-0.000 000 000 742 147 676 632 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 632 2(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
-0.000 000 000 742 147 676 632 2(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:

|-0.000 000 000 742 147 676 632 2| = 0.000 000 000 742 147 676 632 2


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: 0.000 000 000 742 147 676 632 2.

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) 0.000 000 000 742 147 676 632 2 × 2 = 0 + 0.000 000 001 484 295 353 264 4;
  • 2) 0.000 000 001 484 295 353 264 4 × 2 = 0 + 0.000 000 002 968 590 706 528 8;
  • 3) 0.000 000 002 968 590 706 528 8 × 2 = 0 + 0.000 000 005 937 181 413 057 6;
  • 4) 0.000 000 005 937 181 413 057 6 × 2 = 0 + 0.000 000 011 874 362 826 115 2;
  • 5) 0.000 000 011 874 362 826 115 2 × 2 = 0 + 0.000 000 023 748 725 652 230 4;
  • 6) 0.000 000 023 748 725 652 230 4 × 2 = 0 + 0.000 000 047 497 451 304 460 8;
  • 7) 0.000 000 047 497 451 304 460 8 × 2 = 0 + 0.000 000 094 994 902 608 921 6;
  • 8) 0.000 000 094 994 902 608 921 6 × 2 = 0 + 0.000 000 189 989 805 217 843 2;
  • 9) 0.000 000 189 989 805 217 843 2 × 2 = 0 + 0.000 000 379 979 610 435 686 4;
  • 10) 0.000 000 379 979 610 435 686 4 × 2 = 0 + 0.000 000 759 959 220 871 372 8;
  • 11) 0.000 000 759 959 220 871 372 8 × 2 = 0 + 0.000 001 519 918 441 742 745 6;
  • 12) 0.000 001 519 918 441 742 745 6 × 2 = 0 + 0.000 003 039 836 883 485 491 2;
  • 13) 0.000 003 039 836 883 485 491 2 × 2 = 0 + 0.000 006 079 673 766 970 982 4;
  • 14) 0.000 006 079 673 766 970 982 4 × 2 = 0 + 0.000 012 159 347 533 941 964 8;
  • 15) 0.000 012 159 347 533 941 964 8 × 2 = 0 + 0.000 024 318 695 067 883 929 6;
  • 16) 0.000 024 318 695 067 883 929 6 × 2 = 0 + 0.000 048 637 390 135 767 859 2;
  • 17) 0.000 048 637 390 135 767 859 2 × 2 = 0 + 0.000 097 274 780 271 535 718 4;
  • 18) 0.000 097 274 780 271 535 718 4 × 2 = 0 + 0.000 194 549 560 543 071 436 8;
  • 19) 0.000 194 549 560 543 071 436 8 × 2 = 0 + 0.000 389 099 121 086 142 873 6;
  • 20) 0.000 389 099 121 086 142 873 6 × 2 = 0 + 0.000 778 198 242 172 285 747 2;
  • 21) 0.000 778 198 242 172 285 747 2 × 2 = 0 + 0.001 556 396 484 344 571 494 4;
  • 22) 0.001 556 396 484 344 571 494 4 × 2 = 0 + 0.003 112 792 968 689 142 988 8;
  • 23) 0.003 112 792 968 689 142 988 8 × 2 = 0 + 0.006 225 585 937 378 285 977 6;
  • 24) 0.006 225 585 937 378 285 977 6 × 2 = 0 + 0.012 451 171 874 756 571 955 2;
  • 25) 0.012 451 171 874 756 571 955 2 × 2 = 0 + 0.024 902 343 749 513 143 910 4;
  • 26) 0.024 902 343 749 513 143 910 4 × 2 = 0 + 0.049 804 687 499 026 287 820 8;
  • 27) 0.049 804 687 499 026 287 820 8 × 2 = 0 + 0.099 609 374 998 052 575 641 6;
  • 28) 0.099 609 374 998 052 575 641 6 × 2 = 0 + 0.199 218 749 996 105 151 283 2;
  • 29) 0.199 218 749 996 105 151 283 2 × 2 = 0 + 0.398 437 499 992 210 302 566 4;
  • 30) 0.398 437 499 992 210 302 566 4 × 2 = 0 + 0.796 874 999 984 420 605 132 8;
  • 31) 0.796 874 999 984 420 605 132 8 × 2 = 1 + 0.593 749 999 968 841 210 265 6;
  • 32) 0.593 749 999 968 841 210 265 6 × 2 = 1 + 0.187 499 999 937 682 420 531 2;
  • 33) 0.187 499 999 937 682 420 531 2 × 2 = 0 + 0.374 999 999 875 364 841 062 4;
  • 34) 0.374 999 999 875 364 841 062 4 × 2 = 0 + 0.749 999 999 750 729 682 124 8;
  • 35) 0.749 999 999 750 729 682 124 8 × 2 = 1 + 0.499 999 999 501 459 364 249 6;
  • 36) 0.499 999 999 501 459 364 249 6 × 2 = 0 + 0.999 999 999 002 918 728 499 2;
  • 37) 0.999 999 999 002 918 728 499 2 × 2 = 1 + 0.999 999 998 005 837 456 998 4;
  • 38) 0.999 999 998 005 837 456 998 4 × 2 = 1 + 0.999 999 996 011 674 913 996 8;
  • 39) 0.999 999 996 011 674 913 996 8 × 2 = 1 + 0.999 999 992 023 349 827 993 6;
  • 40) 0.999 999 992 023 349 827 993 6 × 2 = 1 + 0.999 999 984 046 699 655 987 2;
  • 41) 0.999 999 984 046 699 655 987 2 × 2 = 1 + 0.999 999 968 093 399 311 974 4;
  • 42) 0.999 999 968 093 399 311 974 4 × 2 = 1 + 0.999 999 936 186 798 623 948 8;
  • 43) 0.999 999 936 186 798 623 948 8 × 2 = 1 + 0.999 999 872 373 597 247 897 6;
  • 44) 0.999 999 872 373 597 247 897 6 × 2 = 1 + 0.999 999 744 747 194 495 795 2;
  • 45) 0.999 999 744 747 194 495 795 2 × 2 = 1 + 0.999 999 489 494 388 991 590 4;
  • 46) 0.999 999 489 494 388 991 590 4 × 2 = 1 + 0.999 998 978 988 777 983 180 8;
  • 47) 0.999 998 978 988 777 983 180 8 × 2 = 1 + 0.999 997 957 977 555 966 361 6;
  • 48) 0.999 997 957 977 555 966 361 6 × 2 = 1 + 0.999 995 915 955 111 932 723 2;
  • 49) 0.999 995 915 955 111 932 723 2 × 2 = 1 + 0.999 991 831 910 223 865 446 4;
  • 50) 0.999 991 831 910 223 865 446 4 × 2 = 1 + 0.999 983 663 820 447 730 892 8;
  • 51) 0.999 983 663 820 447 730 892 8 × 2 = 1 + 0.999 967 327 640 895 461 785 6;
  • 52) 0.999 967 327 640 895 461 785 6 × 2 = 1 + 0.999 934 655 281 790 923 571 2;
  • 53) 0.999 934 655 281 790 923 571 2 × 2 = 1 + 0.999 869 310 563 581 847 142 4;
  • 54) 0.999 869 310 563 581 847 142 4 × 2 = 1 + 0.999 738 621 127 163 694 284 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:


0.000 000 000 742 147 676 632 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 632 2(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:


0.000 000 000 742 147 676 632 2(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 -0.000 000 000 742 147 676 632 2 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