-0.000 000 000 742 147 675 92 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 92(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 92(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 92| = 0.000 000 000 742 147 675 92


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 92.

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 92 × 2 = 0 + 0.000 000 001 484 295 351 84;
  • 2) 0.000 000 001 484 295 351 84 × 2 = 0 + 0.000 000 002 968 590 703 68;
  • 3) 0.000 000 002 968 590 703 68 × 2 = 0 + 0.000 000 005 937 181 407 36;
  • 4) 0.000 000 005 937 181 407 36 × 2 = 0 + 0.000 000 011 874 362 814 72;
  • 5) 0.000 000 011 874 362 814 72 × 2 = 0 + 0.000 000 023 748 725 629 44;
  • 6) 0.000 000 023 748 725 629 44 × 2 = 0 + 0.000 000 047 497 451 258 88;
  • 7) 0.000 000 047 497 451 258 88 × 2 = 0 + 0.000 000 094 994 902 517 76;
  • 8) 0.000 000 094 994 902 517 76 × 2 = 0 + 0.000 000 189 989 805 035 52;
  • 9) 0.000 000 189 989 805 035 52 × 2 = 0 + 0.000 000 379 979 610 071 04;
  • 10) 0.000 000 379 979 610 071 04 × 2 = 0 + 0.000 000 759 959 220 142 08;
  • 11) 0.000 000 759 959 220 142 08 × 2 = 0 + 0.000 001 519 918 440 284 16;
  • 12) 0.000 001 519 918 440 284 16 × 2 = 0 + 0.000 003 039 836 880 568 32;
  • 13) 0.000 003 039 836 880 568 32 × 2 = 0 + 0.000 006 079 673 761 136 64;
  • 14) 0.000 006 079 673 761 136 64 × 2 = 0 + 0.000 012 159 347 522 273 28;
  • 15) 0.000 012 159 347 522 273 28 × 2 = 0 + 0.000 024 318 695 044 546 56;
  • 16) 0.000 024 318 695 044 546 56 × 2 = 0 + 0.000 048 637 390 089 093 12;
  • 17) 0.000 048 637 390 089 093 12 × 2 = 0 + 0.000 097 274 780 178 186 24;
  • 18) 0.000 097 274 780 178 186 24 × 2 = 0 + 0.000 194 549 560 356 372 48;
  • 19) 0.000 194 549 560 356 372 48 × 2 = 0 + 0.000 389 099 120 712 744 96;
  • 20) 0.000 389 099 120 712 744 96 × 2 = 0 + 0.000 778 198 241 425 489 92;
  • 21) 0.000 778 198 241 425 489 92 × 2 = 0 + 0.001 556 396 482 850 979 84;
  • 22) 0.001 556 396 482 850 979 84 × 2 = 0 + 0.003 112 792 965 701 959 68;
  • 23) 0.003 112 792 965 701 959 68 × 2 = 0 + 0.006 225 585 931 403 919 36;
  • 24) 0.006 225 585 931 403 919 36 × 2 = 0 + 0.012 451 171 862 807 838 72;
  • 25) 0.012 451 171 862 807 838 72 × 2 = 0 + 0.024 902 343 725 615 677 44;
  • 26) 0.024 902 343 725 615 677 44 × 2 = 0 + 0.049 804 687 451 231 354 88;
  • 27) 0.049 804 687 451 231 354 88 × 2 = 0 + 0.099 609 374 902 462 709 76;
  • 28) 0.099 609 374 902 462 709 76 × 2 = 0 + 0.199 218 749 804 925 419 52;
  • 29) 0.199 218 749 804 925 419 52 × 2 = 0 + 0.398 437 499 609 850 839 04;
  • 30) 0.398 437 499 609 850 839 04 × 2 = 0 + 0.796 874 999 219 701 678 08;
  • 31) 0.796 874 999 219 701 678 08 × 2 = 1 + 0.593 749 998 439 403 356 16;
  • 32) 0.593 749 998 439 403 356 16 × 2 = 1 + 0.187 499 996 878 806 712 32;
  • 33) 0.187 499 996 878 806 712 32 × 2 = 0 + 0.374 999 993 757 613 424 64;
  • 34) 0.374 999 993 757 613 424 64 × 2 = 0 + 0.749 999 987 515 226 849 28;
  • 35) 0.749 999 987 515 226 849 28 × 2 = 1 + 0.499 999 975 030 453 698 56;
  • 36) 0.499 999 975 030 453 698 56 × 2 = 0 + 0.999 999 950 060 907 397 12;
  • 37) 0.999 999 950 060 907 397 12 × 2 = 1 + 0.999 999 900 121 814 794 24;
  • 38) 0.999 999 900 121 814 794 24 × 2 = 1 + 0.999 999 800 243 629 588 48;
  • 39) 0.999 999 800 243 629 588 48 × 2 = 1 + 0.999 999 600 487 259 176 96;
  • 40) 0.999 999 600 487 259 176 96 × 2 = 1 + 0.999 999 200 974 518 353 92;
  • 41) 0.999 999 200 974 518 353 92 × 2 = 1 + 0.999 998 401 949 036 707 84;
  • 42) 0.999 998 401 949 036 707 84 × 2 = 1 + 0.999 996 803 898 073 415 68;
  • 43) 0.999 996 803 898 073 415 68 × 2 = 1 + 0.999 993 607 796 146 831 36;
  • 44) 0.999 993 607 796 146 831 36 × 2 = 1 + 0.999 987 215 592 293 662 72;
  • 45) 0.999 987 215 592 293 662 72 × 2 = 1 + 0.999 974 431 184 587 325 44;
  • 46) 0.999 974 431 184 587 325 44 × 2 = 1 + 0.999 948 862 369 174 650 88;
  • 47) 0.999 948 862 369 174 650 88 × 2 = 1 + 0.999 897 724 738 349 301 76;
  • 48) 0.999 897 724 738 349 301 76 × 2 = 1 + 0.999 795 449 476 698 603 52;
  • 49) 0.999 795 449 476 698 603 52 × 2 = 1 + 0.999 590 898 953 397 207 04;
  • 50) 0.999 590 898 953 397 207 04 × 2 = 1 + 0.999 181 797 906 794 414 08;
  • 51) 0.999 181 797 906 794 414 08 × 2 = 1 + 0.998 363 595 813 588 828 16;
  • 52) 0.998 363 595 813 588 828 16 × 2 = 1 + 0.996 727 191 627 177 656 32;
  • 53) 0.996 727 191 627 177 656 32 × 2 = 1 + 0.993 454 383 254 355 312 64;
  • 54) 0.993 454 383 254 355 312 64 × 2 = 1 + 0.986 908 766 508 710 625 28;

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 92(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 92(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 92(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 92 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