Convert -0.000 000 000 012 1 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

-0.000 000 000 012 1(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. Start with the positive version of the number:

|-0.000 000 000 012 1| = 0.000 000 000 012 1

2. 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;

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 the binary (base 2) the fractional part: 0.000 000 000 012 1.

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) 0.000 000 000 012 1 × 2 = 0 + 0.000 000 000 024 2;
  • 2) 0.000 000 000 024 2 × 2 = 0 + 0.000 000 000 048 4;
  • 3) 0.000 000 000 048 4 × 2 = 0 + 0.000 000 000 096 8;
  • 4) 0.000 000 000 096 8 × 2 = 0 + 0.000 000 000 193 6;
  • 5) 0.000 000 000 193 6 × 2 = 0 + 0.000 000 000 387 2;
  • 6) 0.000 000 000 387 2 × 2 = 0 + 0.000 000 000 774 4;
  • 7) 0.000 000 000 774 4 × 2 = 0 + 0.000 000 001 548 8;
  • 8) 0.000 000 001 548 8 × 2 = 0 + 0.000 000 003 097 6;
  • 9) 0.000 000 003 097 6 × 2 = 0 + 0.000 000 006 195 2;
  • 10) 0.000 000 006 195 2 × 2 = 0 + 0.000 000 012 390 4;
  • 11) 0.000 000 012 390 4 × 2 = 0 + 0.000 000 024 780 8;
  • 12) 0.000 000 024 780 8 × 2 = 0 + 0.000 000 049 561 6;
  • 13) 0.000 000 049 561 6 × 2 = 0 + 0.000 000 099 123 2;
  • 14) 0.000 000 099 123 2 × 2 = 0 + 0.000 000 198 246 4;
  • 15) 0.000 000 198 246 4 × 2 = 0 + 0.000 000 396 492 8;
  • 16) 0.000 000 396 492 8 × 2 = 0 + 0.000 000 792 985 6;
  • 17) 0.000 000 792 985 6 × 2 = 0 + 0.000 001 585 971 2;
  • 18) 0.000 001 585 971 2 × 2 = 0 + 0.000 003 171 942 4;
  • 19) 0.000 003 171 942 4 × 2 = 0 + 0.000 006 343 884 8;
  • 20) 0.000 006 343 884 8 × 2 = 0 + 0.000 012 687 769 6;
  • 21) 0.000 012 687 769 6 × 2 = 0 + 0.000 025 375 539 2;
  • 22) 0.000 025 375 539 2 × 2 = 0 + 0.000 050 751 078 4;
  • 23) 0.000 050 751 078 4 × 2 = 0 + 0.000 101 502 156 8;
  • 24) 0.000 101 502 156 8 × 2 = 0 + 0.000 203 004 313 6;
  • 25) 0.000 203 004 313 6 × 2 = 0 + 0.000 406 008 627 2;
  • 26) 0.000 406 008 627 2 × 2 = 0 + 0.000 812 017 254 4;
  • 27) 0.000 812 017 254 4 × 2 = 0 + 0.001 624 034 508 8;
  • 28) 0.001 624 034 508 8 × 2 = 0 + 0.003 248 069 017 6;
  • 29) 0.003 248 069 017 6 × 2 = 0 + 0.006 496 138 035 2;
  • 30) 0.006 496 138 035 2 × 2 = 0 + 0.012 992 276 070 4;
  • 31) 0.012 992 276 070 4 × 2 = 0 + 0.025 984 552 140 8;
  • 32) 0.025 984 552 140 8 × 2 = 0 + 0.051 969 104 281 6;
  • 33) 0.051 969 104 281 6 × 2 = 0 + 0.103 938 208 563 2;
  • 34) 0.103 938 208 563 2 × 2 = 0 + 0.207 876 417 126 4;
  • 35) 0.207 876 417 126 4 × 2 = 0 + 0.415 752 834 252 8;
  • 36) 0.415 752 834 252 8 × 2 = 0 + 0.831 505 668 505 6;
  • 37) 0.831 505 668 505 6 × 2 = 1 + 0.663 011 337 011 2;
  • 38) 0.663 011 337 011 2 × 2 = 1 + 0.326 022 674 022 4;
  • 39) 0.326 022 674 022 4 × 2 = 0 + 0.652 045 348 044 8;
  • 40) 0.652 045 348 044 8 × 2 = 1 + 0.304 090 696 089 6;
  • 41) 0.304 090 696 089 6 × 2 = 0 + 0.608 181 392 179 2;
  • 42) 0.608 181 392 179 2 × 2 = 1 + 0.216 362 784 358 4;
  • 43) 0.216 362 784 358 4 × 2 = 0 + 0.432 725 568 716 8;
  • 44) 0.432 725 568 716 8 × 2 = 0 + 0.865 451 137 433 6;
  • 45) 0.865 451 137 433 6 × 2 = 1 + 0.730 902 274 867 2;
  • 46) 0.730 902 274 867 2 × 2 = 1 + 0.461 804 549 734 4;
  • 47) 0.461 804 549 734 4 × 2 = 0 + 0.923 609 099 468 8;
  • 48) 0.923 609 099 468 8 × 2 = 1 + 0.847 218 198 937 6;
  • 49) 0.847 218 198 937 6 × 2 = 1 + 0.694 436 397 875 2;
  • 50) 0.694 436 397 875 2 × 2 = 1 + 0.388 872 795 750 4;
  • 51) 0.388 872 795 750 4 × 2 = 0 + 0.777 745 591 500 8;
  • 52) 0.777 745 591 500 8 × 2 = 1 + 0.555 491 183 001 6;
  • 53) 0.555 491 183 001 6 × 2 = 1 + 0.110 982 366 003 2;
  • 54) 0.110 982 366 003 2 × 2 = 0 + 0.221 964 732 006 4;
  • 55) 0.221 964 732 006 4 × 2 = 0 + 0.443 929 464 012 8;
  • 56) 0.443 929 464 012 8 × 2 = 0 + 0.887 858 928 025 6;
  • 57) 0.887 858 928 025 6 × 2 = 1 + 0.775 717 856 051 2;
  • 58) 0.775 717 856 051 2 × 2 = 1 + 0.551 435 712 102 4;
  • 59) 0.551 435 712 102 4 × 2 = 1 + 0.102 871 424 204 8;
  • 60) 0.102 871 424 204 8 × 2 = 0 + 0.205 742 848 409 6;

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...)


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:

0.000 000 000 012 1(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1101 0100 1101 1101 1000 1110(2)


6. Positive number before normalization:

0.000 000 000 012 1(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1101 0100 1101 1101 1000 1110(2)


7. Normalize the binary representation of the number.

Shift the decimal mark 37 positions to the right so that only one non zero digit remains to the left of it:

0.000 000 000 012 1(10) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1101 0100 1101 1101 1000 1110(2) =


0.0000 0000 0000 0000 0000 0000 0000 0000 0000 1101 0100 1101 1101 1000 1110(2) × 20 =


1.1010 1001 1011 1011 0001 110(2) × 2-37


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): -37


Mantissa (not normalized):
1.1010 1001 1011 1011 0001 110


9. Adjust the exponent.

Use the 8 bit excess/bias notation:

Exponent (adjusted) =


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


-37 + 2(8-1) - 1 =


(-37 + 127)(10) =


90(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;
  • 90 ÷ 2 = 45 + 0;
  • 45 ÷ 2 = 22 + 1;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 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) =


90(10) =


0101 1010(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. 101 0100 1101 1101 1000 1110 =


101 0100 1101 1101 1000 1110


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) =
0101 1010


Mantissa (23 bits) =
101 0100 1101 1101 1000 1110


Number -0.000 000 000 012 1 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
1 - 0101 1010 - 101 0100 1101 1101 1000 1110

(32 bits IEEE 754)
  • Sign (1 bit):

    • 1

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 0

      28
    • 1

      27
    • 1

      26
    • 0

      25
    • 1

      24
    • 0

      23
  • Mantissa (23 bits):

    • 1

      22
    • 0

      21
    • 1

      20
    • 0

      19
    • 1

      18
    • 0

      17
    • 0

      16
    • 1

      15
    • 1

      14
    • 0

      13
    • 1

      12
    • 1

      11
    • 1

      10
    • 0

      9
    • 1

      8
    • 1

      7
    • 0

      6
    • 0

      5
    • 0

      4
    • 1

      3
    • 1

      2
    • 1

      1
    • 0

      0

More operations of this kind:

-0.000 000 000 012 2 = ? ... -0.000 000 000 012 = ?


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)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

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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