354 459 182.742 069 228 471 053 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 354 459 182.742 069 228 471 053 1(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
354 459 182.742 069 228 471 053 1(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: 354 459 182.
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;
  • 354 459 182 ÷ 2 = 177 229 591 + 0;
  • 177 229 591 ÷ 2 = 88 614 795 + 1;
  • 88 614 795 ÷ 2 = 44 307 397 + 1;
  • 44 307 397 ÷ 2 = 22 153 698 + 1;
  • 22 153 698 ÷ 2 = 11 076 849 + 0;
  • 11 076 849 ÷ 2 = 5 538 424 + 1;
  • 5 538 424 ÷ 2 = 2 769 212 + 0;
  • 2 769 212 ÷ 2 = 1 384 606 + 0;
  • 1 384 606 ÷ 2 = 692 303 + 0;
  • 692 303 ÷ 2 = 346 151 + 1;
  • 346 151 ÷ 2 = 173 075 + 1;
  • 173 075 ÷ 2 = 86 537 + 1;
  • 86 537 ÷ 2 = 43 268 + 1;
  • 43 268 ÷ 2 = 21 634 + 0;
  • 21 634 ÷ 2 = 10 817 + 0;
  • 10 817 ÷ 2 = 5 408 + 1;
  • 5 408 ÷ 2 = 2 704 + 0;
  • 2 704 ÷ 2 = 1 352 + 0;
  • 1 352 ÷ 2 = 676 + 0;
  • 676 ÷ 2 = 338 + 0;
  • 338 ÷ 2 = 169 + 0;
  • 169 ÷ 2 = 84 + 1;
  • 84 ÷ 2 = 42 + 0;
  • 42 ÷ 2 = 21 + 0;
  • 21 ÷ 2 = 10 + 1;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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.

354 459 182(10) =

1 0101 0010 0000 1001 1110 0010 1110(2)


3. Convert to binary (base 2) the fractional part: 0.742 069 228 471 053 1.

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) 0.742 069 228 471 053 1 × 2 = 1 + 0.484 138 456 942 106 2;
  • 2) 0.484 138 456 942 106 2 × 2 = 0 + 0.968 276 913 884 212 4;
  • 3) 0.968 276 913 884 212 4 × 2 = 1 + 0.936 553 827 768 424 8;
  • 4) 0.936 553 827 768 424 8 × 2 = 1 + 0.873 107 655 536 849 6;
  • 5) 0.873 107 655 536 849 6 × 2 = 1 + 0.746 215 311 073 699 2;
  • 6) 0.746 215 311 073 699 2 × 2 = 1 + 0.492 430 622 147 398 4;
  • 7) 0.492 430 622 147 398 4 × 2 = 0 + 0.984 861 244 294 796 8;
  • 8) 0.984 861 244 294 796 8 × 2 = 1 + 0.969 722 488 589 593 6;
  • 9) 0.969 722 488 589 593 6 × 2 = 1 + 0.939 444 977 179 187 2;
  • 10) 0.939 444 977 179 187 2 × 2 = 1 + 0.878 889 954 358 374 4;
  • 11) 0.878 889 954 358 374 4 × 2 = 1 + 0.757 779 908 716 748 8;
  • 12) 0.757 779 908 716 748 8 × 2 = 1 + 0.515 559 817 433 497 6;
  • 13) 0.515 559 817 433 497 6 × 2 = 1 + 0.031 119 634 866 995 2;
  • 14) 0.031 119 634 866 995 2 × 2 = 0 + 0.062 239 269 733 990 4;
  • 15) 0.062 239 269 733 990 4 × 2 = 0 + 0.124 478 539 467 980 8;
  • 16) 0.124 478 539 467 980 8 × 2 = 0 + 0.248 957 078 935 961 6;
  • 17) 0.248 957 078 935 961 6 × 2 = 0 + 0.497 914 157 871 923 2;
  • 18) 0.497 914 157 871 923 2 × 2 = 0 + 0.995 828 315 743 846 4;
  • 19) 0.995 828 315 743 846 4 × 2 = 1 + 0.991 656 631 487 692 8;
  • 20) 0.991 656 631 487 692 8 × 2 = 1 + 0.983 313 262 975 385 6;
  • 21) 0.983 313 262 975 385 6 × 2 = 1 + 0.966 626 525 950 771 2;
  • 22) 0.966 626 525 950 771 2 × 2 = 1 + 0.933 253 051 901 542 4;
  • 23) 0.933 253 051 901 542 4 × 2 = 1 + 0.866 506 103 803 084 8;
  • 24) 0.866 506 103 803 084 8 × 2 = 1 + 0.733 012 207 606 169 6;

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:


0.742 069 228 471 053 1(10) =

0.1011 1101 1111 1000 0011 1111(2)

5. Positive number before normalization:

354 459 182.742 069 228 471 053 1(10) =

1 0101 0010 0000 1001 1110 0010 1110.1011 1101 1111 1000 0011 1111(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the left, so that only one non zero digit remains to the left of it:


354 459 182.742 069 228 471 053 1(10) =

1 0101 0010 0000 1001 1110 0010 1110.1011 1101 1111 1000 0011 1111(2) =

1 0101 0010 0000 1001 1110 0010 1110.1011 1101 1111 1000 0011 1111(2) × 20 =

1.0101 0010 0000 1001 1110 0010 1110 1011 1101 1111 1000 0011 1111(2) × 228


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): 28

Mantissa (not normalized):
1.0101 0010 0000 1001 1110 0010 1110 1011 1101 1111 1000 0011 1111


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

28 + 2(8-1) - 1 =

(28 + 127)(10) =

155(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;
  • 155 ÷ 2 = 77 + 1;
  • 77 ÷ 2 = 38 + 1;
  • 38 ÷ 2 = 19 + 0;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 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) =

155(10) =

1001 1011(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. 010 1001 0000 0100 1111 0001 0 1110 1011 1101 1111 1000 0011 1111 =

010 1001 0000 0100 1111 0001


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) =
1001 1011

Mantissa (23 bits) =
010 1001 0000 0100 1111 0001


Decimal number 354 459 182.742 069 228 471 053 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 1001 1011 - 010 1001 0000 0100 1111 0001

» Convert decimal 354 459 182.742 069 228 471 056 9 to 32 bit single precision IEEE 754 binary floating point representation standard

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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