-0.000 000 000 742 147 675 76 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 76(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
-0.000 000 000 742 147 675 76(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:

|-0.000 000 000 742 147 675 76| = 0.000 000 000 742 147 675 76


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: 0.000 000 000 742 147 675 76.

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) 0.000 000 000 742 147 675 76 × 2 = 0 + 0.000 000 001 484 295 351 52;
  • 2) 0.000 000 001 484 295 351 52 × 2 = 0 + 0.000 000 002 968 590 703 04;
  • 3) 0.000 000 002 968 590 703 04 × 2 = 0 + 0.000 000 005 937 181 406 08;
  • 4) 0.000 000 005 937 181 406 08 × 2 = 0 + 0.000 000 011 874 362 812 16;
  • 5) 0.000 000 011 874 362 812 16 × 2 = 0 + 0.000 000 023 748 725 624 32;
  • 6) 0.000 000 023 748 725 624 32 × 2 = 0 + 0.000 000 047 497 451 248 64;
  • 7) 0.000 000 047 497 451 248 64 × 2 = 0 + 0.000 000 094 994 902 497 28;
  • 8) 0.000 000 094 994 902 497 28 × 2 = 0 + 0.000 000 189 989 804 994 56;
  • 9) 0.000 000 189 989 804 994 56 × 2 = 0 + 0.000 000 379 979 609 989 12;
  • 10) 0.000 000 379 979 609 989 12 × 2 = 0 + 0.000 000 759 959 219 978 24;
  • 11) 0.000 000 759 959 219 978 24 × 2 = 0 + 0.000 001 519 918 439 956 48;
  • 12) 0.000 001 519 918 439 956 48 × 2 = 0 + 0.000 003 039 836 879 912 96;
  • 13) 0.000 003 039 836 879 912 96 × 2 = 0 + 0.000 006 079 673 759 825 92;
  • 14) 0.000 006 079 673 759 825 92 × 2 = 0 + 0.000 012 159 347 519 651 84;
  • 15) 0.000 012 159 347 519 651 84 × 2 = 0 + 0.000 024 318 695 039 303 68;
  • 16) 0.000 024 318 695 039 303 68 × 2 = 0 + 0.000 048 637 390 078 607 36;
  • 17) 0.000 048 637 390 078 607 36 × 2 = 0 + 0.000 097 274 780 157 214 72;
  • 18) 0.000 097 274 780 157 214 72 × 2 = 0 + 0.000 194 549 560 314 429 44;
  • 19) 0.000 194 549 560 314 429 44 × 2 = 0 + 0.000 389 099 120 628 858 88;
  • 20) 0.000 389 099 120 628 858 88 × 2 = 0 + 0.000 778 198 241 257 717 76;
  • 21) 0.000 778 198 241 257 717 76 × 2 = 0 + 0.001 556 396 482 515 435 52;
  • 22) 0.001 556 396 482 515 435 52 × 2 = 0 + 0.003 112 792 965 030 871 04;
  • 23) 0.003 112 792 965 030 871 04 × 2 = 0 + 0.006 225 585 930 061 742 08;
  • 24) 0.006 225 585 930 061 742 08 × 2 = 0 + 0.012 451 171 860 123 484 16;
  • 25) 0.012 451 171 860 123 484 16 × 2 = 0 + 0.024 902 343 720 246 968 32;
  • 26) 0.024 902 343 720 246 968 32 × 2 = 0 + 0.049 804 687 440 493 936 64;
  • 27) 0.049 804 687 440 493 936 64 × 2 = 0 + 0.099 609 374 880 987 873 28;
  • 28) 0.099 609 374 880 987 873 28 × 2 = 0 + 0.199 218 749 761 975 746 56;
  • 29) 0.199 218 749 761 975 746 56 × 2 = 0 + 0.398 437 499 523 951 493 12;
  • 30) 0.398 437 499 523 951 493 12 × 2 = 0 + 0.796 874 999 047 902 986 24;
  • 31) 0.796 874 999 047 902 986 24 × 2 = 1 + 0.593 749 998 095 805 972 48;
  • 32) 0.593 749 998 095 805 972 48 × 2 = 1 + 0.187 499 996 191 611 944 96;
  • 33) 0.187 499 996 191 611 944 96 × 2 = 0 + 0.374 999 992 383 223 889 92;
  • 34) 0.374 999 992 383 223 889 92 × 2 = 0 + 0.749 999 984 766 447 779 84;
  • 35) 0.749 999 984 766 447 779 84 × 2 = 1 + 0.499 999 969 532 895 559 68;
  • 36) 0.499 999 969 532 895 559 68 × 2 = 0 + 0.999 999 939 065 791 119 36;
  • 37) 0.999 999 939 065 791 119 36 × 2 = 1 + 0.999 999 878 131 582 238 72;
  • 38) 0.999 999 878 131 582 238 72 × 2 = 1 + 0.999 999 756 263 164 477 44;
  • 39) 0.999 999 756 263 164 477 44 × 2 = 1 + 0.999 999 512 526 328 954 88;
  • 40) 0.999 999 512 526 328 954 88 × 2 = 1 + 0.999 999 025 052 657 909 76;
  • 41) 0.999 999 025 052 657 909 76 × 2 = 1 + 0.999 998 050 105 315 819 52;
  • 42) 0.999 998 050 105 315 819 52 × 2 = 1 + 0.999 996 100 210 631 639 04;
  • 43) 0.999 996 100 210 631 639 04 × 2 = 1 + 0.999 992 200 421 263 278 08;
  • 44) 0.999 992 200 421 263 278 08 × 2 = 1 + 0.999 984 400 842 526 556 16;
  • 45) 0.999 984 400 842 526 556 16 × 2 = 1 + 0.999 968 801 685 053 112 32;
  • 46) 0.999 968 801 685 053 112 32 × 2 = 1 + 0.999 937 603 370 106 224 64;
  • 47) 0.999 937 603 370 106 224 64 × 2 = 1 + 0.999 875 206 740 212 449 28;
  • 48) 0.999 875 206 740 212 449 28 × 2 = 1 + 0.999 750 413 480 424 898 56;
  • 49) 0.999 750 413 480 424 898 56 × 2 = 1 + 0.999 500 826 960 849 797 12;
  • 50) 0.999 500 826 960 849 797 12 × 2 = 1 + 0.999 001 653 921 699 594 24;
  • 51) 0.999 001 653 921 699 594 24 × 2 = 1 + 0.998 003 307 843 399 188 48;
  • 52) 0.998 003 307 843 399 188 48 × 2 = 1 + 0.996 006 615 686 798 376 96;
  • 53) 0.996 006 615 686 798 376 96 × 2 = 1 + 0.992 013 231 373 596 753 92;
  • 54) 0.992 013 231 373 596 753 92 × 2 = 1 + 0.984 026 462 747 193 507 84;

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:


0.000 000 000 742 147 675 76(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 76(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:


0.000 000 000 742 147 675 76(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 -0.000 000 000 742 147 675 76 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