0.000 000 000 266 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 266(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
0.000 000 000 266(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: 0.000 000 000 266.

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) 0.000 000 000 266 × 2 = 0 + 0.000 000 000 532;
  • 2) 0.000 000 000 532 × 2 = 0 + 0.000 000 001 064;
  • 3) 0.000 000 001 064 × 2 = 0 + 0.000 000 002 128;
  • 4) 0.000 000 002 128 × 2 = 0 + 0.000 000 004 256;
  • 5) 0.000 000 004 256 × 2 = 0 + 0.000 000 008 512;
  • 6) 0.000 000 008 512 × 2 = 0 + 0.000 000 017 024;
  • 7) 0.000 000 017 024 × 2 = 0 + 0.000 000 034 048;
  • 8) 0.000 000 034 048 × 2 = 0 + 0.000 000 068 096;
  • 9) 0.000 000 068 096 × 2 = 0 + 0.000 000 136 192;
  • 10) 0.000 000 136 192 × 2 = 0 + 0.000 000 272 384;
  • 11) 0.000 000 272 384 × 2 = 0 + 0.000 000 544 768;
  • 12) 0.000 000 544 768 × 2 = 0 + 0.000 001 089 536;
  • 13) 0.000 001 089 536 × 2 = 0 + 0.000 002 179 072;
  • 14) 0.000 002 179 072 × 2 = 0 + 0.000 004 358 144;
  • 15) 0.000 004 358 144 × 2 = 0 + 0.000 008 716 288;
  • 16) 0.000 008 716 288 × 2 = 0 + 0.000 017 432 576;
  • 17) 0.000 017 432 576 × 2 = 0 + 0.000 034 865 152;
  • 18) 0.000 034 865 152 × 2 = 0 + 0.000 069 730 304;
  • 19) 0.000 069 730 304 × 2 = 0 + 0.000 139 460 608;
  • 20) 0.000 139 460 608 × 2 = 0 + 0.000 278 921 216;
  • 21) 0.000 278 921 216 × 2 = 0 + 0.000 557 842 432;
  • 22) 0.000 557 842 432 × 2 = 0 + 0.001 115 684 864;
  • 23) 0.001 115 684 864 × 2 = 0 + 0.002 231 369 728;
  • 24) 0.002 231 369 728 × 2 = 0 + 0.004 462 739 456;
  • 25) 0.004 462 739 456 × 2 = 0 + 0.008 925 478 912;
  • 26) 0.008 925 478 912 × 2 = 0 + 0.017 850 957 824;
  • 27) 0.017 850 957 824 × 2 = 0 + 0.035 701 915 648;
  • 28) 0.035 701 915 648 × 2 = 0 + 0.071 403 831 296;
  • 29) 0.071 403 831 296 × 2 = 0 + 0.142 807 662 592;
  • 30) 0.142 807 662 592 × 2 = 0 + 0.285 615 325 184;
  • 31) 0.285 615 325 184 × 2 = 0 + 0.571 230 650 368;
  • 32) 0.571 230 650 368 × 2 = 1 + 0.142 461 300 736;
  • 33) 0.142 461 300 736 × 2 = 0 + 0.284 922 601 472;
  • 34) 0.284 922 601 472 × 2 = 0 + 0.569 845 202 944;
  • 35) 0.569 845 202 944 × 2 = 1 + 0.139 690 405 888;
  • 36) 0.139 690 405 888 × 2 = 0 + 0.279 380 811 776;
  • 37) 0.279 380 811 776 × 2 = 0 + 0.558 761 623 552;
  • 38) 0.558 761 623 552 × 2 = 1 + 0.117 523 247 104;
  • 39) 0.117 523 247 104 × 2 = 0 + 0.235 046 494 208;
  • 40) 0.235 046 494 208 × 2 = 0 + 0.470 092 988 416;
  • 41) 0.470 092 988 416 × 2 = 0 + 0.940 185 976 832;
  • 42) 0.940 185 976 832 × 2 = 1 + 0.880 371 953 664;
  • 43) 0.880 371 953 664 × 2 = 1 + 0.760 743 907 328;
  • 44) 0.760 743 907 328 × 2 = 1 + 0.521 487 814 656;
  • 45) 0.521 487 814 656 × 2 = 1 + 0.042 975 629 312;
  • 46) 0.042 975 629 312 × 2 = 0 + 0.085 951 258 624;
  • 47) 0.085 951 258 624 × 2 = 0 + 0.171 902 517 248;
  • 48) 0.171 902 517 248 × 2 = 0 + 0.343 805 034 496;
  • 49) 0.343 805 034 496 × 2 = 0 + 0.687 610 068 992;
  • 50) 0.687 610 068 992 × 2 = 1 + 0.375 220 137 984;
  • 51) 0.375 220 137 984 × 2 = 0 + 0.750 440 275 968;
  • 52) 0.750 440 275 968 × 2 = 1 + 0.500 880 551 936;
  • 53) 0.500 880 551 936 × 2 = 1 + 0.001 761 103 872;
  • 54) 0.001 761 103 872 × 2 = 0 + 0.003 522 207 744;
  • 55) 0.003 522 207 744 × 2 = 0 + 0.007 044 415 488;

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:


0.000 000 000 266(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0010 0100 0111 1000 0101 100(2)

5. Positive number before normalization:

0.000 000 000 266(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0010 0100 0111 1000 0101 100(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 266(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0010 0100 0111 1000 0101 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 0010 0100 0111 1000 0101 100(2) × 20 =

1.0010 0100 0111 1000 0101 100(2) × 2-32


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): -32

Mantissa (not normalized):
1.0010 0100 0111 1000 0101 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(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;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

95(10) =

0101 1111(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. 001 0010 0011 1100 0010 1100 =

001 0010 0011 1100 0010 1100


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) =
0101 1111

Mantissa (23 bits) =
001 0010 0011 1100 0010 1100


Decimal number 0.000 000 000 266 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0101 1111 - 001 0010 0011 1100 0010 1100

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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