0.002 702 702 702 702 702 702 702 702 702 612 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.002 702 702 702 702 702 702 702 702 702 612 6(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
0.002 702 702 702 702 702 702 702 702 702 612 6(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: 0.002 702 702 702 702 702 702 702 702 702 612 6.

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) 0.002 702 702 702 702 702 702 702 702 702 612 6 × 2 = 0 + 0.005 405 405 405 405 405 405 405 405 405 225 2;
  • 2) 0.005 405 405 405 405 405 405 405 405 405 225 2 × 2 = 0 + 0.010 810 810 810 810 810 810 810 810 810 450 4;
  • 3) 0.010 810 810 810 810 810 810 810 810 810 450 4 × 2 = 0 + 0.021 621 621 621 621 621 621 621 621 620 900 8;
  • 4) 0.021 621 621 621 621 621 621 621 621 620 900 8 × 2 = 0 + 0.043 243 243 243 243 243 243 243 243 241 801 6;
  • 5) 0.043 243 243 243 243 243 243 243 243 241 801 6 × 2 = 0 + 0.086 486 486 486 486 486 486 486 486 483 603 2;
  • 6) 0.086 486 486 486 486 486 486 486 486 483 603 2 × 2 = 0 + 0.172 972 972 972 972 972 972 972 972 967 206 4;
  • 7) 0.172 972 972 972 972 972 972 972 972 967 206 4 × 2 = 0 + 0.345 945 945 945 945 945 945 945 945 934 412 8;
  • 8) 0.345 945 945 945 945 945 945 945 945 934 412 8 × 2 = 0 + 0.691 891 891 891 891 891 891 891 891 868 825 6;
  • 9) 0.691 891 891 891 891 891 891 891 891 868 825 6 × 2 = 1 + 0.383 783 783 783 783 783 783 783 783 737 651 2;
  • 10) 0.383 783 783 783 783 783 783 783 783 737 651 2 × 2 = 0 + 0.767 567 567 567 567 567 567 567 567 475 302 4;
  • 11) 0.767 567 567 567 567 567 567 567 567 475 302 4 × 2 = 1 + 0.535 135 135 135 135 135 135 135 134 950 604 8;
  • 12) 0.535 135 135 135 135 135 135 135 134 950 604 8 × 2 = 1 + 0.070 270 270 270 270 270 270 270 269 901 209 6;
  • 13) 0.070 270 270 270 270 270 270 270 269 901 209 6 × 2 = 0 + 0.140 540 540 540 540 540 540 540 539 802 419 2;
  • 14) 0.140 540 540 540 540 540 540 540 539 802 419 2 × 2 = 0 + 0.281 081 081 081 081 081 081 081 079 604 838 4;
  • 15) 0.281 081 081 081 081 081 081 081 079 604 838 4 × 2 = 0 + 0.562 162 162 162 162 162 162 162 159 209 676 8;
  • 16) 0.562 162 162 162 162 162 162 162 159 209 676 8 × 2 = 1 + 0.124 324 324 324 324 324 324 324 318 419 353 6;
  • 17) 0.124 324 324 324 324 324 324 324 318 419 353 6 × 2 = 0 + 0.248 648 648 648 648 648 648 648 636 838 707 2;
  • 18) 0.248 648 648 648 648 648 648 648 636 838 707 2 × 2 = 0 + 0.497 297 297 297 297 297 297 297 273 677 414 4;
  • 19) 0.497 297 297 297 297 297 297 297 273 677 414 4 × 2 = 0 + 0.994 594 594 594 594 594 594 594 547 354 828 8;
  • 20) 0.994 594 594 594 594 594 594 594 547 354 828 8 × 2 = 1 + 0.989 189 189 189 189 189 189 189 094 709 657 6;
  • 21) 0.989 189 189 189 189 189 189 189 094 709 657 6 × 2 = 1 + 0.978 378 378 378 378 378 378 378 189 419 315 2;
  • 22) 0.978 378 378 378 378 378 378 378 189 419 315 2 × 2 = 1 + 0.956 756 756 756 756 756 756 756 378 838 630 4;
  • 23) 0.956 756 756 756 756 756 756 756 378 838 630 4 × 2 = 1 + 0.913 513 513 513 513 513 513 512 757 677 260 8;
  • 24) 0.913 513 513 513 513 513 513 512 757 677 260 8 × 2 = 1 + 0.827 027 027 027 027 027 027 025 515 354 521 6;
  • 25) 0.827 027 027 027 027 027 027 025 515 354 521 6 × 2 = 1 + 0.654 054 054 054 054 054 054 051 030 709 043 2;
  • 26) 0.654 054 054 054 054 054 054 051 030 709 043 2 × 2 = 1 + 0.308 108 108 108 108 108 108 102 061 418 086 4;
  • 27) 0.308 108 108 108 108 108 108 102 061 418 086 4 × 2 = 0 + 0.616 216 216 216 216 216 216 204 122 836 172 8;
  • 28) 0.616 216 216 216 216 216 216 204 122 836 172 8 × 2 = 1 + 0.232 432 432 432 432 432 432 408 245 672 345 6;
  • 29) 0.232 432 432 432 432 432 432 408 245 672 345 6 × 2 = 0 + 0.464 864 864 864 864 864 864 816 491 344 691 2;
  • 30) 0.464 864 864 864 864 864 864 816 491 344 691 2 × 2 = 0 + 0.929 729 729 729 729 729 729 632 982 689 382 4;
  • 31) 0.929 729 729 729 729 729 729 632 982 689 382 4 × 2 = 1 + 0.859 459 459 459 459 459 459 265 965 378 764 8;
  • 32) 0.859 459 459 459 459 459 459 265 965 378 764 8 × 2 = 1 + 0.718 918 918 918 918 918 918 531 930 757 529 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:


0.002 702 702 702 702 702 702 702 702 702 612 6(10) =

0.0000 0000 1011 0001 0001 1111 1101 0011(2)

5. Positive number before normalization:

0.002 702 702 702 702 702 702 702 702 702 612 6(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:


0.002 702 702 702 702 702 702 702 702 702 612 6(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 0.002 702 702 702 702 702 702 702 702 702 612 6 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