-0.000 000 000 636 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 636(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
-0.000 000 000 636(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. Start with the positive version of the number:

|-0.000 000 000 636| = 0.000 000 000 636


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 000 636.

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) 0.000 000 000 636 × 2 = 0 + 0.000 000 001 272;
  • 2) 0.000 000 001 272 × 2 = 0 + 0.000 000 002 544;
  • 3) 0.000 000 002 544 × 2 = 0 + 0.000 000 005 088;
  • 4) 0.000 000 005 088 × 2 = 0 + 0.000 000 010 176;
  • 5) 0.000 000 010 176 × 2 = 0 + 0.000 000 020 352;
  • 6) 0.000 000 020 352 × 2 = 0 + 0.000 000 040 704;
  • 7) 0.000 000 040 704 × 2 = 0 + 0.000 000 081 408;
  • 8) 0.000 000 081 408 × 2 = 0 + 0.000 000 162 816;
  • 9) 0.000 000 162 816 × 2 = 0 + 0.000 000 325 632;
  • 10) 0.000 000 325 632 × 2 = 0 + 0.000 000 651 264;
  • 11) 0.000 000 651 264 × 2 = 0 + 0.000 001 302 528;
  • 12) 0.000 001 302 528 × 2 = 0 + 0.000 002 605 056;
  • 13) 0.000 002 605 056 × 2 = 0 + 0.000 005 210 112;
  • 14) 0.000 005 210 112 × 2 = 0 + 0.000 010 420 224;
  • 15) 0.000 010 420 224 × 2 = 0 + 0.000 020 840 448;
  • 16) 0.000 020 840 448 × 2 = 0 + 0.000 041 680 896;
  • 17) 0.000 041 680 896 × 2 = 0 + 0.000 083 361 792;
  • 18) 0.000 083 361 792 × 2 = 0 + 0.000 166 723 584;
  • 19) 0.000 166 723 584 × 2 = 0 + 0.000 333 447 168;
  • 20) 0.000 333 447 168 × 2 = 0 + 0.000 666 894 336;
  • 21) 0.000 666 894 336 × 2 = 0 + 0.001 333 788 672;
  • 22) 0.001 333 788 672 × 2 = 0 + 0.002 667 577 344;
  • 23) 0.002 667 577 344 × 2 = 0 + 0.005 335 154 688;
  • 24) 0.005 335 154 688 × 2 = 0 + 0.010 670 309 376;
  • 25) 0.010 670 309 376 × 2 = 0 + 0.021 340 618 752;
  • 26) 0.021 340 618 752 × 2 = 0 + 0.042 681 237 504;
  • 27) 0.042 681 237 504 × 2 = 0 + 0.085 362 475 008;
  • 28) 0.085 362 475 008 × 2 = 0 + 0.170 724 950 016;
  • 29) 0.170 724 950 016 × 2 = 0 + 0.341 449 900 032;
  • 30) 0.341 449 900 032 × 2 = 0 + 0.682 899 800 064;
  • 31) 0.682 899 800 064 × 2 = 1 + 0.365 799 600 128;
  • 32) 0.365 799 600 128 × 2 = 0 + 0.731 599 200 256;
  • 33) 0.731 599 200 256 × 2 = 1 + 0.463 198 400 512;
  • 34) 0.463 198 400 512 × 2 = 0 + 0.926 396 801 024;
  • 35) 0.926 396 801 024 × 2 = 1 + 0.852 793 602 048;
  • 36) 0.852 793 602 048 × 2 = 1 + 0.705 587 204 096;
  • 37) 0.705 587 204 096 × 2 = 1 + 0.411 174 408 192;
  • 38) 0.411 174 408 192 × 2 = 0 + 0.822 348 816 384;
  • 39) 0.822 348 816 384 × 2 = 1 + 0.644 697 632 768;
  • 40) 0.644 697 632 768 × 2 = 1 + 0.289 395 265 536;
  • 41) 0.289 395 265 536 × 2 = 0 + 0.578 790 531 072;
  • 42) 0.578 790 531 072 × 2 = 1 + 0.157 581 062 144;
  • 43) 0.157 581 062 144 × 2 = 0 + 0.315 162 124 288;
  • 44) 0.315 162 124 288 × 2 = 0 + 0.630 324 248 576;
  • 45) 0.630 324 248 576 × 2 = 1 + 0.260 648 497 152;
  • 46) 0.260 648 497 152 × 2 = 0 + 0.521 296 994 304;
  • 47) 0.521 296 994 304 × 2 = 1 + 0.042 593 988 608;
  • 48) 0.042 593 988 608 × 2 = 0 + 0.085 187 977 216;
  • 49) 0.085 187 977 216 × 2 = 0 + 0.170 375 954 432;
  • 50) 0.170 375 954 432 × 2 = 0 + 0.340 751 908 864;
  • 51) 0.340 751 908 864 × 2 = 0 + 0.681 503 817 728;
  • 52) 0.681 503 817 728 × 2 = 1 + 0.363 007 635 456;
  • 53) 0.363 007 635 456 × 2 = 0 + 0.726 015 270 912;
  • 54) 0.726 015 270 912 × 2 = 1 + 0.452 030 541 824;

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).


5. 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:


0.000 000 000 636(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 1011 0100 1010 0001 01(2)

6. Positive number before normalization:

0.000 000 000 636(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 1011 0100 1010 0001 01(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 636(10) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 1011 0100 1010 0001 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0010 1011 1011 0100 1010 0001 01(2) × 20 =

1.0101 1101 1010 0101 0000 101(2) × 2-31


8. 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 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.0101 1101 1010 0101 0000 101


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. 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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

96(10) =

0110 0000(2)


12. 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. 010 1110 1101 0010 1000 0101 =

010 1110 1101 0010 1000 0101


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
010 1110 1101 0010 1000 0101


Decimal number -0.000 000 000 636 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 010 1110 1101 0010 1000 0101

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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