3.821 146 326 489 802 928 74 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.821 146 326 489 802 928 74(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
3.821 146 326 489 802 928 74(10) to 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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.821 146 326 489 802 928 74.

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) 0.821 146 326 489 802 928 74 × 2 = 1 + 0.642 292 652 979 605 857 48;
  • 2) 0.642 292 652 979 605 857 48 × 2 = 1 + 0.284 585 305 959 211 714 96;
  • 3) 0.284 585 305 959 211 714 96 × 2 = 0 + 0.569 170 611 918 423 429 92;
  • 4) 0.569 170 611 918 423 429 92 × 2 = 1 + 0.138 341 223 836 846 859 84;
  • 5) 0.138 341 223 836 846 859 84 × 2 = 0 + 0.276 682 447 673 693 719 68;
  • 6) 0.276 682 447 673 693 719 68 × 2 = 0 + 0.553 364 895 347 387 439 36;
  • 7) 0.553 364 895 347 387 439 36 × 2 = 1 + 0.106 729 790 694 774 878 72;
  • 8) 0.106 729 790 694 774 878 72 × 2 = 0 + 0.213 459 581 389 549 757 44;
  • 9) 0.213 459 581 389 549 757 44 × 2 = 0 + 0.426 919 162 779 099 514 88;
  • 10) 0.426 919 162 779 099 514 88 × 2 = 0 + 0.853 838 325 558 199 029 76;
  • 11) 0.853 838 325 558 199 029 76 × 2 = 1 + 0.707 676 651 116 398 059 52;
  • 12) 0.707 676 651 116 398 059 52 × 2 = 1 + 0.415 353 302 232 796 119 04;
  • 13) 0.415 353 302 232 796 119 04 × 2 = 0 + 0.830 706 604 465 592 238 08;
  • 14) 0.830 706 604 465 592 238 08 × 2 = 1 + 0.661 413 208 931 184 476 16;
  • 15) 0.661 413 208 931 184 476 16 × 2 = 1 + 0.322 826 417 862 368 952 32;
  • 16) 0.322 826 417 862 368 952 32 × 2 = 0 + 0.645 652 835 724 737 904 64;
  • 17) 0.645 652 835 724 737 904 64 × 2 = 1 + 0.291 305 671 449 475 809 28;
  • 18) 0.291 305 671 449 475 809 28 × 2 = 0 + 0.582 611 342 898 951 618 56;
  • 19) 0.582 611 342 898 951 618 56 × 2 = 1 + 0.165 222 685 797 903 237 12;
  • 20) 0.165 222 685 797 903 237 12 × 2 = 0 + 0.330 445 371 595 806 474 24;
  • 21) 0.330 445 371 595 806 474 24 × 2 = 0 + 0.660 890 743 191 612 948 48;
  • 22) 0.660 890 743 191 612 948 48 × 2 = 1 + 0.321 781 486 383 225 896 96;
  • 23) 0.321 781 486 383 225 896 96 × 2 = 0 + 0.643 562 972 766 451 793 92;
  • 24) 0.643 562 972 766 451 793 92 × 2 = 1 + 0.287 125 945 532 903 587 84;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


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:


0.821 146 326 489 802 928 74(10) =

0.1101 0010 0011 0110 1010 0101(2)

5. Positive number before normalization:

3.821 146 326 489 802 928 74(10) =

11.1101 0010 0011 0110 1010 0101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the left, so that only one non zero digit remains to the left of it:


3.821 146 326 489 802 928 74(10) =

11.1101 0010 0011 0110 1010 0101(2) =

11.1101 0010 0011 0110 1010 0101(2) × 20 =

1.1110 1001 0001 1011 0101 0010 1(2) × 21


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): 1

Mantissa (not normalized):
1.1110 1001 0001 1011 0101 0010 1


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

1 + 2(8-1) - 1 =

(1 + 127)(10) =

128(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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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) =

128(10) =

1000 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, 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. 111 0100 1000 1101 1010 1001 01 =

111 0100 1000 1101 1010 1001


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) =
1000 0000

Mantissa (23 bits) =
111 0100 1000 1101 1010 1001


Decimal number 3.821 146 326 489 802 928 74 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1000 0000 - 111 0100 1000 1101 1010 1001

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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