0.000 000 110 019 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 019 7(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
0.000 000 110 019 7(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: 0.000 000 110 019 7.

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) 0.000 000 110 019 7 × 2 = 0 + 0.000 000 220 039 4;
  • 2) 0.000 000 220 039 4 × 2 = 0 + 0.000 000 440 078 8;
  • 3) 0.000 000 440 078 8 × 2 = 0 + 0.000 000 880 157 6;
  • 4) 0.000 000 880 157 6 × 2 = 0 + 0.000 001 760 315 2;
  • 5) 0.000 001 760 315 2 × 2 = 0 + 0.000 003 520 630 4;
  • 6) 0.000 003 520 630 4 × 2 = 0 + 0.000 007 041 260 8;
  • 7) 0.000 007 041 260 8 × 2 = 0 + 0.000 014 082 521 6;
  • 8) 0.000 014 082 521 6 × 2 = 0 + 0.000 028 165 043 2;
  • 9) 0.000 028 165 043 2 × 2 = 0 + 0.000 056 330 086 4;
  • 10) 0.000 056 330 086 4 × 2 = 0 + 0.000 112 660 172 8;
  • 11) 0.000 112 660 172 8 × 2 = 0 + 0.000 225 320 345 6;
  • 12) 0.000 225 320 345 6 × 2 = 0 + 0.000 450 640 691 2;
  • 13) 0.000 450 640 691 2 × 2 = 0 + 0.000 901 281 382 4;
  • 14) 0.000 901 281 382 4 × 2 = 0 + 0.001 802 562 764 8;
  • 15) 0.001 802 562 764 8 × 2 = 0 + 0.003 605 125 529 6;
  • 16) 0.003 605 125 529 6 × 2 = 0 + 0.007 210 251 059 2;
  • 17) 0.007 210 251 059 2 × 2 = 0 + 0.014 420 502 118 4;
  • 18) 0.014 420 502 118 4 × 2 = 0 + 0.028 841 004 236 8;
  • 19) 0.028 841 004 236 8 × 2 = 0 + 0.057 682 008 473 6;
  • 20) 0.057 682 008 473 6 × 2 = 0 + 0.115 364 016 947 2;
  • 21) 0.115 364 016 947 2 × 2 = 0 + 0.230 728 033 894 4;
  • 22) 0.230 728 033 894 4 × 2 = 0 + 0.461 456 067 788 8;
  • 23) 0.461 456 067 788 8 × 2 = 0 + 0.922 912 135 577 6;
  • 24) 0.922 912 135 577 6 × 2 = 1 + 0.845 824 271 155 2;
  • 25) 0.845 824 271 155 2 × 2 = 1 + 0.691 648 542 310 4;
  • 26) 0.691 648 542 310 4 × 2 = 1 + 0.383 297 084 620 8;
  • 27) 0.383 297 084 620 8 × 2 = 0 + 0.766 594 169 241 6;
  • 28) 0.766 594 169 241 6 × 2 = 1 + 0.533 188 338 483 2;
  • 29) 0.533 188 338 483 2 × 2 = 1 + 0.066 376 676 966 4;
  • 30) 0.066 376 676 966 4 × 2 = 0 + 0.132 753 353 932 8;
  • 31) 0.132 753 353 932 8 × 2 = 0 + 0.265 506 707 865 6;
  • 32) 0.265 506 707 865 6 × 2 = 0 + 0.531 013 415 731 2;
  • 33) 0.531 013 415 731 2 × 2 = 1 + 0.062 026 831 462 4;
  • 34) 0.062 026 831 462 4 × 2 = 0 + 0.124 053 662 924 8;
  • 35) 0.124 053 662 924 8 × 2 = 0 + 0.248 107 325 849 6;
  • 36) 0.248 107 325 849 6 × 2 = 0 + 0.496 214 651 699 2;
  • 37) 0.496 214 651 699 2 × 2 = 0 + 0.992 429 303 398 4;
  • 38) 0.992 429 303 398 4 × 2 = 1 + 0.984 858 606 796 8;
  • 39) 0.984 858 606 796 8 × 2 = 1 + 0.969 717 213 593 6;
  • 40) 0.969 717 213 593 6 × 2 = 1 + 0.939 434 427 187 2;
  • 41) 0.939 434 427 187 2 × 2 = 1 + 0.878 868 854 374 4;
  • 42) 0.878 868 854 374 4 × 2 = 1 + 0.757 737 708 748 8;
  • 43) 0.757 737 708 748 8 × 2 = 1 + 0.515 475 417 497 6;
  • 44) 0.515 475 417 497 6 × 2 = 1 + 0.030 950 834 995 2;
  • 45) 0.030 950 834 995 2 × 2 = 0 + 0.061 901 669 990 4;
  • 46) 0.061 901 669 990 4 × 2 = 0 + 0.123 803 339 980 8;
  • 47) 0.123 803 339 980 8 × 2 = 0 + 0.247 606 679 961 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:


0.000 000 110 019 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0111 1111 000(2)

5. Positive number before normalization:

0.000 000 110 019 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0111 1111 000(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:


0.000 000 110 019 7(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0111 1111 000(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 1000 0111 1111 000(2) × 20 =

1.1101 1000 1000 0111 1111 000(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 1000 0111 1111 000


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 0100 0011 1111 1000 =

110 1100 0100 0011 1111 1000


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 0100 0011 1111 1000


Decimal number 0.000 000 110 019 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0100 0011 1111 1000

Share

» Convert decimal 0.000 000 110 012 6 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: Show, Don’t Tell, the Strategic Power of Action. Gemini Notebook YouTube Video of 9.21 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