-0.000 000 000 742 147 42 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 42(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
-0.000 000 000 742 147 42(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:

|-0.000 000 000 742 147 42| = 0.000 000 000 742 147 42


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: 0.000 000 000 742 147 42.

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) 0.000 000 000 742 147 42 × 2 = 0 + 0.000 000 001 484 294 84;
  • 2) 0.000 000 001 484 294 84 × 2 = 0 + 0.000 000 002 968 589 68;
  • 3) 0.000 000 002 968 589 68 × 2 = 0 + 0.000 000 005 937 179 36;
  • 4) 0.000 000 005 937 179 36 × 2 = 0 + 0.000 000 011 874 358 72;
  • 5) 0.000 000 011 874 358 72 × 2 = 0 + 0.000 000 023 748 717 44;
  • 6) 0.000 000 023 748 717 44 × 2 = 0 + 0.000 000 047 497 434 88;
  • 7) 0.000 000 047 497 434 88 × 2 = 0 + 0.000 000 094 994 869 76;
  • 8) 0.000 000 094 994 869 76 × 2 = 0 + 0.000 000 189 989 739 52;
  • 9) 0.000 000 189 989 739 52 × 2 = 0 + 0.000 000 379 979 479 04;
  • 10) 0.000 000 379 979 479 04 × 2 = 0 + 0.000 000 759 958 958 08;
  • 11) 0.000 000 759 958 958 08 × 2 = 0 + 0.000 001 519 917 916 16;
  • 12) 0.000 001 519 917 916 16 × 2 = 0 + 0.000 003 039 835 832 32;
  • 13) 0.000 003 039 835 832 32 × 2 = 0 + 0.000 006 079 671 664 64;
  • 14) 0.000 006 079 671 664 64 × 2 = 0 + 0.000 012 159 343 329 28;
  • 15) 0.000 012 159 343 329 28 × 2 = 0 + 0.000 024 318 686 658 56;
  • 16) 0.000 024 318 686 658 56 × 2 = 0 + 0.000 048 637 373 317 12;
  • 17) 0.000 048 637 373 317 12 × 2 = 0 + 0.000 097 274 746 634 24;
  • 18) 0.000 097 274 746 634 24 × 2 = 0 + 0.000 194 549 493 268 48;
  • 19) 0.000 194 549 493 268 48 × 2 = 0 + 0.000 389 098 986 536 96;
  • 20) 0.000 389 098 986 536 96 × 2 = 0 + 0.000 778 197 973 073 92;
  • 21) 0.000 778 197 973 073 92 × 2 = 0 + 0.001 556 395 946 147 84;
  • 22) 0.001 556 395 946 147 84 × 2 = 0 + 0.003 112 791 892 295 68;
  • 23) 0.003 112 791 892 295 68 × 2 = 0 + 0.006 225 583 784 591 36;
  • 24) 0.006 225 583 784 591 36 × 2 = 0 + 0.012 451 167 569 182 72;
  • 25) 0.012 451 167 569 182 72 × 2 = 0 + 0.024 902 335 138 365 44;
  • 26) 0.024 902 335 138 365 44 × 2 = 0 + 0.049 804 670 276 730 88;
  • 27) 0.049 804 670 276 730 88 × 2 = 0 + 0.099 609 340 553 461 76;
  • 28) 0.099 609 340 553 461 76 × 2 = 0 + 0.199 218 681 106 923 52;
  • 29) 0.199 218 681 106 923 52 × 2 = 0 + 0.398 437 362 213 847 04;
  • 30) 0.398 437 362 213 847 04 × 2 = 0 + 0.796 874 724 427 694 08;
  • 31) 0.796 874 724 427 694 08 × 2 = 1 + 0.593 749 448 855 388 16;
  • 32) 0.593 749 448 855 388 16 × 2 = 1 + 0.187 498 897 710 776 32;
  • 33) 0.187 498 897 710 776 32 × 2 = 0 + 0.374 997 795 421 552 64;
  • 34) 0.374 997 795 421 552 64 × 2 = 0 + 0.749 995 590 843 105 28;
  • 35) 0.749 995 590 843 105 28 × 2 = 1 + 0.499 991 181 686 210 56;
  • 36) 0.499 991 181 686 210 56 × 2 = 0 + 0.999 982 363 372 421 12;
  • 37) 0.999 982 363 372 421 12 × 2 = 1 + 0.999 964 726 744 842 24;
  • 38) 0.999 964 726 744 842 24 × 2 = 1 + 0.999 929 453 489 684 48;
  • 39) 0.999 929 453 489 684 48 × 2 = 1 + 0.999 858 906 979 368 96;
  • 40) 0.999 858 906 979 368 96 × 2 = 1 + 0.999 717 813 958 737 92;
  • 41) 0.999 717 813 958 737 92 × 2 = 1 + 0.999 435 627 917 475 84;
  • 42) 0.999 435 627 917 475 84 × 2 = 1 + 0.998 871 255 834 951 68;
  • 43) 0.998 871 255 834 951 68 × 2 = 1 + 0.997 742 511 669 903 36;
  • 44) 0.997 742 511 669 903 36 × 2 = 1 + 0.995 485 023 339 806 72;
  • 45) 0.995 485 023 339 806 72 × 2 = 1 + 0.990 970 046 679 613 44;
  • 46) 0.990 970 046 679 613 44 × 2 = 1 + 0.981 940 093 359 226 88;
  • 47) 0.981 940 093 359 226 88 × 2 = 1 + 0.963 880 186 718 453 76;
  • 48) 0.963 880 186 718 453 76 × 2 = 1 + 0.927 760 373 436 907 52;
  • 49) 0.927 760 373 436 907 52 × 2 = 1 + 0.855 520 746 873 815 04;
  • 50) 0.855 520 746 873 815 04 × 2 = 1 + 0.711 041 493 747 630 08;
  • 51) 0.711 041 493 747 630 08 × 2 = 1 + 0.422 082 987 495 260 16;
  • 52) 0.422 082 987 495 260 16 × 2 = 0 + 0.844 165 974 990 520 32;
  • 53) 0.844 165 974 990 520 32 × 2 = 1 + 0.688 331 949 981 040 64;
  • 54) 0.688 331 949 981 040 64 × 2 = 1 + 0.376 663 899 962 081 28;

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:


0.000 000 000 742 147 42(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 42(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(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:


0.000 000 000 742 147 42(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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):
1.1001 0111 1111 1111 1111 011


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. 100 1011 1111 1111 1111 1011 =

100 1011 1111 1111 1111 1011


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) =
100 1011 1111 1111 1111 1011


Decimal number -0.000 000 000 742 147 42 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

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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