0.292 360 388 906 672 596 931 457 519 409 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.292 360 388 906 672 596 931 457 519 409(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
0.292 360 388 906 672 596 931 457 519 409(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: 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: 0.292 360 388 906 672 596 931 457 519 409.

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) 0.292 360 388 906 672 596 931 457 519 409 × 2 = 0 + 0.584 720 777 813 345 193 862 915 038 818;
  • 2) 0.584 720 777 813 345 193 862 915 038 818 × 2 = 1 + 0.169 441 555 626 690 387 725 830 077 636;
  • 3) 0.169 441 555 626 690 387 725 830 077 636 × 2 = 0 + 0.338 883 111 253 380 775 451 660 155 272;
  • 4) 0.338 883 111 253 380 775 451 660 155 272 × 2 = 0 + 0.677 766 222 506 761 550 903 320 310 544;
  • 5) 0.677 766 222 506 761 550 903 320 310 544 × 2 = 1 + 0.355 532 445 013 523 101 806 640 621 088;
  • 6) 0.355 532 445 013 523 101 806 640 621 088 × 2 = 0 + 0.711 064 890 027 046 203 613 281 242 176;
  • 7) 0.711 064 890 027 046 203 613 281 242 176 × 2 = 1 + 0.422 129 780 054 092 407 226 562 484 352;
  • 8) 0.422 129 780 054 092 407 226 562 484 352 × 2 = 0 + 0.844 259 560 108 184 814 453 124 968 704;
  • 9) 0.844 259 560 108 184 814 453 124 968 704 × 2 = 1 + 0.688 519 120 216 369 628 906 249 937 408;
  • 10) 0.688 519 120 216 369 628 906 249 937 408 × 2 = 1 + 0.377 038 240 432 739 257 812 499 874 816;
  • 11) 0.377 038 240 432 739 257 812 499 874 816 × 2 = 0 + 0.754 076 480 865 478 515 624 999 749 632;
  • 12) 0.754 076 480 865 478 515 624 999 749 632 × 2 = 1 + 0.508 152 961 730 957 031 249 999 499 264;
  • 13) 0.508 152 961 730 957 031 249 999 499 264 × 2 = 1 + 0.016 305 923 461 914 062 499 998 998 528;
  • 14) 0.016 305 923 461 914 062 499 998 998 528 × 2 = 0 + 0.032 611 846 923 828 124 999 997 997 056;
  • 15) 0.032 611 846 923 828 124 999 997 997 056 × 2 = 0 + 0.065 223 693 847 656 249 999 995 994 112;
  • 16) 0.065 223 693 847 656 249 999 995 994 112 × 2 = 0 + 0.130 447 387 695 312 499 999 991 988 224;
  • 17) 0.130 447 387 695 312 499 999 991 988 224 × 2 = 0 + 0.260 894 775 390 624 999 999 983 976 448;
  • 18) 0.260 894 775 390 624 999 999 983 976 448 × 2 = 0 + 0.521 789 550 781 249 999 999 967 952 896;
  • 19) 0.521 789 550 781 249 999 999 967 952 896 × 2 = 1 + 0.043 579 101 562 499 999 999 935 905 792;
  • 20) 0.043 579 101 562 499 999 999 935 905 792 × 2 = 0 + 0.087 158 203 124 999 999 999 871 811 584;
  • 21) 0.087 158 203 124 999 999 999 871 811 584 × 2 = 0 + 0.174 316 406 249 999 999 999 743 623 168;
  • 22) 0.174 316 406 249 999 999 999 743 623 168 × 2 = 0 + 0.348 632 812 499 999 999 999 487 246 336;
  • 23) 0.348 632 812 499 999 999 999 487 246 336 × 2 = 0 + 0.697 265 624 999 999 999 998 974 492 672;
  • 24) 0.697 265 624 999 999 999 998 974 492 672 × 2 = 1 + 0.394 531 249 999 999 999 997 948 985 344;
  • 25) 0.394 531 249 999 999 999 997 948 985 344 × 2 = 0 + 0.789 062 499 999 999 999 995 897 970 688;

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:


0.292 360 388 906 672 596 931 457 519 409(10) =

0.0100 1010 1101 1000 0010 0001 0(2)

5. Positive number before normalization:

0.292 360 388 906 672 596 931 457 519 409(10) =

0.0100 1010 1101 1000 0010 0001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.292 360 388 906 672 596 931 457 519 409(10) =

0.0100 1010 1101 1000 0010 0001 0(2) =

0.0100 1010 1101 1000 0010 0001 0(2) × 20 =

1.0010 1011 0110 0000 1000 010(2) × 2-2


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): -2

Mantissa (not normalized):
1.0010 1011 0110 0000 1000 010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-2 + 127)(10) =

125(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;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 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) =

125(10) =

0111 1101(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. 001 0101 1011 0000 0100 0010 =

001 0101 1011 0000 0100 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 1101

Mantissa (23 bits) =
001 0101 1011 0000 0100 0010


Decimal number 0.292 360 388 906 672 596 931 457 519 409 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1101 - 001 0101 1011 0000 0100 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