-0.000 000 000 742 147 675 81 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 81(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 81(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 81| = 0.000 000 000 742 147 675 81


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 81.

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 81 × 2 = 0 + 0.000 000 001 484 295 351 62;
  • 2) 0.000 000 001 484 295 351 62 × 2 = 0 + 0.000 000 002 968 590 703 24;
  • 3) 0.000 000 002 968 590 703 24 × 2 = 0 + 0.000 000 005 937 181 406 48;
  • 4) 0.000 000 005 937 181 406 48 × 2 = 0 + 0.000 000 011 874 362 812 96;
  • 5) 0.000 000 011 874 362 812 96 × 2 = 0 + 0.000 000 023 748 725 625 92;
  • 6) 0.000 000 023 748 725 625 92 × 2 = 0 + 0.000 000 047 497 451 251 84;
  • 7) 0.000 000 047 497 451 251 84 × 2 = 0 + 0.000 000 094 994 902 503 68;
  • 8) 0.000 000 094 994 902 503 68 × 2 = 0 + 0.000 000 189 989 805 007 36;
  • 9) 0.000 000 189 989 805 007 36 × 2 = 0 + 0.000 000 379 979 610 014 72;
  • 10) 0.000 000 379 979 610 014 72 × 2 = 0 + 0.000 000 759 959 220 029 44;
  • 11) 0.000 000 759 959 220 029 44 × 2 = 0 + 0.000 001 519 918 440 058 88;
  • 12) 0.000 001 519 918 440 058 88 × 2 = 0 + 0.000 003 039 836 880 117 76;
  • 13) 0.000 003 039 836 880 117 76 × 2 = 0 + 0.000 006 079 673 760 235 52;
  • 14) 0.000 006 079 673 760 235 52 × 2 = 0 + 0.000 012 159 347 520 471 04;
  • 15) 0.000 012 159 347 520 471 04 × 2 = 0 + 0.000 024 318 695 040 942 08;
  • 16) 0.000 024 318 695 040 942 08 × 2 = 0 + 0.000 048 637 390 081 884 16;
  • 17) 0.000 048 637 390 081 884 16 × 2 = 0 + 0.000 097 274 780 163 768 32;
  • 18) 0.000 097 274 780 163 768 32 × 2 = 0 + 0.000 194 549 560 327 536 64;
  • 19) 0.000 194 549 560 327 536 64 × 2 = 0 + 0.000 389 099 120 655 073 28;
  • 20) 0.000 389 099 120 655 073 28 × 2 = 0 + 0.000 778 198 241 310 146 56;
  • 21) 0.000 778 198 241 310 146 56 × 2 = 0 + 0.001 556 396 482 620 293 12;
  • 22) 0.001 556 396 482 620 293 12 × 2 = 0 + 0.003 112 792 965 240 586 24;
  • 23) 0.003 112 792 965 240 586 24 × 2 = 0 + 0.006 225 585 930 481 172 48;
  • 24) 0.006 225 585 930 481 172 48 × 2 = 0 + 0.012 451 171 860 962 344 96;
  • 25) 0.012 451 171 860 962 344 96 × 2 = 0 + 0.024 902 343 721 924 689 92;
  • 26) 0.024 902 343 721 924 689 92 × 2 = 0 + 0.049 804 687 443 849 379 84;
  • 27) 0.049 804 687 443 849 379 84 × 2 = 0 + 0.099 609 374 887 698 759 68;
  • 28) 0.099 609 374 887 698 759 68 × 2 = 0 + 0.199 218 749 775 397 519 36;
  • 29) 0.199 218 749 775 397 519 36 × 2 = 0 + 0.398 437 499 550 795 038 72;
  • 30) 0.398 437 499 550 795 038 72 × 2 = 0 + 0.796 874 999 101 590 077 44;
  • 31) 0.796 874 999 101 590 077 44 × 2 = 1 + 0.593 749 998 203 180 154 88;
  • 32) 0.593 749 998 203 180 154 88 × 2 = 1 + 0.187 499 996 406 360 309 76;
  • 33) 0.187 499 996 406 360 309 76 × 2 = 0 + 0.374 999 992 812 720 619 52;
  • 34) 0.374 999 992 812 720 619 52 × 2 = 0 + 0.749 999 985 625 441 239 04;
  • 35) 0.749 999 985 625 441 239 04 × 2 = 1 + 0.499 999 971 250 882 478 08;
  • 36) 0.499 999 971 250 882 478 08 × 2 = 0 + 0.999 999 942 501 764 956 16;
  • 37) 0.999 999 942 501 764 956 16 × 2 = 1 + 0.999 999 885 003 529 912 32;
  • 38) 0.999 999 885 003 529 912 32 × 2 = 1 + 0.999 999 770 007 059 824 64;
  • 39) 0.999 999 770 007 059 824 64 × 2 = 1 + 0.999 999 540 014 119 649 28;
  • 40) 0.999 999 540 014 119 649 28 × 2 = 1 + 0.999 999 080 028 239 298 56;
  • 41) 0.999 999 080 028 239 298 56 × 2 = 1 + 0.999 998 160 056 478 597 12;
  • 42) 0.999 998 160 056 478 597 12 × 2 = 1 + 0.999 996 320 112 957 194 24;
  • 43) 0.999 996 320 112 957 194 24 × 2 = 1 + 0.999 992 640 225 914 388 48;
  • 44) 0.999 992 640 225 914 388 48 × 2 = 1 + 0.999 985 280 451 828 776 96;
  • 45) 0.999 985 280 451 828 776 96 × 2 = 1 + 0.999 970 560 903 657 553 92;
  • 46) 0.999 970 560 903 657 553 92 × 2 = 1 + 0.999 941 121 807 315 107 84;
  • 47) 0.999 941 121 807 315 107 84 × 2 = 1 + 0.999 882 243 614 630 215 68;
  • 48) 0.999 882 243 614 630 215 68 × 2 = 1 + 0.999 764 487 229 260 431 36;
  • 49) 0.999 764 487 229 260 431 36 × 2 = 1 + 0.999 528 974 458 520 862 72;
  • 50) 0.999 528 974 458 520 862 72 × 2 = 1 + 0.999 057 948 917 041 725 44;
  • 51) 0.999 057 948 917 041 725 44 × 2 = 1 + 0.998 115 897 834 083 450 88;
  • 52) 0.998 115 897 834 083 450 88 × 2 = 1 + 0.996 231 795 668 166 901 76;
  • 53) 0.996 231 795 668 166 901 76 × 2 = 1 + 0.992 463 591 336 333 803 52;
  • 54) 0.992 463 591 336 333 803 52 × 2 = 1 + 0.984 927 182 672 667 607 04;

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 81(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 81(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 81(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 81 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