-0.000 000 000 742 137 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 137 9(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
-0.000 000 000 742 137 9(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:

|-0.000 000 000 742 137 9| = 0.000 000 000 742 137 9


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: 0.000 000 000 742 137 9.

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) 0.000 000 000 742 137 9 × 2 = 0 + 0.000 000 001 484 275 8;
  • 2) 0.000 000 001 484 275 8 × 2 = 0 + 0.000 000 002 968 551 6;
  • 3) 0.000 000 002 968 551 6 × 2 = 0 + 0.000 000 005 937 103 2;
  • 4) 0.000 000 005 937 103 2 × 2 = 0 + 0.000 000 011 874 206 4;
  • 5) 0.000 000 011 874 206 4 × 2 = 0 + 0.000 000 023 748 412 8;
  • 6) 0.000 000 023 748 412 8 × 2 = 0 + 0.000 000 047 496 825 6;
  • 7) 0.000 000 047 496 825 6 × 2 = 0 + 0.000 000 094 993 651 2;
  • 8) 0.000 000 094 993 651 2 × 2 = 0 + 0.000 000 189 987 302 4;
  • 9) 0.000 000 189 987 302 4 × 2 = 0 + 0.000 000 379 974 604 8;
  • 10) 0.000 000 379 974 604 8 × 2 = 0 + 0.000 000 759 949 209 6;
  • 11) 0.000 000 759 949 209 6 × 2 = 0 + 0.000 001 519 898 419 2;
  • 12) 0.000 001 519 898 419 2 × 2 = 0 + 0.000 003 039 796 838 4;
  • 13) 0.000 003 039 796 838 4 × 2 = 0 + 0.000 006 079 593 676 8;
  • 14) 0.000 006 079 593 676 8 × 2 = 0 + 0.000 012 159 187 353 6;
  • 15) 0.000 012 159 187 353 6 × 2 = 0 + 0.000 024 318 374 707 2;
  • 16) 0.000 024 318 374 707 2 × 2 = 0 + 0.000 048 636 749 414 4;
  • 17) 0.000 048 636 749 414 4 × 2 = 0 + 0.000 097 273 498 828 8;
  • 18) 0.000 097 273 498 828 8 × 2 = 0 + 0.000 194 546 997 657 6;
  • 19) 0.000 194 546 997 657 6 × 2 = 0 + 0.000 389 093 995 315 2;
  • 20) 0.000 389 093 995 315 2 × 2 = 0 + 0.000 778 187 990 630 4;
  • 21) 0.000 778 187 990 630 4 × 2 = 0 + 0.001 556 375 981 260 8;
  • 22) 0.001 556 375 981 260 8 × 2 = 0 + 0.003 112 751 962 521 6;
  • 23) 0.003 112 751 962 521 6 × 2 = 0 + 0.006 225 503 925 043 2;
  • 24) 0.006 225 503 925 043 2 × 2 = 0 + 0.012 451 007 850 086 4;
  • 25) 0.012 451 007 850 086 4 × 2 = 0 + 0.024 902 015 700 172 8;
  • 26) 0.024 902 015 700 172 8 × 2 = 0 + 0.049 804 031 400 345 6;
  • 27) 0.049 804 031 400 345 6 × 2 = 0 + 0.099 608 062 800 691 2;
  • 28) 0.099 608 062 800 691 2 × 2 = 0 + 0.199 216 125 601 382 4;
  • 29) 0.199 216 125 601 382 4 × 2 = 0 + 0.398 432 251 202 764 8;
  • 30) 0.398 432 251 202 764 8 × 2 = 0 + 0.796 864 502 405 529 6;
  • 31) 0.796 864 502 405 529 6 × 2 = 1 + 0.593 729 004 811 059 2;
  • 32) 0.593 729 004 811 059 2 × 2 = 1 + 0.187 458 009 622 118 4;
  • 33) 0.187 458 009 622 118 4 × 2 = 0 + 0.374 916 019 244 236 8;
  • 34) 0.374 916 019 244 236 8 × 2 = 0 + 0.749 832 038 488 473 6;
  • 35) 0.749 832 038 488 473 6 × 2 = 1 + 0.499 664 076 976 947 2;
  • 36) 0.499 664 076 976 947 2 × 2 = 0 + 0.999 328 153 953 894 4;
  • 37) 0.999 328 153 953 894 4 × 2 = 1 + 0.998 656 307 907 788 8;
  • 38) 0.998 656 307 907 788 8 × 2 = 1 + 0.997 312 615 815 577 6;
  • 39) 0.997 312 615 815 577 6 × 2 = 1 + 0.994 625 231 631 155 2;
  • 40) 0.994 625 231 631 155 2 × 2 = 1 + 0.989 250 463 262 310 4;
  • 41) 0.989 250 463 262 310 4 × 2 = 1 + 0.978 500 926 524 620 8;
  • 42) 0.978 500 926 524 620 8 × 2 = 1 + 0.957 001 853 049 241 6;
  • 43) 0.957 001 853 049 241 6 × 2 = 1 + 0.914 003 706 098 483 2;
  • 44) 0.914 003 706 098 483 2 × 2 = 1 + 0.828 007 412 196 966 4;
  • 45) 0.828 007 412 196 966 4 × 2 = 1 + 0.656 014 824 393 932 8;
  • 46) 0.656 014 824 393 932 8 × 2 = 1 + 0.312 029 648 787 865 6;
  • 47) 0.312 029 648 787 865 6 × 2 = 0 + 0.624 059 297 575 731 2;
  • 48) 0.624 059 297 575 731 2 × 2 = 1 + 0.248 118 595 151 462 4;
  • 49) 0.248 118 595 151 462 4 × 2 = 0 + 0.496 237 190 302 924 8;
  • 50) 0.496 237 190 302 924 8 × 2 = 0 + 0.992 474 380 605 849 6;
  • 51) 0.992 474 380 605 849 6 × 2 = 1 + 0.984 948 761 211 699 2;
  • 52) 0.984 948 761 211 699 2 × 2 = 1 + 0.969 897 522 423 398 4;
  • 53) 0.969 897 522 423 398 4 × 2 = 1 + 0.939 795 044 846 796 8;
  • 54) 0.939 795 044 846 796 8 × 2 = 1 + 0.879 590 089 693 593 6;

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:


0.000 000 000 742 137 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0011 11(2)

6. Positive number before normalization:

0.000 000 000 742 137 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0011 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:


0.000 000 000 742 137 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0011 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1101 0011 11(2) × 20 =

1.1001 0111 1111 1110 1001 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 1110 1001 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 0100 1111 =

100 1011 1111 1111 0100 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 0100 1111


Decimal number -0.000 000 000 742 137 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 0100 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