32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 1.234 567 890 112 342 342 342 323 532 4 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 1.234 567 890 112 342 342 342 323 532 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: 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;
  • 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.


1(10) =


1(2)


3. Convert to binary (base 2) the fractional part: 0.234 567 890 112 342 342 342 323 532 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) 0.234 567 890 112 342 342 342 323 532 4 × 2 = 0 + 0.469 135 780 224 684 684 684 647 064 8;
  • 2) 0.469 135 780 224 684 684 684 647 064 8 × 2 = 0 + 0.938 271 560 449 369 369 369 294 129 6;
  • 3) 0.938 271 560 449 369 369 369 294 129 6 × 2 = 1 + 0.876 543 120 898 738 738 738 588 259 2;
  • 4) 0.876 543 120 898 738 738 738 588 259 2 × 2 = 1 + 0.753 086 241 797 477 477 477 176 518 4;
  • 5) 0.753 086 241 797 477 477 477 176 518 4 × 2 = 1 + 0.506 172 483 594 954 954 954 353 036 8;
  • 6) 0.506 172 483 594 954 954 954 353 036 8 × 2 = 1 + 0.012 344 967 189 909 909 908 706 073 6;
  • 7) 0.012 344 967 189 909 909 908 706 073 6 × 2 = 0 + 0.024 689 934 379 819 819 817 412 147 2;
  • 8) 0.024 689 934 379 819 819 817 412 147 2 × 2 = 0 + 0.049 379 868 759 639 639 634 824 294 4;
  • 9) 0.049 379 868 759 639 639 634 824 294 4 × 2 = 0 + 0.098 759 737 519 279 279 269 648 588 8;
  • 10) 0.098 759 737 519 279 279 269 648 588 8 × 2 = 0 + 0.197 519 475 038 558 558 539 297 177 6;
  • 11) 0.197 519 475 038 558 558 539 297 177 6 × 2 = 0 + 0.395 038 950 077 117 117 078 594 355 2;
  • 12) 0.395 038 950 077 117 117 078 594 355 2 × 2 = 0 + 0.790 077 900 154 234 234 157 188 710 4;
  • 13) 0.790 077 900 154 234 234 157 188 710 4 × 2 = 1 + 0.580 155 800 308 468 468 314 377 420 8;
  • 14) 0.580 155 800 308 468 468 314 377 420 8 × 2 = 1 + 0.160 311 600 616 936 936 628 754 841 6;
  • 15) 0.160 311 600 616 936 936 628 754 841 6 × 2 = 0 + 0.320 623 201 233 873 873 257 509 683 2;
  • 16) 0.320 623 201 233 873 873 257 509 683 2 × 2 = 0 + 0.641 246 402 467 747 746 515 019 366 4;
  • 17) 0.641 246 402 467 747 746 515 019 366 4 × 2 = 1 + 0.282 492 804 935 495 493 030 038 732 8;
  • 18) 0.282 492 804 935 495 493 030 038 732 8 × 2 = 0 + 0.564 985 609 870 990 986 060 077 465 6;
  • 19) 0.564 985 609 870 990 986 060 077 465 6 × 2 = 1 + 0.129 971 219 741 981 972 120 154 931 2;
  • 20) 0.129 971 219 741 981 972 120 154 931 2 × 2 = 0 + 0.259 942 439 483 963 944 240 309 862 4;
  • 21) 0.259 942 439 483 963 944 240 309 862 4 × 2 = 0 + 0.519 884 878 967 927 888 480 619 724 8;
  • 22) 0.519 884 878 967 927 888 480 619 724 8 × 2 = 1 + 0.039 769 757 935 855 776 961 239 449 6;
  • 23) 0.039 769 757 935 855 776 961 239 449 6 × 2 = 0 + 0.079 539 515 871 711 553 922 478 899 2;
  • 24) 0.079 539 515 871 711 553 922 478 899 2 × 2 = 0 + 0.159 079 031 743 423 107 844 957 798 4;

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:


0.234 567 890 112 342 342 342 323 532 4(10) =


0.0011 1100 0000 1100 1010 0100(2)


5. Positive number before normalization:

1.234 567 890 112 342 342 342 323 532 4(10) =


1.0011 1100 0000 1100 1010 0100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.234 567 890 112 342 342 342 323 532 4(10) =


1.0011 1100 0000 1100 1010 0100(2) =


1.0011 1100 0000 1100 1010 0100(2) × 20


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): 0


Mantissa (not normalized):
1.0011 1100 0000 1100 1010 0100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


0 + 2(8-1) - 1 =


(0 + 127)(10) =


127(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;
  • 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;

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) =


127(10) =


0111 1111(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. 001 1110 0000 0110 0101 0010 0 =


001 1110 0000 0110 0101 0010


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 1111


Mantissa (23 bits) =
001 1110 0000 0110 0101 0010


The base ten decimal number 1.234 567 890 112 342 342 342 323 532 4 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 1111 - 001 1110 0000 0110 0101 0010

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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