0.000 000 637 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 637(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
0.000 000 637(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: 0.000 000 637.

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) 0.000 000 637 × 2 = 0 + 0.000 001 274;
  • 2) 0.000 001 274 × 2 = 0 + 0.000 002 548;
  • 3) 0.000 002 548 × 2 = 0 + 0.000 005 096;
  • 4) 0.000 005 096 × 2 = 0 + 0.000 010 192;
  • 5) 0.000 010 192 × 2 = 0 + 0.000 020 384;
  • 6) 0.000 020 384 × 2 = 0 + 0.000 040 768;
  • 7) 0.000 040 768 × 2 = 0 + 0.000 081 536;
  • 8) 0.000 081 536 × 2 = 0 + 0.000 163 072;
  • 9) 0.000 163 072 × 2 = 0 + 0.000 326 144;
  • 10) 0.000 326 144 × 2 = 0 + 0.000 652 288;
  • 11) 0.000 652 288 × 2 = 0 + 0.001 304 576;
  • 12) 0.001 304 576 × 2 = 0 + 0.002 609 152;
  • 13) 0.002 609 152 × 2 = 0 + 0.005 218 304;
  • 14) 0.005 218 304 × 2 = 0 + 0.010 436 608;
  • 15) 0.010 436 608 × 2 = 0 + 0.020 873 216;
  • 16) 0.020 873 216 × 2 = 0 + 0.041 746 432;
  • 17) 0.041 746 432 × 2 = 0 + 0.083 492 864;
  • 18) 0.083 492 864 × 2 = 0 + 0.166 985 728;
  • 19) 0.166 985 728 × 2 = 0 + 0.333 971 456;
  • 20) 0.333 971 456 × 2 = 0 + 0.667 942 912;
  • 21) 0.667 942 912 × 2 = 1 + 0.335 885 824;
  • 22) 0.335 885 824 × 2 = 0 + 0.671 771 648;
  • 23) 0.671 771 648 × 2 = 1 + 0.343 543 296;
  • 24) 0.343 543 296 × 2 = 0 + 0.687 086 592;
  • 25) 0.687 086 592 × 2 = 1 + 0.374 173 184;
  • 26) 0.374 173 184 × 2 = 0 + 0.748 346 368;
  • 27) 0.748 346 368 × 2 = 1 + 0.496 692 736;
  • 28) 0.496 692 736 × 2 = 0 + 0.993 385 472;
  • 29) 0.993 385 472 × 2 = 1 + 0.986 770 944;
  • 30) 0.986 770 944 × 2 = 1 + 0.973 541 888;
  • 31) 0.973 541 888 × 2 = 1 + 0.947 083 776;
  • 32) 0.947 083 776 × 2 = 1 + 0.894 167 552;
  • 33) 0.894 167 552 × 2 = 1 + 0.788 335 104;
  • 34) 0.788 335 104 × 2 = 1 + 0.576 670 208;
  • 35) 0.576 670 208 × 2 = 1 + 0.153 340 416;
  • 36) 0.153 340 416 × 2 = 0 + 0.306 680 832;
  • 37) 0.306 680 832 × 2 = 0 + 0.613 361 664;
  • 38) 0.613 361 664 × 2 = 1 + 0.226 723 328;
  • 39) 0.226 723 328 × 2 = 0 + 0.453 446 656;
  • 40) 0.453 446 656 × 2 = 0 + 0.906 893 312;
  • 41) 0.906 893 312 × 2 = 1 + 0.813 786 624;
  • 42) 0.813 786 624 × 2 = 1 + 0.627 573 248;
  • 43) 0.627 573 248 × 2 = 1 + 0.255 146 496;
  • 44) 0.255 146 496 × 2 = 0 + 0.510 292 992;

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:


0.000 000 637(10) =

0.0000 0000 0000 0000 0000 1010 1010 1111 1110 0100 1110(2)

5. Positive number before normalization:

0.000 000 637(10) =

0.0000 0000 0000 0000 0000 1010 1010 1111 1110 0100 1110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 637(10) =

0.0000 0000 0000 0000 0000 1010 1010 1111 1110 0100 1110(2) =

0.0000 0000 0000 0000 0000 1010 1010 1111 1110 0100 1110(2) × 20 =

1.0101 0101 1111 1100 1001 110(2) × 2-21


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): -21

Mantissa (not normalized):
1.0101 0101 1111 1100 1001 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-21 + 2(8-1) - 1 =

(-21 + 127)(10) =

106(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;
  • 106 ÷ 2 = 53 + 0;
  • 53 ÷ 2 = 26 + 1;
  • 26 ÷ 2 = 13 + 0;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

106(10) =

0110 1010(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. 010 1010 1111 1110 0100 1110 =

010 1010 1111 1110 0100 1110


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 1010

Mantissa (23 bits) =
010 1010 1111 1110 0100 1110


Decimal number 0.000 000 637 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 1010 - 010 1010 1111 1110 0100 1110

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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