32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 111 641 111 682 112 114 352 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 111 641 111 682 112 114 352(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;
  • 111 641 111 682 112 114 352 ÷ 2 = 55 820 555 841 056 057 176 + 0;
  • 55 820 555 841 056 057 176 ÷ 2 = 27 910 277 920 528 028 588 + 0;
  • 27 910 277 920 528 028 588 ÷ 2 = 13 955 138 960 264 014 294 + 0;
  • 13 955 138 960 264 014 294 ÷ 2 = 6 977 569 480 132 007 147 + 0;
  • 6 977 569 480 132 007 147 ÷ 2 = 3 488 784 740 066 003 573 + 1;
  • 3 488 784 740 066 003 573 ÷ 2 = 1 744 392 370 033 001 786 + 1;
  • 1 744 392 370 033 001 786 ÷ 2 = 872 196 185 016 500 893 + 0;
  • 872 196 185 016 500 893 ÷ 2 = 436 098 092 508 250 446 + 1;
  • 436 098 092 508 250 446 ÷ 2 = 218 049 046 254 125 223 + 0;
  • 218 049 046 254 125 223 ÷ 2 = 109 024 523 127 062 611 + 1;
  • 109 024 523 127 062 611 ÷ 2 = 54 512 261 563 531 305 + 1;
  • 54 512 261 563 531 305 ÷ 2 = 27 256 130 781 765 652 + 1;
  • 27 256 130 781 765 652 ÷ 2 = 13 628 065 390 882 826 + 0;
  • 13 628 065 390 882 826 ÷ 2 = 6 814 032 695 441 413 + 0;
  • 6 814 032 695 441 413 ÷ 2 = 3 407 016 347 720 706 + 1;
  • 3 407 016 347 720 706 ÷ 2 = 1 703 508 173 860 353 + 0;
  • 1 703 508 173 860 353 ÷ 2 = 851 754 086 930 176 + 1;
  • 851 754 086 930 176 ÷ 2 = 425 877 043 465 088 + 0;
  • 425 877 043 465 088 ÷ 2 = 212 938 521 732 544 + 0;
  • 212 938 521 732 544 ÷ 2 = 106 469 260 866 272 + 0;
  • 106 469 260 866 272 ÷ 2 = 53 234 630 433 136 + 0;
  • 53 234 630 433 136 ÷ 2 = 26 617 315 216 568 + 0;
  • 26 617 315 216 568 ÷ 2 = 13 308 657 608 284 + 0;
  • 13 308 657 608 284 ÷ 2 = 6 654 328 804 142 + 0;
  • 6 654 328 804 142 ÷ 2 = 3 327 164 402 071 + 0;
  • 3 327 164 402 071 ÷ 2 = 1 663 582 201 035 + 1;
  • 1 663 582 201 035 ÷ 2 = 831 791 100 517 + 1;
  • 831 791 100 517 ÷ 2 = 415 895 550 258 + 1;
  • 415 895 550 258 ÷ 2 = 207 947 775 129 + 0;
  • 207 947 775 129 ÷ 2 = 103 973 887 564 + 1;
  • 103 973 887 564 ÷ 2 = 51 986 943 782 + 0;
  • 51 986 943 782 ÷ 2 = 25 993 471 891 + 0;
  • 25 993 471 891 ÷ 2 = 12 996 735 945 + 1;
  • 12 996 735 945 ÷ 2 = 6 498 367 972 + 1;
  • 6 498 367 972 ÷ 2 = 3 249 183 986 + 0;
  • 3 249 183 986 ÷ 2 = 1 624 591 993 + 0;
  • 1 624 591 993 ÷ 2 = 812 295 996 + 1;
  • 812 295 996 ÷ 2 = 406 147 998 + 0;
  • 406 147 998 ÷ 2 = 203 073 999 + 0;
  • 203 073 999 ÷ 2 = 101 536 999 + 1;
  • 101 536 999 ÷ 2 = 50 768 499 + 1;
  • 50 768 499 ÷ 2 = 25 384 249 + 1;
  • 25 384 249 ÷ 2 = 12 692 124 + 1;
  • 12 692 124 ÷ 2 = 6 346 062 + 0;
  • 6 346 062 ÷ 2 = 3 173 031 + 0;
  • 3 173 031 ÷ 2 = 1 586 515 + 1;
  • 1 586 515 ÷ 2 = 793 257 + 1;
  • 793 257 ÷ 2 = 396 628 + 1;
  • 396 628 ÷ 2 = 198 314 + 0;
  • 198 314 ÷ 2 = 99 157 + 0;
  • 99 157 ÷ 2 = 49 578 + 1;
  • 49 578 ÷ 2 = 24 789 + 0;
  • 24 789 ÷ 2 = 12 394 + 1;
  • 12 394 ÷ 2 = 6 197 + 0;
  • 6 197 ÷ 2 = 3 098 + 1;
  • 3 098 ÷ 2 = 1 549 + 0;
  • 1 549 ÷ 2 = 774 + 1;
  • 774 ÷ 2 = 387 + 0;
  • 387 ÷ 2 = 193 + 1;
  • 193 ÷ 2 = 96 + 1;
  • 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;

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

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


111 641 111 682 112 114 352(10) =


110 0000 1101 0101 0100 1110 0111 1001 0011 0010 1110 0000 0001 0100 1110 1011 0000(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 66 positions to the left, so that only one non zero digit remains to the left of it:


111 641 111 682 112 114 352(10) =


110 0000 1101 0101 0100 1110 0111 1001 0011 0010 1110 0000 0001 0100 1110 1011 0000(2) =


110 0000 1101 0101 0100 1110 0111 1001 0011 0010 1110 0000 0001 0100 1110 1011 0000(2) × 20 =


1.1000 0011 0101 0101 0011 1001 1110 0100 1100 1011 1000 0000 0101 0011 1010 1100 00(2) × 266


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): 66


Mantissa (not normalized):
1.1000 0011 0101 0101 0011 1001 1110 0100 1100 1011 1000 0000 0101 0011 1010 1100 00


5. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =


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


66 + 2(8-1) - 1 =


(66 + 127)(10) =


193(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;
  • 193 ÷ 2 = 96 + 1;
  • 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;

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) =


193(10) =


1100 0001(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. 100 0001 1010 1010 1001 1100 111 1001 0011 0010 1110 0000 0001 0100 1110 1011 0000 =


100 0001 1010 1010 1001 1100


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) =
1100 0001


Mantissa (23 bits) =
100 0001 1010 1010 1001 1100


The base ten decimal number 111 641 111 682 112 114 352 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 1100 0001 - 100 0001 1010 1010 1001 1100

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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