 081 216 × 2 = 0 + 0.006 835 937 510 162 432;
  • 24) 0.006 835 937 510 162 432 × 2 = 0 + 0.013 671 875 020 324 864;
  • 25) 0.013 671 875 020 324 864 × 2 = 0 + 0.027 343 750 040 649 728;
  • 26) 0.027 343 750 040 649 728 × 2 = 0 + 0.054 687 500 081 299 456;
  • 27) 0.054 687 500 081 299 456 × 2 = 0 + 0.109 375 000 162 598 912;
  • 28) 0.109 375 000 162 598 912 × 2 = 0 + 0.218 750 000 325 197 824;
  • 29) 0.218 750 000 325 197 824 × 2 = 0 + 0.437 500 000 650 395 648;
  • 30) 0.437 500 000 650 395 648 × 2 = 0 + 0.875 000 001 300 791 296;
  • 31) 0.875 000 001 300 791 296 × 2 = 1 + 0.750 000 002 601 582 592;
  • 32) 0.750 000 002 601 582 592 × 2 = 1 + 0.500 000 005 203 165 184;
  • 33) 0.500 000 005 203 165 184 × 2 = 1 + 0.000 000 010 406 330 368;
  • 34) 0.000 000 010 406 330 368 × 2 = 0 + 0.000 000 020 812 660 736;
  • 35) 0.000 000 020 812 660 736 × 2 = 0 + 0.000 000 041 625 321 472;
  • 36) 0.000 000 041 625 321 472 × 2 = 0 + 0.000 000 083 250 642 944;
  • 37) 0.000 000 083 250 642 944 × 2 = 0 + 0.000 000 166 501 285 888;
  • 38) 0.000 000 166 501 285 888 × 2 = 0 + 0.000 000 333 002 571 776;
  • 39) 0.000 000 333 002 571 776 × 2 = 0 + 0.000 000 666 005 143 552;
  • 40) 0.000 000 666 005 143 552 × 2 = 0 + 0.000 001 332 010 287 104;
  • 41) 0.000 001 332 010 287 104 × 2 = 0 + 0.000 002 664 020 574 208;
  • 42) 0.000 002 664 020 574 208 × 2 = 0 + 0.000 005 328 041 148 416;
  • 43) 0.000 005 328 041 148 416 × 2 = 0 + 0.000 010 656 082 296 832;
  • 44) 0.000 010 656 082 296 832 × 2 = 0 + 0.000 021 312 164 593 664;
  • 45) 0.000 021 312 164 593 664 × 2 = 0 + 0.000 042 624 329 187 328;
  • 46) 0.000 042 624 329 187 328 × 2 = 0 + 0.000 085 248 658 374 656;
  • 47) 0.000 085 248 658 374 656 × 2 = 0 + 0.000 170 497 316 749 312;
  • 48) 0.000 170 497 316 749 312 × 2 = 0 + 0.000 340 994 633 498 624;
  • 49) 0.000 340 994 633 498 624 × 2 = 0 + 0.000 681 989 266 997 248;
  • 50) 0.000 681 989 266 997 248 × 2 = 0 + 0.001 363 978 533 994 496;
  • 51) 0.001 363 978 533 994 496 × 2 = 0 + 0.002 727 957 067 988 992;
  • 52) 0.002 727 957 067 988 992 × 2 = 0 + 0.005 455 914 135 977 984;
  • 53) 0.005 455 914 135 977 984 × 2 = 0 + 0.010 911 828 271 955 968;
  • 54) 0.010 911 828 271 955 968 × 2 = 0 + 0.021 823 656 543 911 936;

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:


0.000 000 000 814 907 254(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 0000 0000 0000 0000 00(2)


5. Positive number before normalization:

0.000 000 000 814 907 254(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 0000 0000 0000 0000 00(2)

6. 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:


0.000 000 000 814 907 254(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 1000 0000 0000 0000 0000 00(2) × 20 =

1.1100 0000 0000 0000 0000 000(2) × 2-31


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): -31

Mantissa (not normalized):
1.1100 0000 0000 0000 0000 000


8. 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)


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;
  • 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;

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) =

96(10) =

0110 0000(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. 110 0000 0000 0000 0000 0000 =

110 0000 0000 0000 0000 0000


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) =
0110 0000

Mantissa (23 bits) =
110 0000 0000 0000 0000 0000


The base ten decimal number 0.000 000 000 814 907 254 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0110 0000 - 110 0000 0000 0000 0000 0000

More operations with decimal numbers converted 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