-0.000 000 110 943 797 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 110 943 797 6(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
-0.000 000 110 943 797 6(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:

|-0.000 000 110 943 797 6| = 0.000 000 110 943 797 6


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: 0.000 000 110 943 797 6.

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) 0.000 000 110 943 797 6 × 2 = 0 + 0.000 000 221 887 595 2;
  • 2) 0.000 000 221 887 595 2 × 2 = 0 + 0.000 000 443 775 190 4;
  • 3) 0.000 000 443 775 190 4 × 2 = 0 + 0.000 000 887 550 380 8;
  • 4) 0.000 000 887 550 380 8 × 2 = 0 + 0.000 001 775 100 761 6;
  • 5) 0.000 001 775 100 761 6 × 2 = 0 + 0.000 003 550 201 523 2;
  • 6) 0.000 003 550 201 523 2 × 2 = 0 + 0.000 007 100 403 046 4;
  • 7) 0.000 007 100 403 046 4 × 2 = 0 + 0.000 014 200 806 092 8;
  • 8) 0.000 014 200 806 092 8 × 2 = 0 + 0.000 028 401 612 185 6;
  • 9) 0.000 028 401 612 185 6 × 2 = 0 + 0.000 056 803 224 371 2;
  • 10) 0.000 056 803 224 371 2 × 2 = 0 + 0.000 113 606 448 742 4;
  • 11) 0.000 113 606 448 742 4 × 2 = 0 + 0.000 227 212 897 484 8;
  • 12) 0.000 227 212 897 484 8 × 2 = 0 + 0.000 454 425 794 969 6;
  • 13) 0.000 454 425 794 969 6 × 2 = 0 + 0.000 908 851 589 939 2;
  • 14) 0.000 908 851 589 939 2 × 2 = 0 + 0.001 817 703 179 878 4;
  • 15) 0.001 817 703 179 878 4 × 2 = 0 + 0.003 635 406 359 756 8;
  • 16) 0.003 635 406 359 756 8 × 2 = 0 + 0.007 270 812 719 513 6;
  • 17) 0.007 270 812 719 513 6 × 2 = 0 + 0.014 541 625 439 027 2;
  • 18) 0.014 541 625 439 027 2 × 2 = 0 + 0.029 083 250 878 054 4;
  • 19) 0.029 083 250 878 054 4 × 2 = 0 + 0.058 166 501 756 108 8;
  • 20) 0.058 166 501 756 108 8 × 2 = 0 + 0.116 333 003 512 217 6;
  • 21) 0.116 333 003 512 217 6 × 2 = 0 + 0.232 666 007 024 435 2;
  • 22) 0.232 666 007 024 435 2 × 2 = 0 + 0.465 332 014 048 870 4;
  • 23) 0.465 332 014 048 870 4 × 2 = 0 + 0.930 664 028 097 740 8;
  • 24) 0.930 664 028 097 740 8 × 2 = 1 + 0.861 328 056 195 481 6;
  • 25) 0.861 328 056 195 481 6 × 2 = 1 + 0.722 656 112 390 963 2;
  • 26) 0.722 656 112 390 963 2 × 2 = 1 + 0.445 312 224 781 926 4;
  • 27) 0.445 312 224 781 926 4 × 2 = 0 + 0.890 624 449 563 852 8;
  • 28) 0.890 624 449 563 852 8 × 2 = 1 + 0.781 248 899 127 705 6;
  • 29) 0.781 248 899 127 705 6 × 2 = 1 + 0.562 497 798 255 411 2;
  • 30) 0.562 497 798 255 411 2 × 2 = 1 + 0.124 995 596 510 822 4;
  • 31) 0.124 995 596 510 822 4 × 2 = 0 + 0.249 991 193 021 644 8;
  • 32) 0.249 991 193 021 644 8 × 2 = 0 + 0.499 982 386 043 289 6;
  • 33) 0.499 982 386 043 289 6 × 2 = 0 + 0.999 964 772 086 579 2;
  • 34) 0.999 964 772 086 579 2 × 2 = 1 + 0.999 929 544 173 158 4;
  • 35) 0.999 929 544 173 158 4 × 2 = 1 + 0.999 859 088 346 316 8;
  • 36) 0.999 859 088 346 316 8 × 2 = 1 + 0.999 718 176 692 633 6;
  • 37) 0.999 718 176 692 633 6 × 2 = 1 + 0.999 436 353 385 267 2;
  • 38) 0.999 436 353 385 267 2 × 2 = 1 + 0.998 872 706 770 534 4;
  • 39) 0.998 872 706 770 534 4 × 2 = 1 + 0.997 745 413 541 068 8;
  • 40) 0.997 745 413 541 068 8 × 2 = 1 + 0.995 490 827 082 137 6;
  • 41) 0.995 490 827 082 137 6 × 2 = 1 + 0.990 981 654 164 275 2;
  • 42) 0.990 981 654 164 275 2 × 2 = 1 + 0.981 963 308 328 550 4;
  • 43) 0.981 963 308 328 550 4 × 2 = 1 + 0.963 926 616 657 100 8;
  • 44) 0.963 926 616 657 100 8 × 2 = 1 + 0.927 853 233 314 201 6;
  • 45) 0.927 853 233 314 201 6 × 2 = 1 + 0.855 706 466 628 403 2;
  • 46) 0.855 706 466 628 403 2 × 2 = 1 + 0.711 412 933 256 806 4;
  • 47) 0.711 412 933 256 806 4 × 2 = 1 + 0.422 825 866 513 612 8;

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:


0.000 000 110 943 797 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1100 0111 1111 1111 111(2)

6. Positive number before normalization:

0.000 000 110 943 797 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1100 0111 1111 1111 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 110 943 797 6(10) =

0.0000 0000 0000 0000 0000 0001 1101 1100 0111 1111 1111 111(2) =

0.0000 0000 0000 0000 0000 0001 1101 1100 0111 1111 1111 111(2) × 20 =

1.1101 1100 0111 1111 1111 111(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.1101 1100 0111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 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) =

103(10) =

0110 0111(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. 110 1110 0011 1111 1111 1111 =

110 1110 0011 1111 1111 1111


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 0111

Mantissa (23 bits) =
110 1110 0011 1111 1111 1111


Decimal number -0.000 000 110 943 797 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 110 1110 0011 1111 1111 1111

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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