32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 000 000 000 019 317 4 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 000 000 000 019 317 4(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: 0.000 000 000 000 019 317 4.

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) 0.000 000 000 000 019 317 4 × 2 = 0 + 0.000 000 000 000 038 634 8;
  • 2) 0.000 000 000 000 038 634 8 × 2 = 0 + 0.000 000 000 000 077 269 6;
  • 3) 0.000 000 000 000 077 269 6 × 2 = 0 + 0.000 000 000 000 154 539 2;
  • 4) 0.000 000 000 000 154 539 2 × 2 = 0 + 0.000 000 000 000 309 078 4;
  • 5) 0.000 000 000 000 309 078 4 × 2 = 0 + 0.000 000 000 000 618 156 8;
  • 6) 0.000 000 000 000 618 156 8 × 2 = 0 + 0.000 000 000 001 236 313 6;
  • 7) 0.000 000 000 001 236 313 6 × 2 = 0 + 0.000 000 000 002 472 627 2;
  • 8) 0.000 000 000 002 472 627 2 × 2 = 0 + 0.000 000 000 004 945 254 4;
  • 9) 0.000 000 000 004 945 254 4 × 2 = 0 + 0.000 000 000 009 890 508 8;
  • 10) 0.000 000 000 009 890 508 8 × 2 = 0 + 0.000 000 000 019 781 017 6;
  • 11) 0.000 000 000 019 781 017 6 × 2 = 0 + 0.000 000 000 039 562 035 2;
  • 12) 0.000 000 000 039 562 035 2 × 2 = 0 + 0.000 000 000 079 124 070 4;
  • 13) 0.000 000 000 079 124 070 4 × 2 = 0 + 0.000 000 000 158 248 140 8;
  • 14) 0.000 000 000 158 248 140 8 × 2 = 0 + 0.000 000 000 316 496 281 6;
  • 15) 0.000 000 000 316 496 281 6 × 2 = 0 + 0.000 000 000 632 992 563 2;
  • 16) 0.000 000 000 632 992 563 2 × 2 = 0 + 0.000 000 001 265 985 126 4;
  • 17) 0.000 000 001 265 985 126 4 × 2 = 0 + 0.000 000 002 531 970 252 8;
  • 18) 0.000 000 002 531 970 252 8 × 2 = 0 + 0.000 000 005 063 940 505 6;
  • 19) 0.000 000 005 063 940 505 6 × 2 = 0 + 0.000 000 010 127 881 011 2;
  • 20) 0.000 000 010 127 881 011 2 × 2 = 0 + 0.000 000 020 255 762 022 4;
  • 21) 0.000 000 020 255 762 022 4 × 2 = 0 + 0.000 000 040 511 524 044 8;
  • 22) 0.000 000 040 511 524 044 8 × 2 = 0 + 0.000 000 081 023 048 089 6;
  • 23) 0.000 000 081 023 048 089 6 × 2 = 0 + 0.000 000 162 046 096 179 2;
  • 24) 0.000 000 162 046 096 179 2 × 2 = 0 + 0.000 000 324 092 192 358 4;
  • 25) 0.000 000 324 092 192 358 4 × 2 = 0 + 0.000 000 648 184 384 716 8;
  • 26) 0.000 000 648 184 384 716 8 × 2 = 0 + 0.000 001 296 368 769 433 6;
  • 27) 0.000 001 296 368 769 433 6 × 2 = 0 + 0.000 002 592 737 538 867 2;
  • 28) 0.000 002 592 737 538 867 2 × 2 = 0 + 0.000 005 185 475 077 734 4;
  • 29) 0.000 005 185 475 077 734 4 × 2 = 0 + 0.000 010 370 950 155 468 8;
  • 30) 0.000 010 370 950 155 468 8 × 2 = 0 + 0.000 020 741 900 310 937 6;
  • 31) 0.000 020 741 900 310 937 6 × 2 = 0 + 0.000 041 483 800 621 875 2;
  • 32) 0.000 041 483 800 621 875 2 × 2 = 0 + 0.000 082 967 601 243 750 4;
  • 33) 0.000 082 967 601 243 750 4 × 2 = 0 + 0.000 165 935 202 487 500 8;
  • 34) 0.000 165 935 202 487 500 8 × 2 = 0 + 0.000 331 870 404 975 001 6;
  • 35) 0.000 331 870 404 975 001 6 × 2 = 0 + 0.000 663 740 809 950 003 2;
  • 36) 0.000 663 740 809 950 003 2 × 2 = 0 + 0.001 327 481 619 900 006 4;
  • 37) 0.001 327 481 619 900 006 4 × 2 = 0 + 0.002 654 963 239 800 012 8;
  • 38) 0.002 654 963 239 800 012 8 × 2 = 0 + 0.005 309 926 479 600 025 6;
  • 39) 0.005 309 926 479 600 025 6 × 2 = 0 + 0.010 619 852 959 200 051 2;
  • 40) 0.010 619 852 959 200 051 2 × 2 = 0 + 0.021 239 705 918 400 102 4;
  • 41) 0.021 239 705 918 400 102 4 × 2 = 0 + 0.042 479 411 836 800 204 8;
  • 42) 0.042 479 411 836 800 204 8 × 2 = 0 + 0.084 958 823 673 600 409 6;
  • 43) 0.084 958 823 673 600 409 6 × 2 = 0 + 0.169 917 647 347 200 819 2;
  • 44) 0.169 917 647 347 200 819 2 × 2 = 0 + 0.339 835 294 694 401 638 4;
  • 45) 0.339 835 294 694 401 638 4 × 2 = 0 + 0.679 670 589 388 803 276 8;
  • 46) 0.679 670 589 388 803 276 8 × 2 = 1 + 0.359 341 178 777 606 553 6;
  • 47) 0.359 341 178 777 606 553 6 × 2 = 0 + 0.718 682 357 555 213 107 2;
  • 48) 0.718 682 357 555 213 107 2 × 2 = 1 + 0.437 364 715 110 426 214 4;
  • 49) 0.437 364 715 110 426 214 4 × 2 = 0 + 0.874 729 430 220 852 428 8;
  • 50) 0.874 729 430 220 852 428 8 × 2 = 1 + 0.749 458 860 441 704 857 6;
  • 51) 0.749 458 860 441 704 857 6 × 2 = 1 + 0.498 917 720 883 409 715 2;
  • 52) 0.498 917 720 883 409 715 2 × 2 = 0 + 0.997 835 441 766 819 430 4;
  • 53) 0.997 835 441 766 819 430 4 × 2 = 1 + 0.995 670 883 533 638 860 8;
  • 54) 0.995 670 883 533 638 860 8 × 2 = 1 + 0.991 341 767 067 277 721 6;
  • 55) 0.991 341 767 067 277 721 6 × 2 = 1 + 0.982 683 534 134 555 443 2;
  • 56) 0.982 683 534 134 555 443 2 × 2 = 1 + 0.965 367 068 269 110 886 4;
  • 57) 0.965 367 068 269 110 886 4 × 2 = 1 + 0.930 734 136 538 221 772 8;
  • 58) 0.930 734 136 538 221 772 8 × 2 = 1 + 0.861 468 273 076 443 545 6;
  • 59) 0.861 468 273 076 443 545 6 × 2 = 1 + 0.722 936 546 152 887 091 2;
  • 60) 0.722 936 546 152 887 091 2 × 2 = 1 + 0.445 873 092 305 774 182 4;
  • 61) 0.445 873 092 305 774 182 4 × 2 = 0 + 0.891 746 184 611 548 364 8;
  • 62) 0.891 746 184 611 548 364 8 × 2 = 1 + 0.783 492 369 223 096 729 6;
  • 63) 0.783 492 369 223 096 729 6 × 2 = 1 + 0.566 984 738 446 193 459 2;
  • 64) 0.566 984 738 446 193 459 2 × 2 = 1 + 0.133 969 476 892 386 918 4;
  • 65) 0.133 969 476 892 386 918 4 × 2 = 0 + 0.267 938 953 784 773 836 8;
  • 66) 0.267 938 953 784 773 836 8 × 2 = 0 + 0.535 877 907 569 547 673 6;
  • 67) 0.535 877 907 569 547 673 6 × 2 = 1 + 0.071 755 815 139 095 347 2;
  • 68) 0.071 755 815 139 095 347 2 × 2 = 0 + 0.143 511 630 278 190 694 4;
  • 69) 0.143 511 630 278 190 694 4 × 2 = 0 + 0.287 023 260 556 381 388 8;

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:


0.000 000 000 000 019 317 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0110 1111 1111 0111 0010 0(2)


5. Positive number before normalization:

0.000 000 000 000 019 317 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0110 1111 1111 0111 0010 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 019 317 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0110 1111 1111 0111 0010 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 0110 1111 1111 0111 0010 0(2) × 20 =

1.0101 1011 1111 1101 1100 100(2) × 2-46


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): -46

Mantissa (not normalized):
1.0101 1011 1111 1101 1100 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-46 + 2(8-1) - 1 =

(-46 + 127)(10) =

81(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;
  • 81 ÷ 2 = 40 + 1;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 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) =

81(10) =

0101 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. 010 1101 1111 1110 1110 0100 =

010 1101 1111 1110 1110 0100


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) =
0101 0001

Mantissa (23 bits) =
010 1101 1111 1110 1110 0100


The base ten decimal number 0.000 000 000 000 019 317 4 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0101 0001 - 010 1101 1111 1110 1110 0100

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