0.000 000 109 989 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 109 989 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
0.000 000 109 989 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: 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: 0.000 000 109 989 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) 0.000 000 109 989 1 × 2 = 0 + 0.000 000 219 978 2;
  • 2) 0.000 000 219 978 2 × 2 = 0 + 0.000 000 439 956 4;
  • 3) 0.000 000 439 956 4 × 2 = 0 + 0.000 000 879 912 8;
  • 4) 0.000 000 879 912 8 × 2 = 0 + 0.000 001 759 825 6;
  • 5) 0.000 001 759 825 6 × 2 = 0 + 0.000 003 519 651 2;
  • 6) 0.000 003 519 651 2 × 2 = 0 + 0.000 007 039 302 4;
  • 7) 0.000 007 039 302 4 × 2 = 0 + 0.000 014 078 604 8;
  • 8) 0.000 014 078 604 8 × 2 = 0 + 0.000 028 157 209 6;
  • 9) 0.000 028 157 209 6 × 2 = 0 + 0.000 056 314 419 2;
  • 10) 0.000 056 314 419 2 × 2 = 0 + 0.000 112 628 838 4;
  • 11) 0.000 112 628 838 4 × 2 = 0 + 0.000 225 257 676 8;
  • 12) 0.000 225 257 676 8 × 2 = 0 + 0.000 450 515 353 6;
  • 13) 0.000 450 515 353 6 × 2 = 0 + 0.000 901 030 707 2;
  • 14) 0.000 901 030 707 2 × 2 = 0 + 0.001 802 061 414 4;
  • 15) 0.001 802 061 414 4 × 2 = 0 + 0.003 604 122 828 8;
  • 16) 0.003 604 122 828 8 × 2 = 0 + 0.007 208 245 657 6;
  • 17) 0.007 208 245 657 6 × 2 = 0 + 0.014 416 491 315 2;
  • 18) 0.014 416 491 315 2 × 2 = 0 + 0.028 832 982 630 4;
  • 19) 0.028 832 982 630 4 × 2 = 0 + 0.057 665 965 260 8;
  • 20) 0.057 665 965 260 8 × 2 = 0 + 0.115 331 930 521 6;
  • 21) 0.115 331 930 521 6 × 2 = 0 + 0.230 663 861 043 2;
  • 22) 0.230 663 861 043 2 × 2 = 0 + 0.461 327 722 086 4;
  • 23) 0.461 327 722 086 4 × 2 = 0 + 0.922 655 444 172 8;
  • 24) 0.922 655 444 172 8 × 2 = 1 + 0.845 310 888 345 6;
  • 25) 0.845 310 888 345 6 × 2 = 1 + 0.690 621 776 691 2;
  • 26) 0.690 621 776 691 2 × 2 = 1 + 0.381 243 553 382 4;
  • 27) 0.381 243 553 382 4 × 2 = 0 + 0.762 487 106 764 8;
  • 28) 0.762 487 106 764 8 × 2 = 1 + 0.524 974 213 529 6;
  • 29) 0.524 974 213 529 6 × 2 = 1 + 0.049 948 427 059 2;
  • 30) 0.049 948 427 059 2 × 2 = 0 + 0.099 896 854 118 4;
  • 31) 0.099 896 854 118 4 × 2 = 0 + 0.199 793 708 236 8;
  • 32) 0.199 793 708 236 8 × 2 = 0 + 0.399 587 416 473 6;
  • 33) 0.399 587 416 473 6 × 2 = 0 + 0.799 174 832 947 2;
  • 34) 0.799 174 832 947 2 × 2 = 1 + 0.598 349 665 894 4;
  • 35) 0.598 349 665 894 4 × 2 = 1 + 0.196 699 331 788 8;
  • 36) 0.196 699 331 788 8 × 2 = 0 + 0.393 398 663 577 6;
  • 37) 0.393 398 663 577 6 × 2 = 0 + 0.786 797 327 155 2;
  • 38) 0.786 797 327 155 2 × 2 = 1 + 0.573 594 654 310 4;
  • 39) 0.573 594 654 310 4 × 2 = 1 + 0.147 189 308 620 8;
  • 40) 0.147 189 308 620 8 × 2 = 0 + 0.294 378 617 241 6;
  • 41) 0.294 378 617 241 6 × 2 = 0 + 0.588 757 234 483 2;
  • 42) 0.588 757 234 483 2 × 2 = 1 + 0.177 514 468 966 4;
  • 43) 0.177 514 468 966 4 × 2 = 0 + 0.355 028 937 932 8;
  • 44) 0.355 028 937 932 8 × 2 = 0 + 0.710 057 875 865 6;
  • 45) 0.710 057 875 865 6 × 2 = 1 + 0.420 115 751 731 2;
  • 46) 0.420 115 751 731 2 × 2 = 0 + 0.840 231 503 462 4;
  • 47) 0.840 231 503 462 4 × 2 = 1 + 0.680 463 006 924 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:


0.000 000 109 989 1(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0110 0100 101(2)

5. Positive number before normalization:

0.000 000 109 989 1(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0110 0100 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:


0.000 000 109 989 1(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0110 0100 101(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0110 0110 0100 101(2) × 20 =

1.1101 1000 0110 0110 0100 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 0110 0110 0100 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 0011 0011 0010 0101 =

110 1100 0011 0011 0010 0101


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 0011 0011 0010 0101


Decimal number 0.000 000 109 989 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 0011 0010 0101

Share

» Convert decimal 0.000 000 109 982 7 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 minutes

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