-0.000 000 000 742 147 675 95 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 95(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 95(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 95| = 0.000 000 000 742 147 675 95


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 95.

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 95 × 2 = 0 + 0.000 000 001 484 295 351 9;
  • 2) 0.000 000 001 484 295 351 9 × 2 = 0 + 0.000 000 002 968 590 703 8;
  • 3) 0.000 000 002 968 590 703 8 × 2 = 0 + 0.000 000 005 937 181 407 6;
  • 4) 0.000 000 005 937 181 407 6 × 2 = 0 + 0.000 000 011 874 362 815 2;
  • 5) 0.000 000 011 874 362 815 2 × 2 = 0 + 0.000 000 023 748 725 630 4;
  • 6) 0.000 000 023 748 725 630 4 × 2 = 0 + 0.000 000 047 497 451 260 8;
  • 7) 0.000 000 047 497 451 260 8 × 2 = 0 + 0.000 000 094 994 902 521 6;
  • 8) 0.000 000 094 994 902 521 6 × 2 = 0 + 0.000 000 189 989 805 043 2;
  • 9) 0.000 000 189 989 805 043 2 × 2 = 0 + 0.000 000 379 979 610 086 4;
  • 10) 0.000 000 379 979 610 086 4 × 2 = 0 + 0.000 000 759 959 220 172 8;
  • 11) 0.000 000 759 959 220 172 8 × 2 = 0 + 0.000 001 519 918 440 345 6;
  • 12) 0.000 001 519 918 440 345 6 × 2 = 0 + 0.000 003 039 836 880 691 2;
  • 13) 0.000 003 039 836 880 691 2 × 2 = 0 + 0.000 006 079 673 761 382 4;
  • 14) 0.000 006 079 673 761 382 4 × 2 = 0 + 0.000 012 159 347 522 764 8;
  • 15) 0.000 012 159 347 522 764 8 × 2 = 0 + 0.000 024 318 695 045 529 6;
  • 16) 0.000 024 318 695 045 529 6 × 2 = 0 + 0.000 048 637 390 091 059 2;
  • 17) 0.000 048 637 390 091 059 2 × 2 = 0 + 0.000 097 274 780 182 118 4;
  • 18) 0.000 097 274 780 182 118 4 × 2 = 0 + 0.000 194 549 560 364 236 8;
  • 19) 0.000 194 549 560 364 236 8 × 2 = 0 + 0.000 389 099 120 728 473 6;
  • 20) 0.000 389 099 120 728 473 6 × 2 = 0 + 0.000 778 198 241 456 947 2;
  • 21) 0.000 778 198 241 456 947 2 × 2 = 0 + 0.001 556 396 482 913 894 4;
  • 22) 0.001 556 396 482 913 894 4 × 2 = 0 + 0.003 112 792 965 827 788 8;
  • 23) 0.003 112 792 965 827 788 8 × 2 = 0 + 0.006 225 585 931 655 577 6;
  • 24) 0.006 225 585 931 655 577 6 × 2 = 0 + 0.012 451 171 863 311 155 2;
  • 25) 0.012 451 171 863 311 155 2 × 2 = 0 + 0.024 902 343 726 622 310 4;
  • 26) 0.024 902 343 726 622 310 4 × 2 = 0 + 0.049 804 687 453 244 620 8;
  • 27) 0.049 804 687 453 244 620 8 × 2 = 0 + 0.099 609 374 906 489 241 6;
  • 28) 0.099 609 374 906 489 241 6 × 2 = 0 + 0.199 218 749 812 978 483 2;
  • 29) 0.199 218 749 812 978 483 2 × 2 = 0 + 0.398 437 499 625 956 966 4;
  • 30) 0.398 437 499 625 956 966 4 × 2 = 0 + 0.796 874 999 251 913 932 8;
  • 31) 0.796 874 999 251 913 932 8 × 2 = 1 + 0.593 749 998 503 827 865 6;
  • 32) 0.593 749 998 503 827 865 6 × 2 = 1 + 0.187 499 997 007 655 731 2;
  • 33) 0.187 499 997 007 655 731 2 × 2 = 0 + 0.374 999 994 015 311 462 4;
  • 34) 0.374 999 994 015 311 462 4 × 2 = 0 + 0.749 999 988 030 622 924 8;
  • 35) 0.749 999 988 030 622 924 8 × 2 = 1 + 0.499 999 976 061 245 849 6;
  • 36) 0.499 999 976 061 245 849 6 × 2 = 0 + 0.999 999 952 122 491 699 2;
  • 37) 0.999 999 952 122 491 699 2 × 2 = 1 + 0.999 999 904 244 983 398 4;
  • 38) 0.999 999 904 244 983 398 4 × 2 = 1 + 0.999 999 808 489 966 796 8;
  • 39) 0.999 999 808 489 966 796 8 × 2 = 1 + 0.999 999 616 979 933 593 6;
  • 40) 0.999 999 616 979 933 593 6 × 2 = 1 + 0.999 999 233 959 867 187 2;
  • 41) 0.999 999 233 959 867 187 2 × 2 = 1 + 0.999 998 467 919 734 374 4;
  • 42) 0.999 998 467 919 734 374 4 × 2 = 1 + 0.999 996 935 839 468 748 8;
  • 43) 0.999 996 935 839 468 748 8 × 2 = 1 + 0.999 993 871 678 937 497 6;
  • 44) 0.999 993 871 678 937 497 6 × 2 = 1 + 0.999 987 743 357 874 995 2;
  • 45) 0.999 987 743 357 874 995 2 × 2 = 1 + 0.999 975 486 715 749 990 4;
  • 46) 0.999 975 486 715 749 990 4 × 2 = 1 + 0.999 950 973 431 499 980 8;
  • 47) 0.999 950 973 431 499 980 8 × 2 = 1 + 0.999 901 946 862 999 961 6;
  • 48) 0.999 901 946 862 999 961 6 × 2 = 1 + 0.999 803 893 725 999 923 2;
  • 49) 0.999 803 893 725 999 923 2 × 2 = 1 + 0.999 607 787 451 999 846 4;
  • 50) 0.999 607 787 451 999 846 4 × 2 = 1 + 0.999 215 574 903 999 692 8;
  • 51) 0.999 215 574 903 999 692 8 × 2 = 1 + 0.998 431 149 807 999 385 6;
  • 52) 0.998 431 149 807 999 385 6 × 2 = 1 + 0.996 862 299 615 998 771 2;
  • 53) 0.996 862 299 615 998 771 2 × 2 = 1 + 0.993 724 599 231 997 542 4;
  • 54) 0.993 724 599 231 997 542 4 × 2 = 1 + 0.987 449 198 463 995 084 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:


0.000 000 000 742 147 675 95(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 95(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 95(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 95 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