0.000 000 109 953 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 109 953(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
0.000 000 109 953(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: 0.000 000 109 953.

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) 0.000 000 109 953 × 2 = 0 + 0.000 000 219 906;
  • 2) 0.000 000 219 906 × 2 = 0 + 0.000 000 439 812;
  • 3) 0.000 000 439 812 × 2 = 0 + 0.000 000 879 624;
  • 4) 0.000 000 879 624 × 2 = 0 + 0.000 001 759 248;
  • 5) 0.000 001 759 248 × 2 = 0 + 0.000 003 518 496;
  • 6) 0.000 003 518 496 × 2 = 0 + 0.000 007 036 992;
  • 7) 0.000 007 036 992 × 2 = 0 + 0.000 014 073 984;
  • 8) 0.000 014 073 984 × 2 = 0 + 0.000 028 147 968;
  • 9) 0.000 028 147 968 × 2 = 0 + 0.000 056 295 936;
  • 10) 0.000 056 295 936 × 2 = 0 + 0.000 112 591 872;
  • 11) 0.000 112 591 872 × 2 = 0 + 0.000 225 183 744;
  • 12) 0.000 225 183 744 × 2 = 0 + 0.000 450 367 488;
  • 13) 0.000 450 367 488 × 2 = 0 + 0.000 900 734 976;
  • 14) 0.000 900 734 976 × 2 = 0 + 0.001 801 469 952;
  • 15) 0.001 801 469 952 × 2 = 0 + 0.003 602 939 904;
  • 16) 0.003 602 939 904 × 2 = 0 + 0.007 205 879 808;
  • 17) 0.007 205 879 808 × 2 = 0 + 0.014 411 759 616;
  • 18) 0.014 411 759 616 × 2 = 0 + 0.028 823 519 232;
  • 19) 0.028 823 519 232 × 2 = 0 + 0.057 647 038 464;
  • 20) 0.057 647 038 464 × 2 = 0 + 0.115 294 076 928;
  • 21) 0.115 294 076 928 × 2 = 0 + 0.230 588 153 856;
  • 22) 0.230 588 153 856 × 2 = 0 + 0.461 176 307 712;
  • 23) 0.461 176 307 712 × 2 = 0 + 0.922 352 615 424;
  • 24) 0.922 352 615 424 × 2 = 1 + 0.844 705 230 848;
  • 25) 0.844 705 230 848 × 2 = 1 + 0.689 410 461 696;
  • 26) 0.689 410 461 696 × 2 = 1 + 0.378 820 923 392;
  • 27) 0.378 820 923 392 × 2 = 0 + 0.757 641 846 784;
  • 28) 0.757 641 846 784 × 2 = 1 + 0.515 283 693 568;
  • 29) 0.515 283 693 568 × 2 = 1 + 0.030 567 387 136;
  • 30) 0.030 567 387 136 × 2 = 0 + 0.061 134 774 272;
  • 31) 0.061 134 774 272 × 2 = 0 + 0.122 269 548 544;
  • 32) 0.122 269 548 544 × 2 = 0 + 0.244 539 097 088;
  • 33) 0.244 539 097 088 × 2 = 0 + 0.489 078 194 176;
  • 34) 0.489 078 194 176 × 2 = 0 + 0.978 156 388 352;
  • 35) 0.978 156 388 352 × 2 = 1 + 0.956 312 776 704;
  • 36) 0.956 312 776 704 × 2 = 1 + 0.912 625 553 408;
  • 37) 0.912 625 553 408 × 2 = 1 + 0.825 251 106 816;
  • 38) 0.825 251 106 816 × 2 = 1 + 0.650 502 213 632;
  • 39) 0.650 502 213 632 × 2 = 1 + 0.301 004 427 264;
  • 40) 0.301 004 427 264 × 2 = 0 + 0.602 008 854 528;
  • 41) 0.602 008 854 528 × 2 = 1 + 0.204 017 709 056;
  • 42) 0.204 017 709 056 × 2 = 0 + 0.408 035 418 112;
  • 43) 0.408 035 418 112 × 2 = 0 + 0.816 070 836 224;
  • 44) 0.816 070 836 224 × 2 = 1 + 0.632 141 672 448;
  • 45) 0.632 141 672 448 × 2 = 1 + 0.264 283 344 896;
  • 46) 0.264 283 344 896 × 2 = 0 + 0.528 566 689 792;
  • 47) 0.528 566 689 792 × 2 = 1 + 0.057 133 379 584;

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:


0.000 000 109 953(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0011 1110 1001 101(2)

5. Positive number before normalization:

0.000 000 109 953(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0011 1110 1001 101(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 109 953(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0011 1110 1001 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0011 1110 1001 101(2) × 20 =

1.1101 1000 0011 1110 1001 101(2) × 2-24


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): -24

Mantissa (not normalized):
1.1101 1000 0011 1110 1001 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

103(10) =

0110 0111(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. 110 1100 0001 1111 0100 1101 =

110 1100 0001 1111 0100 1101


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) =
0110 0111

Mantissa (23 bits) =
110 1100 0001 1111 0100 1101


Decimal number 0.000 000 109 953 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0001 1111 0100 1101

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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