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

Convert decimal -0.000 000 000 742 147 676 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 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 2| = 0.000 000 000 742 147 676 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 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 2 × 2 = 0 + 0.000 000 001 484 295 352 4;
  • 2) 0.000 000 001 484 295 352 4 × 2 = 0 + 0.000 000 002 968 590 704 8;
  • 3) 0.000 000 002 968 590 704 8 × 2 = 0 + 0.000 000 005 937 181 409 6;
  • 4) 0.000 000 005 937 181 409 6 × 2 = 0 + 0.000 000 011 874 362 819 2;
  • 5) 0.000 000 011 874 362 819 2 × 2 = 0 + 0.000 000 023 748 725 638 4;
  • 6) 0.000 000 023 748 725 638 4 × 2 = 0 + 0.000 000 047 497 451 276 8;
  • 7) 0.000 000 047 497 451 276 8 × 2 = 0 + 0.000 000 094 994 902 553 6;
  • 8) 0.000 000 094 994 902 553 6 × 2 = 0 + 0.000 000 189 989 805 107 2;
  • 9) 0.000 000 189 989 805 107 2 × 2 = 0 + 0.000 000 379 979 610 214 4;
  • 10) 0.000 000 379 979 610 214 4 × 2 = 0 + 0.000 000 759 959 220 428 8;
  • 11) 0.000 000 759 959 220 428 8 × 2 = 0 + 0.000 001 519 918 440 857 6;
  • 12) 0.000 001 519 918 440 857 6 × 2 = 0 + 0.000 003 039 836 881 715 2;
  • 13) 0.000 003 039 836 881 715 2 × 2 = 0 + 0.000 006 079 673 763 430 4;
  • 14) 0.000 006 079 673 763 430 4 × 2 = 0 + 0.000 012 159 347 526 860 8;
  • 15) 0.000 012 159 347 526 860 8 × 2 = 0 + 0.000 024 318 695 053 721 6;
  • 16) 0.000 024 318 695 053 721 6 × 2 = 0 + 0.000 048 637 390 107 443 2;
  • 17) 0.000 048 637 390 107 443 2 × 2 = 0 + 0.000 097 274 780 214 886 4;
  • 18) 0.000 097 274 780 214 886 4 × 2 = 0 + 0.000 194 549 560 429 772 8;
  • 19) 0.000 194 549 560 429 772 8 × 2 = 0 + 0.000 389 099 120 859 545 6;
  • 20) 0.000 389 099 120 859 545 6 × 2 = 0 + 0.000 778 198 241 719 091 2;
  • 21) 0.000 778 198 241 719 091 2 × 2 = 0 + 0.001 556 396 483 438 182 4;
  • 22) 0.001 556 396 483 438 182 4 × 2 = 0 + 0.003 112 792 966 876 364 8;
  • 23) 0.003 112 792 966 876 364 8 × 2 = 0 + 0.006 225 585 933 752 729 6;
  • 24) 0.006 225 585 933 752 729 6 × 2 = 0 + 0.012 451 171 867 505 459 2;
  • 25) 0.012 451 171 867 505 459 2 × 2 = 0 + 0.024 902 343 735 010 918 4;
  • 26) 0.024 902 343 735 010 918 4 × 2 = 0 + 0.049 804 687 470 021 836 8;
  • 27) 0.049 804 687 470 021 836 8 × 2 = 0 + 0.099 609 374 940 043 673 6;
  • 28) 0.099 609 374 940 043 673 6 × 2 = 0 + 0.199 218 749 880 087 347 2;
  • 29) 0.199 218 749 880 087 347 2 × 2 = 0 + 0.398 437 499 760 174 694 4;
  • 30) 0.398 437 499 760 174 694 4 × 2 = 0 + 0.796 874 999 520 349 388 8;
  • 31) 0.796 874 999 520 349 388 8 × 2 = 1 + 0.593 749 999 040 698 777 6;
  • 32) 0.593 749 999 040 698 777 6 × 2 = 1 + 0.187 499 998 081 397 555 2;
  • 33) 0.187 499 998 081 397 555 2 × 2 = 0 + 0.374 999 996 162 795 110 4;
  • 34) 0.374 999 996 162 795 110 4 × 2 = 0 + 0.749 999 992 325 590 220 8;
  • 35) 0.749 999 992 325 590 220 8 × 2 = 1 + 0.499 999 984 651 180 441 6;
  • 36) 0.499 999 984 651 180 441 6 × 2 = 0 + 0.999 999 969 302 360 883 2;
  • 37) 0.999 999 969 302 360 883 2 × 2 = 1 + 0.999 999 938 604 721 766 4;
  • 38) 0.999 999 938 604 721 766 4 × 2 = 1 + 0.999 999 877 209 443 532 8;
  • 39) 0.999 999 877 209 443 532 8 × 2 = 1 + 0.999 999 754 418 887 065 6;
  • 40) 0.999 999 754 418 887 065 6 × 2 = 1 + 0.999 999 508 837 774 131 2;
  • 41) 0.999 999 508 837 774 131 2 × 2 = 1 + 0.999 999 017 675 548 262 4;
  • 42) 0.999 999 017 675 548 262 4 × 2 = 1 + 0.999 998 035 351 096 524 8;
  • 43) 0.999 998 035 351 096 524 8 × 2 = 1 + 0.999 996 070 702 193 049 6;
  • 44) 0.999 996 070 702 193 049 6 × 2 = 1 + 0.999 992 141 404 386 099 2;
  • 45) 0.999 992 141 404 386 099 2 × 2 = 1 + 0.999 984 282 808 772 198 4;
  • 46) 0.999 984 282 808 772 198 4 × 2 = 1 + 0.999 968 565 617 544 396 8;
  • 47) 0.999 968 565 617 544 396 8 × 2 = 1 + 0.999 937 131 235 088 793 6;
  • 48) 0.999 937 131 235 088 793 6 × 2 = 1 + 0.999 874 262 470 177 587 2;
  • 49) 0.999 874 262 470 177 587 2 × 2 = 1 + 0.999 748 524 940 355 174 4;
  • 50) 0.999 748 524 940 355 174 4 × 2 = 1 + 0.999 497 049 880 710 348 8;
  • 51) 0.999 497 049 880 710 348 8 × 2 = 1 + 0.998 994 099 761 420 697 6;
  • 52) 0.998 994 099 761 420 697 6 × 2 = 1 + 0.997 988 199 522 841 395 2;
  • 53) 0.997 988 199 522 841 395 2 × 2 = 1 + 0.995 976 399 045 682 790 4;
  • 54) 0.995 976 399 045 682 790 4 × 2 = 1 + 0.991 952 798 091 365 580 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 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 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 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 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