0.002 702 702 702 702 702 702 702 698 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.002 702 702 702 702 702 702 702 698 8(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
0.002 702 702 702 702 702 702 702 698 8(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: 0.002 702 702 702 702 702 702 702 698 8.

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) 0.002 702 702 702 702 702 702 702 698 8 × 2 = 0 + 0.005 405 405 405 405 405 405 405 397 6;
  • 2) 0.005 405 405 405 405 405 405 405 397 6 × 2 = 0 + 0.010 810 810 810 810 810 810 810 795 2;
  • 3) 0.010 810 810 810 810 810 810 810 795 2 × 2 = 0 + 0.021 621 621 621 621 621 621 621 590 4;
  • 4) 0.021 621 621 621 621 621 621 621 590 4 × 2 = 0 + 0.043 243 243 243 243 243 243 243 180 8;
  • 5) 0.043 243 243 243 243 243 243 243 180 8 × 2 = 0 + 0.086 486 486 486 486 486 486 486 361 6;
  • 6) 0.086 486 486 486 486 486 486 486 361 6 × 2 = 0 + 0.172 972 972 972 972 972 972 972 723 2;
  • 7) 0.172 972 972 972 972 972 972 972 723 2 × 2 = 0 + 0.345 945 945 945 945 945 945 945 446 4;
  • 8) 0.345 945 945 945 945 945 945 945 446 4 × 2 = 0 + 0.691 891 891 891 891 891 891 890 892 8;
  • 9) 0.691 891 891 891 891 891 891 890 892 8 × 2 = 1 + 0.383 783 783 783 783 783 783 781 785 6;
  • 10) 0.383 783 783 783 783 783 783 781 785 6 × 2 = 0 + 0.767 567 567 567 567 567 567 563 571 2;
  • 11) 0.767 567 567 567 567 567 567 563 571 2 × 2 = 1 + 0.535 135 135 135 135 135 135 127 142 4;
  • 12) 0.535 135 135 135 135 135 135 127 142 4 × 2 = 1 + 0.070 270 270 270 270 270 270 254 284 8;
  • 13) 0.070 270 270 270 270 270 270 254 284 8 × 2 = 0 + 0.140 540 540 540 540 540 540 508 569 6;
  • 14) 0.140 540 540 540 540 540 540 508 569 6 × 2 = 0 + 0.281 081 081 081 081 081 081 017 139 2;
  • 15) 0.281 081 081 081 081 081 081 017 139 2 × 2 = 0 + 0.562 162 162 162 162 162 162 034 278 4;
  • 16) 0.562 162 162 162 162 162 162 034 278 4 × 2 = 1 + 0.124 324 324 324 324 324 324 068 556 8;
  • 17) 0.124 324 324 324 324 324 324 068 556 8 × 2 = 0 + 0.248 648 648 648 648 648 648 137 113 6;
  • 18) 0.248 648 648 648 648 648 648 137 113 6 × 2 = 0 + 0.497 297 297 297 297 297 296 274 227 2;
  • 19) 0.497 297 297 297 297 297 296 274 227 2 × 2 = 0 + 0.994 594 594 594 594 594 592 548 454 4;
  • 20) 0.994 594 594 594 594 594 592 548 454 4 × 2 = 1 + 0.989 189 189 189 189 189 185 096 908 8;
  • 21) 0.989 189 189 189 189 189 185 096 908 8 × 2 = 1 + 0.978 378 378 378 378 378 370 193 817 6;
  • 22) 0.978 378 378 378 378 378 370 193 817 6 × 2 = 1 + 0.956 756 756 756 756 756 740 387 635 2;
  • 23) 0.956 756 756 756 756 756 740 387 635 2 × 2 = 1 + 0.913 513 513 513 513 513 480 775 270 4;
  • 24) 0.913 513 513 513 513 513 480 775 270 4 × 2 = 1 + 0.827 027 027 027 027 026 961 550 540 8;
  • 25) 0.827 027 027 027 027 026 961 550 540 8 × 2 = 1 + 0.654 054 054 054 054 053 923 101 081 6;
  • 26) 0.654 054 054 054 054 053 923 101 081 6 × 2 = 1 + 0.308 108 108 108 108 107 846 202 163 2;
  • 27) 0.308 108 108 108 108 107 846 202 163 2 × 2 = 0 + 0.616 216 216 216 216 215 692 404 326 4;
  • 28) 0.616 216 216 216 216 215 692 404 326 4 × 2 = 1 + 0.232 432 432 432 432 431 384 808 652 8;
  • 29) 0.232 432 432 432 432 431 384 808 652 8 × 2 = 0 + 0.464 864 864 864 864 862 769 617 305 6;
  • 30) 0.464 864 864 864 864 862 769 617 305 6 × 2 = 0 + 0.929 729 729 729 729 725 539 234 611 2;
  • 31) 0.929 729 729 729 729 725 539 234 611 2 × 2 = 1 + 0.859 459 459 459 459 451 078 469 222 4;
  • 32) 0.859 459 459 459 459 451 078 469 222 4 × 2 = 1 + 0.718 918 918 918 918 902 156 938 444 8;

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:


0.002 702 702 702 702 702 702 702 698 8(10) =

0.0000 0000 1011 0001 0001 1111 1101 0011(2)

5. Positive number before normalization:

0.002 702 702 702 702 702 702 702 698 8(10) =

0.0000 0000 1011 0001 0001 1111 1101 0011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the right, so that only one non zero digit remains to the left of it:


0.002 702 702 702 702 702 702 702 698 8(10) =

0.0000 0000 1011 0001 0001 1111 1101 0011(2) =

0.0000 0000 1011 0001 0001 1111 1101 0011(2) × 20 =

1.0110 0010 0011 1111 1010 011(2) × 2-9


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): -9

Mantissa (not normalized):
1.0110 0010 0011 1111 1010 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-9 + 2(8-1) - 1 =

(-9 + 127)(10) =

118(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;
  • 118 ÷ 2 = 59 + 0;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 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) =

118(10) =

0111 0110(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. 011 0001 0001 1111 1101 0011 =

011 0001 0001 1111 1101 0011


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 0110

Mantissa (23 bits) =
011 0001 0001 1111 1101 0011


Decimal number 0.002 702 702 702 702 702 702 702 698 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 0110 - 011 0001 0001 1111 1101 0011

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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