0.000 000 110 009 14 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 110 009 14(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
0.000 000 110 009 14(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: 0.000 000 110 009 14.

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) 0.000 000 110 009 14 × 2 = 0 + 0.000 000 220 018 28;
  • 2) 0.000 000 220 018 28 × 2 = 0 + 0.000 000 440 036 56;
  • 3) 0.000 000 440 036 56 × 2 = 0 + 0.000 000 880 073 12;
  • 4) 0.000 000 880 073 12 × 2 = 0 + 0.000 001 760 146 24;
  • 5) 0.000 001 760 146 24 × 2 = 0 + 0.000 003 520 292 48;
  • 6) 0.000 003 520 292 48 × 2 = 0 + 0.000 007 040 584 96;
  • 7) 0.000 007 040 584 96 × 2 = 0 + 0.000 014 081 169 92;
  • 8) 0.000 014 081 169 92 × 2 = 0 + 0.000 028 162 339 84;
  • 9) 0.000 028 162 339 84 × 2 = 0 + 0.000 056 324 679 68;
  • 10) 0.000 056 324 679 68 × 2 = 0 + 0.000 112 649 359 36;
  • 11) 0.000 112 649 359 36 × 2 = 0 + 0.000 225 298 718 72;
  • 12) 0.000 225 298 718 72 × 2 = 0 + 0.000 450 597 437 44;
  • 13) 0.000 450 597 437 44 × 2 = 0 + 0.000 901 194 874 88;
  • 14) 0.000 901 194 874 88 × 2 = 0 + 0.001 802 389 749 76;
  • 15) 0.001 802 389 749 76 × 2 = 0 + 0.003 604 779 499 52;
  • 16) 0.003 604 779 499 52 × 2 = 0 + 0.007 209 558 999 04;
  • 17) 0.007 209 558 999 04 × 2 = 0 + 0.014 419 117 998 08;
  • 18) 0.014 419 117 998 08 × 2 = 0 + 0.028 838 235 996 16;
  • 19) 0.028 838 235 996 16 × 2 = 0 + 0.057 676 471 992 32;
  • 20) 0.057 676 471 992 32 × 2 = 0 + 0.115 352 943 984 64;
  • 21) 0.115 352 943 984 64 × 2 = 0 + 0.230 705 887 969 28;
  • 22) 0.230 705 887 969 28 × 2 = 0 + 0.461 411 775 938 56;
  • 23) 0.461 411 775 938 56 × 2 = 0 + 0.922 823 551 877 12;
  • 24) 0.922 823 551 877 12 × 2 = 1 + 0.845 647 103 754 24;
  • 25) 0.845 647 103 754 24 × 2 = 1 + 0.691 294 207 508 48;
  • 26) 0.691 294 207 508 48 × 2 = 1 + 0.382 588 415 016 96;
  • 27) 0.382 588 415 016 96 × 2 = 0 + 0.765 176 830 033 92;
  • 28) 0.765 176 830 033 92 × 2 = 1 + 0.530 353 660 067 84;
  • 29) 0.530 353 660 067 84 × 2 = 1 + 0.060 707 320 135 68;
  • 30) 0.060 707 320 135 68 × 2 = 0 + 0.121 414 640 271 36;
  • 31) 0.121 414 640 271 36 × 2 = 0 + 0.242 829 280 542 72;
  • 32) 0.242 829 280 542 72 × 2 = 0 + 0.485 658 561 085 44;
  • 33) 0.485 658 561 085 44 × 2 = 0 + 0.971 317 122 170 88;
  • 34) 0.971 317 122 170 88 × 2 = 1 + 0.942 634 244 341 76;
  • 35) 0.942 634 244 341 76 × 2 = 1 + 0.885 268 488 683 52;
  • 36) 0.885 268 488 683 52 × 2 = 1 + 0.770 536 977 367 04;
  • 37) 0.770 536 977 367 04 × 2 = 1 + 0.541 073 954 734 08;
  • 38) 0.541 073 954 734 08 × 2 = 1 + 0.082 147 909 468 16;
  • 39) 0.082 147 909 468 16 × 2 = 0 + 0.164 295 818 936 32;
  • 40) 0.164 295 818 936 32 × 2 = 0 + 0.328 591 637 872 64;
  • 41) 0.328 591 637 872 64 × 2 = 0 + 0.657 183 275 745 28;
  • 42) 0.657 183 275 745 28 × 2 = 1 + 0.314 366 551 490 56;
  • 43) 0.314 366 551 490 56 × 2 = 0 + 0.628 733 102 981 12;
  • 44) 0.628 733 102 981 12 × 2 = 1 + 0.257 466 205 962 24;
  • 45) 0.257 466 205 962 24 × 2 = 0 + 0.514 932 411 924 48;
  • 46) 0.514 932 411 924 48 × 2 = 1 + 0.029 864 823 848 96;
  • 47) 0.029 864 823 848 96 × 2 = 0 + 0.059 729 647 697 92;

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:


0.000 000 110 009 14(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0101 010(2)

5. Positive number before normalization:

0.000 000 110 009 14(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0101 010(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:


0.000 000 110 009 14(10) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0101 010(2) =

0.0000 0000 0000 0000 0000 0001 1101 1000 0111 1100 0101 010(2) × 20 =

1.1101 1000 0111 1100 0101 010(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 0111 1100 0101 010


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 1110 0010 1010 =

110 1100 0011 1110 0010 1010


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 1110 0010 1010


Decimal number 0.000 000 110 009 14 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0111 - 110 1100 0011 1110 0010 1010

Share

» Convert decimal 0.000 000 110 008 86 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: Reputation: Bridge Between Company's Actions & Market Value. Gemini Notebook YouTube Video of 6.55 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