-0.000 000 000 742 147 676 656 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 656 7(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
-0.000 000 000 742 147 676 656 7(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:

|-0.000 000 000 742 147 676 656 7| = 0.000 000 000 742 147 676 656 7


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: 0.000 000 000 742 147 676 656 7.

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) 0.000 000 000 742 147 676 656 7 × 2 = 0 + 0.000 000 001 484 295 353 313 4;
  • 2) 0.000 000 001 484 295 353 313 4 × 2 = 0 + 0.000 000 002 968 590 706 626 8;
  • 3) 0.000 000 002 968 590 706 626 8 × 2 = 0 + 0.000 000 005 937 181 413 253 6;
  • 4) 0.000 000 005 937 181 413 253 6 × 2 = 0 + 0.000 000 011 874 362 826 507 2;
  • 5) 0.000 000 011 874 362 826 507 2 × 2 = 0 + 0.000 000 023 748 725 653 014 4;
  • 6) 0.000 000 023 748 725 653 014 4 × 2 = 0 + 0.000 000 047 497 451 306 028 8;
  • 7) 0.000 000 047 497 451 306 028 8 × 2 = 0 + 0.000 000 094 994 902 612 057 6;
  • 8) 0.000 000 094 994 902 612 057 6 × 2 = 0 + 0.000 000 189 989 805 224 115 2;
  • 9) 0.000 000 189 989 805 224 115 2 × 2 = 0 + 0.000 000 379 979 610 448 230 4;
  • 10) 0.000 000 379 979 610 448 230 4 × 2 = 0 + 0.000 000 759 959 220 896 460 8;
  • 11) 0.000 000 759 959 220 896 460 8 × 2 = 0 + 0.000 001 519 918 441 792 921 6;
  • 12) 0.000 001 519 918 441 792 921 6 × 2 = 0 + 0.000 003 039 836 883 585 843 2;
  • 13) 0.000 003 039 836 883 585 843 2 × 2 = 0 + 0.000 006 079 673 767 171 686 4;
  • 14) 0.000 006 079 673 767 171 686 4 × 2 = 0 + 0.000 012 159 347 534 343 372 8;
  • 15) 0.000 012 159 347 534 343 372 8 × 2 = 0 + 0.000 024 318 695 068 686 745 6;
  • 16) 0.000 024 318 695 068 686 745 6 × 2 = 0 + 0.000 048 637 390 137 373 491 2;
  • 17) 0.000 048 637 390 137 373 491 2 × 2 = 0 + 0.000 097 274 780 274 746 982 4;
  • 18) 0.000 097 274 780 274 746 982 4 × 2 = 0 + 0.000 194 549 560 549 493 964 8;
  • 19) 0.000 194 549 560 549 493 964 8 × 2 = 0 + 0.000 389 099 121 098 987 929 6;
  • 20) 0.000 389 099 121 098 987 929 6 × 2 = 0 + 0.000 778 198 242 197 975 859 2;
  • 21) 0.000 778 198 242 197 975 859 2 × 2 = 0 + 0.001 556 396 484 395 951 718 4;
  • 22) 0.001 556 396 484 395 951 718 4 × 2 = 0 + 0.003 112 792 968 791 903 436 8;
  • 23) 0.003 112 792 968 791 903 436 8 × 2 = 0 + 0.006 225 585 937 583 806 873 6;
  • 24) 0.006 225 585 937 583 806 873 6 × 2 = 0 + 0.012 451 171 875 167 613 747 2;
  • 25) 0.012 451 171 875 167 613 747 2 × 2 = 0 + 0.024 902 343 750 335 227 494 4;
  • 26) 0.024 902 343 750 335 227 494 4 × 2 = 0 + 0.049 804 687 500 670 454 988 8;
  • 27) 0.049 804 687 500 670 454 988 8 × 2 = 0 + 0.099 609 375 001 340 909 977 6;
  • 28) 0.099 609 375 001 340 909 977 6 × 2 = 0 + 0.199 218 750 002 681 819 955 2;
  • 29) 0.199 218 750 002 681 819 955 2 × 2 = 0 + 0.398 437 500 005 363 639 910 4;
  • 30) 0.398 437 500 005 363 639 910 4 × 2 = 0 + 0.796 875 000 010 727 279 820 8;
  • 31) 0.796 875 000 010 727 279 820 8 × 2 = 1 + 0.593 750 000 021 454 559 641 6;
  • 32) 0.593 750 000 021 454 559 641 6 × 2 = 1 + 0.187 500 000 042 909 119 283 2;
  • 33) 0.187 500 000 042 909 119 283 2 × 2 = 0 + 0.375 000 000 085 818 238 566 4;
  • 34) 0.375 000 000 085 818 238 566 4 × 2 = 0 + 0.750 000 000 171 636 477 132 8;
  • 35) 0.750 000 000 171 636 477 132 8 × 2 = 1 + 0.500 000 000 343 272 954 265 6;
  • 36) 0.500 000 000 343 272 954 265 6 × 2 = 1 + 0.000 000 000 686 545 908 531 2;
  • 37) 0.000 000 000 686 545 908 531 2 × 2 = 0 + 0.000 000 001 373 091 817 062 4;
  • 38) 0.000 000 001 373 091 817 062 4 × 2 = 0 + 0.000 000 002 746 183 634 124 8;
  • 39) 0.000 000 002 746 183 634 124 8 × 2 = 0 + 0.000 000 005 492 367 268 249 6;
  • 40) 0.000 000 005 492 367 268 249 6 × 2 = 0 + 0.000 000 010 984 734 536 499 2;
  • 41) 0.000 000 010 984 734 536 499 2 × 2 = 0 + 0.000 000 021 969 469 072 998 4;
  • 42) 0.000 000 021 969 469 072 998 4 × 2 = 0 + 0.000 000 043 938 938 145 996 8;
  • 43) 0.000 000 043 938 938 145 996 8 × 2 = 0 + 0.000 000 087 877 876 291 993 6;
  • 44) 0.000 000 087 877 876 291 993 6 × 2 = 0 + 0.000 000 175 755 752 583 987 2;
  • 45) 0.000 000 175 755 752 583 987 2 × 2 = 0 + 0.000 000 351 511 505 167 974 4;
  • 46) 0.000 000 351 511 505 167 974 4 × 2 = 0 + 0.000 000 703 023 010 335 948 8;
  • 47) 0.000 000 703 023 010 335 948 8 × 2 = 0 + 0.000 001 406 046 020 671 897 6;
  • 48) 0.000 001 406 046 020 671 897 6 × 2 = 0 + 0.000 002 812 092 041 343 795 2;
  • 49) 0.000 002 812 092 041 343 795 2 × 2 = 0 + 0.000 005 624 184 082 687 590 4;
  • 50) 0.000 005 624 184 082 687 590 4 × 2 = 0 + 0.000 011 248 368 165 375 180 8;
  • 51) 0.000 011 248 368 165 375 180 8 × 2 = 0 + 0.000 022 496 736 330 750 361 6;
  • 52) 0.000 022 496 736 330 750 361 6 × 2 = 0 + 0.000 044 993 472 661 500 723 2;
  • 53) 0.000 044 993 472 661 500 723 2 × 2 = 0 + 0.000 089 986 945 323 001 446 4;
  • 54) 0.000 089 986 945 323 001 446 4 × 2 = 0 + 0.000 179 973 890 646 002 892 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:


0.000 000 000 742 147 676 656 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 656 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 656 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 656 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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