-0.000 000 000 742 147 676 46 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 46(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
-0.000 000 000 742 147 676 46(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:

|-0.000 000 000 742 147 676 46| = 0.000 000 000 742 147 676 46


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: 0.000 000 000 742 147 676 46.

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) 0.000 000 000 742 147 676 46 × 2 = 0 + 0.000 000 001 484 295 352 92;
  • 2) 0.000 000 001 484 295 352 92 × 2 = 0 + 0.000 000 002 968 590 705 84;
  • 3) 0.000 000 002 968 590 705 84 × 2 = 0 + 0.000 000 005 937 181 411 68;
  • 4) 0.000 000 005 937 181 411 68 × 2 = 0 + 0.000 000 011 874 362 823 36;
  • 5) 0.000 000 011 874 362 823 36 × 2 = 0 + 0.000 000 023 748 725 646 72;
  • 6) 0.000 000 023 748 725 646 72 × 2 = 0 + 0.000 000 047 497 451 293 44;
  • 7) 0.000 000 047 497 451 293 44 × 2 = 0 + 0.000 000 094 994 902 586 88;
  • 8) 0.000 000 094 994 902 586 88 × 2 = 0 + 0.000 000 189 989 805 173 76;
  • 9) 0.000 000 189 989 805 173 76 × 2 = 0 + 0.000 000 379 979 610 347 52;
  • 10) 0.000 000 379 979 610 347 52 × 2 = 0 + 0.000 000 759 959 220 695 04;
  • 11) 0.000 000 759 959 220 695 04 × 2 = 0 + 0.000 001 519 918 441 390 08;
  • 12) 0.000 001 519 918 441 390 08 × 2 = 0 + 0.000 003 039 836 882 780 16;
  • 13) 0.000 003 039 836 882 780 16 × 2 = 0 + 0.000 006 079 673 765 560 32;
  • 14) 0.000 006 079 673 765 560 32 × 2 = 0 + 0.000 012 159 347 531 120 64;
  • 15) 0.000 012 159 347 531 120 64 × 2 = 0 + 0.000 024 318 695 062 241 28;
  • 16) 0.000 024 318 695 062 241 28 × 2 = 0 + 0.000 048 637 390 124 482 56;
  • 17) 0.000 048 637 390 124 482 56 × 2 = 0 + 0.000 097 274 780 248 965 12;
  • 18) 0.000 097 274 780 248 965 12 × 2 = 0 + 0.000 194 549 560 497 930 24;
  • 19) 0.000 194 549 560 497 930 24 × 2 = 0 + 0.000 389 099 120 995 860 48;
  • 20) 0.000 389 099 120 995 860 48 × 2 = 0 + 0.000 778 198 241 991 720 96;
  • 21) 0.000 778 198 241 991 720 96 × 2 = 0 + 0.001 556 396 483 983 441 92;
  • 22) 0.001 556 396 483 983 441 92 × 2 = 0 + 0.003 112 792 967 966 883 84;
  • 23) 0.003 112 792 967 966 883 84 × 2 = 0 + 0.006 225 585 935 933 767 68;
  • 24) 0.006 225 585 935 933 767 68 × 2 = 0 + 0.012 451 171 871 867 535 36;
  • 25) 0.012 451 171 871 867 535 36 × 2 = 0 + 0.024 902 343 743 735 070 72;
  • 26) 0.024 902 343 743 735 070 72 × 2 = 0 + 0.049 804 687 487 470 141 44;
  • 27) 0.049 804 687 487 470 141 44 × 2 = 0 + 0.099 609 374 974 940 282 88;
  • 28) 0.099 609 374 974 940 282 88 × 2 = 0 + 0.199 218 749 949 880 565 76;
  • 29) 0.199 218 749 949 880 565 76 × 2 = 0 + 0.398 437 499 899 761 131 52;
  • 30) 0.398 437 499 899 761 131 52 × 2 = 0 + 0.796 874 999 799 522 263 04;
  • 31) 0.796 874 999 799 522 263 04 × 2 = 1 + 0.593 749 999 599 044 526 08;
  • 32) 0.593 749 999 599 044 526 08 × 2 = 1 + 0.187 499 999 198 089 052 16;
  • 33) 0.187 499 999 198 089 052 16 × 2 = 0 + 0.374 999 998 396 178 104 32;
  • 34) 0.374 999 998 396 178 104 32 × 2 = 0 + 0.749 999 996 792 356 208 64;
  • 35) 0.749 999 996 792 356 208 64 × 2 = 1 + 0.499 999 993 584 712 417 28;
  • 36) 0.499 999 993 584 712 417 28 × 2 = 0 + 0.999 999 987 169 424 834 56;
  • 37) 0.999 999 987 169 424 834 56 × 2 = 1 + 0.999 999 974 338 849 669 12;
  • 38) 0.999 999 974 338 849 669 12 × 2 = 1 + 0.999 999 948 677 699 338 24;
  • 39) 0.999 999 948 677 699 338 24 × 2 = 1 + 0.999 999 897 355 398 676 48;
  • 40) 0.999 999 897 355 398 676 48 × 2 = 1 + 0.999 999 794 710 797 352 96;
  • 41) 0.999 999 794 710 797 352 96 × 2 = 1 + 0.999 999 589 421 594 705 92;
  • 42) 0.999 999 589 421 594 705 92 × 2 = 1 + 0.999 999 178 843 189 411 84;
  • 43) 0.999 999 178 843 189 411 84 × 2 = 1 + 0.999 998 357 686 378 823 68;
  • 44) 0.999 998 357 686 378 823 68 × 2 = 1 + 0.999 996 715 372 757 647 36;
  • 45) 0.999 996 715 372 757 647 36 × 2 = 1 + 0.999 993 430 745 515 294 72;
  • 46) 0.999 993 430 745 515 294 72 × 2 = 1 + 0.999 986 861 491 030 589 44;
  • 47) 0.999 986 861 491 030 589 44 × 2 = 1 + 0.999 973 722 982 061 178 88;
  • 48) 0.999 973 722 982 061 178 88 × 2 = 1 + 0.999 947 445 964 122 357 76;
  • 49) 0.999 947 445 964 122 357 76 × 2 = 1 + 0.999 894 891 928 244 715 52;
  • 50) 0.999 894 891 928 244 715 52 × 2 = 1 + 0.999 789 783 856 489 431 04;
  • 51) 0.999 789 783 856 489 431 04 × 2 = 1 + 0.999 579 567 712 978 862 08;
  • 52) 0.999 579 567 712 978 862 08 × 2 = 1 + 0.999 159 135 425 957 724 16;
  • 53) 0.999 159 135 425 957 724 16 × 2 = 1 + 0.998 318 270 851 915 448 32;
  • 54) 0.998 318 270 851 915 448 32 × 2 = 1 + 0.996 636 541 703 830 896 64;

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:


0.000 000 000 742 147 676 46(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 46(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 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:


0.000 000 000 742 147 676 46(10) =

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

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

1.1001 0111 1111 1111 1111 111(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 111


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 1111 =

100 1011 1111 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 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 46 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 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