-0.000 000 000 742 147 676 275 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 275(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
-0.000 000 000 742 147 676 275(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:

|-0.000 000 000 742 147 676 275| = 0.000 000 000 742 147 676 275


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: 0.000 000 000 742 147 676 275.

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) 0.000 000 000 742 147 676 275 × 2 = 0 + 0.000 000 001 484 295 352 55;
  • 2) 0.000 000 001 484 295 352 55 × 2 = 0 + 0.000 000 002 968 590 705 1;
  • 3) 0.000 000 002 968 590 705 1 × 2 = 0 + 0.000 000 005 937 181 410 2;
  • 4) 0.000 000 005 937 181 410 2 × 2 = 0 + 0.000 000 011 874 362 820 4;
  • 5) 0.000 000 011 874 362 820 4 × 2 = 0 + 0.000 000 023 748 725 640 8;
  • 6) 0.000 000 023 748 725 640 8 × 2 = 0 + 0.000 000 047 497 451 281 6;
  • 7) 0.000 000 047 497 451 281 6 × 2 = 0 + 0.000 000 094 994 902 563 2;
  • 8) 0.000 000 094 994 902 563 2 × 2 = 0 + 0.000 000 189 989 805 126 4;
  • 9) 0.000 000 189 989 805 126 4 × 2 = 0 + 0.000 000 379 979 610 252 8;
  • 10) 0.000 000 379 979 610 252 8 × 2 = 0 + 0.000 000 759 959 220 505 6;
  • 11) 0.000 000 759 959 220 505 6 × 2 = 0 + 0.000 001 519 918 441 011 2;
  • 12) 0.000 001 519 918 441 011 2 × 2 = 0 + 0.000 003 039 836 882 022 4;
  • 13) 0.000 003 039 836 882 022 4 × 2 = 0 + 0.000 006 079 673 764 044 8;
  • 14) 0.000 006 079 673 764 044 8 × 2 = 0 + 0.000 012 159 347 528 089 6;
  • 15) 0.000 012 159 347 528 089 6 × 2 = 0 + 0.000 024 318 695 056 179 2;
  • 16) 0.000 024 318 695 056 179 2 × 2 = 0 + 0.000 048 637 390 112 358 4;
  • 17) 0.000 048 637 390 112 358 4 × 2 = 0 + 0.000 097 274 780 224 716 8;
  • 18) 0.000 097 274 780 224 716 8 × 2 = 0 + 0.000 194 549 560 449 433 6;
  • 19) 0.000 194 549 560 449 433 6 × 2 = 0 + 0.000 389 099 120 898 867 2;
  • 20) 0.000 389 099 120 898 867 2 × 2 = 0 + 0.000 778 198 241 797 734 4;
  • 21) 0.000 778 198 241 797 734 4 × 2 = 0 + 0.001 556 396 483 595 468 8;
  • 22) 0.001 556 396 483 595 468 8 × 2 = 0 + 0.003 112 792 967 190 937 6;
  • 23) 0.003 112 792 967 190 937 6 × 2 = 0 + 0.006 225 585 934 381 875 2;
  • 24) 0.006 225 585 934 381 875 2 × 2 = 0 + 0.012 451 171 868 763 750 4;
  • 25) 0.012 451 171 868 763 750 4 × 2 = 0 + 0.024 902 343 737 527 500 8;
  • 26) 0.024 902 343 737 527 500 8 × 2 = 0 + 0.049 804 687 475 055 001 6;
  • 27) 0.049 804 687 475 055 001 6 × 2 = 0 + 0.099 609 374 950 110 003 2;
  • 28) 0.099 609 374 950 110 003 2 × 2 = 0 + 0.199 218 749 900 220 006 4;
  • 29) 0.199 218 749 900 220 006 4 × 2 = 0 + 0.398 437 499 800 440 012 8;
  • 30) 0.398 437 499 800 440 012 8 × 2 = 0 + 0.796 874 999 600 880 025 6;
  • 31) 0.796 874 999 600 880 025 6 × 2 = 1 + 0.593 749 999 201 760 051 2;
  • 32) 0.593 749 999 201 760 051 2 × 2 = 1 + 0.187 499 998 403 520 102 4;
  • 33) 0.187 499 998 403 520 102 4 × 2 = 0 + 0.374 999 996 807 040 204 8;
  • 34) 0.374 999 996 807 040 204 8 × 2 = 0 + 0.749 999 993 614 080 409 6;
  • 35) 0.749 999 993 614 080 409 6 × 2 = 1 + 0.499 999 987 228 160 819 2;
  • 36) 0.499 999 987 228 160 819 2 × 2 = 0 + 0.999 999 974 456 321 638 4;
  • 37) 0.999 999 974 456 321 638 4 × 2 = 1 + 0.999 999 948 912 643 276 8;
  • 38) 0.999 999 948 912 643 276 8 × 2 = 1 + 0.999 999 897 825 286 553 6;
  • 39) 0.999 999 897 825 286 553 6 × 2 = 1 + 0.999 999 795 650 573 107 2;
  • 40) 0.999 999 795 650 573 107 2 × 2 = 1 + 0.999 999 591 301 146 214 4;
  • 41) 0.999 999 591 301 146 214 4 × 2 = 1 + 0.999 999 182 602 292 428 8;
  • 42) 0.999 999 182 602 292 428 8 × 2 = 1 + 0.999 998 365 204 584 857 6;
  • 43) 0.999 998 365 204 584 857 6 × 2 = 1 + 0.999 996 730 409 169 715 2;
  • 44) 0.999 996 730 409 169 715 2 × 2 = 1 + 0.999 993 460 818 339 430 4;
  • 45) 0.999 993 460 818 339 430 4 × 2 = 1 + 0.999 986 921 636 678 860 8;
  • 46) 0.999 986 921 636 678 860 8 × 2 = 1 + 0.999 973 843 273 357 721 6;
  • 47) 0.999 973 843 273 357 721 6 × 2 = 1 + 0.999 947 686 546 715 443 2;
  • 48) 0.999 947 686 546 715 443 2 × 2 = 1 + 0.999 895 373 093 430 886 4;
  • 49) 0.999 895 373 093 430 886 4 × 2 = 1 + 0.999 790 746 186 861 772 8;
  • 50) 0.999 790 746 186 861 772 8 × 2 = 1 + 0.999 581 492 373 723 545 6;
  • 51) 0.999 581 492 373 723 545 6 × 2 = 1 + 0.999 162 984 747 447 091 2;
  • 52) 0.999 162 984 747 447 091 2 × 2 = 1 + 0.998 325 969 494 894 182 4;
  • 53) 0.998 325 969 494 894 182 4 × 2 = 1 + 0.996 651 938 989 788 364 8;
  • 54) 0.996 651 938 989 788 364 8 × 2 = 1 + 0.993 303 877 979 576 729 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:


0.000 000 000 742 147 676 275(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 275(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:


0.000 000 000 742 147 676 275(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 -0.000 000 000 742 147 676 275 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