Convert 0.000 001 907 34 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

0.000 001 907 34(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.000 001 907 34.

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) 0.000 001 907 34 × 2 = 0 + 0.000 003 814 68;
  • 2) 0.000 003 814 68 × 2 = 0 + 0.000 007 629 36;
  • 3) 0.000 007 629 36 × 2 = 0 + 0.000 015 258 72;
  • 4) 0.000 015 258 72 × 2 = 0 + 0.000 030 517 44;
  • 5) 0.000 030 517 44 × 2 = 0 + 0.000 061 034 88;
  • 6) 0.000 061 034 88 × 2 = 0 + 0.000 122 069 76;
  • 7) 0.000 122 069 76 × 2 = 0 + 0.000 244 139 52;
  • 8) 0.000 244 139 52 × 2 = 0 + 0.000 488 279 04;
  • 9) 0.000 488 279 04 × 2 = 0 + 0.000 976 558 08;
  • 10) 0.000 976 558 08 × 2 = 0 + 0.001 953 116 16;
  • 11) 0.001 953 116 16 × 2 = 0 + 0.003 906 232 32;
  • 12) 0.003 906 232 32 × 2 = 0 + 0.007 812 464 64;
  • 13) 0.007 812 464 64 × 2 = 0 + 0.015 624 929 28;
  • 14) 0.015 624 929 28 × 2 = 0 + 0.031 249 858 56;
  • 15) 0.031 249 858 56 × 2 = 0 + 0.062 499 717 12;
  • 16) 0.062 499 717 12 × 2 = 0 + 0.124 999 434 24;
  • 17) 0.124 999 434 24 × 2 = 0 + 0.249 998 868 48;
  • 18) 0.249 998 868 48 × 2 = 0 + 0.499 997 736 96;
  • 19) 0.499 997 736 96 × 2 = 0 + 0.999 995 473 92;
  • 20) 0.999 995 473 92 × 2 = 1 + 0.999 990 947 84;
  • 21) 0.999 990 947 84 × 2 = 1 + 0.999 981 895 68;
  • 22) 0.999 981 895 68 × 2 = 1 + 0.999 963 791 36;
  • 23) 0.999 963 791 36 × 2 = 1 + 0.999 927 582 72;
  • 24) 0.999 927 582 72 × 2 = 1 + 0.999 855 165 44;
  • 25) 0.999 855 165 44 × 2 = 1 + 0.999 710 330 88;
  • 26) 0.999 710 330 88 × 2 = 1 + 0.999 420 661 76;
  • 27) 0.999 420 661 76 × 2 = 1 + 0.998 841 323 52;
  • 28) 0.998 841 323 52 × 2 = 1 + 0.997 682 647 04;
  • 29) 0.997 682 647 04 × 2 = 1 + 0.995 365 294 08;
  • 30) 0.995 365 294 08 × 2 = 1 + 0.990 730 588 16;
  • 31) 0.990 730 588 16 × 2 = 1 + 0.981 461 176 32;
  • 32) 0.981 461 176 32 × 2 = 1 + 0.962 922 352 64;
  • 33) 0.962 922 352 64 × 2 = 1 + 0.925 844 705 28;
  • 34) 0.925 844 705 28 × 2 = 1 + 0.851 689 410 56;
  • 35) 0.851 689 410 56 × 2 = 1 + 0.703 378 821 12;
  • 36) 0.703 378 821 12 × 2 = 1 + 0.406 757 642 24;
  • 37) 0.406 757 642 24 × 2 = 0 + 0.813 515 284 48;
  • 38) 0.813 515 284 48 × 2 = 1 + 0.627 030 568 96;
  • 39) 0.627 030 568 96 × 2 = 1 + 0.254 061 137 92;
  • 40) 0.254 061 137 92 × 2 = 0 + 0.508 122 275 84;
  • 41) 0.508 122 275 84 × 2 = 1 + 0.016 244 551 68;
  • 42) 0.016 244 551 68 × 2 = 0 + 0.032 489 103 36;
  • 43) 0.032 489 103 36 × 2 = 0 + 0.064 978 206 72;

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...)


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:


0.000 001 907 34(10) =

0.0000 0000 0000 0000 0001 1111 1111 1111 1111 0110 100(2)


5. Positive number before normalization:

0.000 001 907 34(10) =

0.0000 0000 0000 0000 0001 1111 1111 1111 1111 0110 100(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 20 positions to the right so that only one non zero digit remains to the left of it:


0.000 001 907 34(10) =

0.0000 0000 0000 0000 0001 1111 1111 1111 1111 0110 100(2) =

0.0000 0000 0000 0000 0001 1111 1111 1111 1111 0110 100(2) × 20 =

1.1111 1111 1111 1111 0110 100(2) × 2-20


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): -20

Mantissa (not normalized):
1.1111 1111 1111 1111 0110 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-20 + 2(8-1) - 1 =

(-20 + 127)(10) =

107(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;
  • 107 ÷ 2 = 53 + 1;
  • 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) =

107(10) =

0110 1011(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. 111 1111 1111 1111 1011 0100 =

111 1111 1111 1111 1011 0100


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 1011

Mantissa (23 bits) =
111 1111 1111 1111 1011 0100


Number 0.000 001 907 34 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 0110 1011 - 111 1111 1111 1111 1011 0100

(32 bits IEEE 754)

More operations of this kind:

0.000 001 907 33 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ?

0.000 001 907 35 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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