-0.000 000 000 742 147 676 617 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 617 9(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
-0.000 000 000 742 147 676 617 9(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:

|-0.000 000 000 742 147 676 617 9| = 0.000 000 000 742 147 676 617 9


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: 0.000 000 000 742 147 676 617 9.

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) 0.000 000 000 742 147 676 617 9 × 2 = 0 + 0.000 000 001 484 295 353 235 8;
  • 2) 0.000 000 001 484 295 353 235 8 × 2 = 0 + 0.000 000 002 968 590 706 471 6;
  • 3) 0.000 000 002 968 590 706 471 6 × 2 = 0 + 0.000 000 005 937 181 412 943 2;
  • 4) 0.000 000 005 937 181 412 943 2 × 2 = 0 + 0.000 000 011 874 362 825 886 4;
  • 5) 0.000 000 011 874 362 825 886 4 × 2 = 0 + 0.000 000 023 748 725 651 772 8;
  • 6) 0.000 000 023 748 725 651 772 8 × 2 = 0 + 0.000 000 047 497 451 303 545 6;
  • 7) 0.000 000 047 497 451 303 545 6 × 2 = 0 + 0.000 000 094 994 902 607 091 2;
  • 8) 0.000 000 094 994 902 607 091 2 × 2 = 0 + 0.000 000 189 989 805 214 182 4;
  • 9) 0.000 000 189 989 805 214 182 4 × 2 = 0 + 0.000 000 379 979 610 428 364 8;
  • 10) 0.000 000 379 979 610 428 364 8 × 2 = 0 + 0.000 000 759 959 220 856 729 6;
  • 11) 0.000 000 759 959 220 856 729 6 × 2 = 0 + 0.000 001 519 918 441 713 459 2;
  • 12) 0.000 001 519 918 441 713 459 2 × 2 = 0 + 0.000 003 039 836 883 426 918 4;
  • 13) 0.000 003 039 836 883 426 918 4 × 2 = 0 + 0.000 006 079 673 766 853 836 8;
  • 14) 0.000 006 079 673 766 853 836 8 × 2 = 0 + 0.000 012 159 347 533 707 673 6;
  • 15) 0.000 012 159 347 533 707 673 6 × 2 = 0 + 0.000 024 318 695 067 415 347 2;
  • 16) 0.000 024 318 695 067 415 347 2 × 2 = 0 + 0.000 048 637 390 134 830 694 4;
  • 17) 0.000 048 637 390 134 830 694 4 × 2 = 0 + 0.000 097 274 780 269 661 388 8;
  • 18) 0.000 097 274 780 269 661 388 8 × 2 = 0 + 0.000 194 549 560 539 322 777 6;
  • 19) 0.000 194 549 560 539 322 777 6 × 2 = 0 + 0.000 389 099 121 078 645 555 2;
  • 20) 0.000 389 099 121 078 645 555 2 × 2 = 0 + 0.000 778 198 242 157 291 110 4;
  • 21) 0.000 778 198 242 157 291 110 4 × 2 = 0 + 0.001 556 396 484 314 582 220 8;
  • 22) 0.001 556 396 484 314 582 220 8 × 2 = 0 + 0.003 112 792 968 629 164 441 6;
  • 23) 0.003 112 792 968 629 164 441 6 × 2 = 0 + 0.006 225 585 937 258 328 883 2;
  • 24) 0.006 225 585 937 258 328 883 2 × 2 = 0 + 0.012 451 171 874 516 657 766 4;
  • 25) 0.012 451 171 874 516 657 766 4 × 2 = 0 + 0.024 902 343 749 033 315 532 8;
  • 26) 0.024 902 343 749 033 315 532 8 × 2 = 0 + 0.049 804 687 498 066 631 065 6;
  • 27) 0.049 804 687 498 066 631 065 6 × 2 = 0 + 0.099 609 374 996 133 262 131 2;
  • 28) 0.099 609 374 996 133 262 131 2 × 2 = 0 + 0.199 218 749 992 266 524 262 4;
  • 29) 0.199 218 749 992 266 524 262 4 × 2 = 0 + 0.398 437 499 984 533 048 524 8;
  • 30) 0.398 437 499 984 533 048 524 8 × 2 = 0 + 0.796 874 999 969 066 097 049 6;
  • 31) 0.796 874 999 969 066 097 049 6 × 2 = 1 + 0.593 749 999 938 132 194 099 2;
  • 32) 0.593 749 999 938 132 194 099 2 × 2 = 1 + 0.187 499 999 876 264 388 198 4;
  • 33) 0.187 499 999 876 264 388 198 4 × 2 = 0 + 0.374 999 999 752 528 776 396 8;
  • 34) 0.374 999 999 752 528 776 396 8 × 2 = 0 + 0.749 999 999 505 057 552 793 6;
  • 35) 0.749 999 999 505 057 552 793 6 × 2 = 1 + 0.499 999 999 010 115 105 587 2;
  • 36) 0.499 999 999 010 115 105 587 2 × 2 = 0 + 0.999 999 998 020 230 211 174 4;
  • 37) 0.999 999 998 020 230 211 174 4 × 2 = 1 + 0.999 999 996 040 460 422 348 8;
  • 38) 0.999 999 996 040 460 422 348 8 × 2 = 1 + 0.999 999 992 080 920 844 697 6;
  • 39) 0.999 999 992 080 920 844 697 6 × 2 = 1 + 0.999 999 984 161 841 689 395 2;
  • 40) 0.999 999 984 161 841 689 395 2 × 2 = 1 + 0.999 999 968 323 683 378 790 4;
  • 41) 0.999 999 968 323 683 378 790 4 × 2 = 1 + 0.999 999 936 647 366 757 580 8;
  • 42) 0.999 999 936 647 366 757 580 8 × 2 = 1 + 0.999 999 873 294 733 515 161 6;
  • 43) 0.999 999 873 294 733 515 161 6 × 2 = 1 + 0.999 999 746 589 467 030 323 2;
  • 44) 0.999 999 746 589 467 030 323 2 × 2 = 1 + 0.999 999 493 178 934 060 646 4;
  • 45) 0.999 999 493 178 934 060 646 4 × 2 = 1 + 0.999 998 986 357 868 121 292 8;
  • 46) 0.999 998 986 357 868 121 292 8 × 2 = 1 + 0.999 997 972 715 736 242 585 6;
  • 47) 0.999 997 972 715 736 242 585 6 × 2 = 1 + 0.999 995 945 431 472 485 171 2;
  • 48) 0.999 995 945 431 472 485 171 2 × 2 = 1 + 0.999 991 890 862 944 970 342 4;
  • 49) 0.999 991 890 862 944 970 342 4 × 2 = 1 + 0.999 983 781 725 889 940 684 8;
  • 50) 0.999 983 781 725 889 940 684 8 × 2 = 1 + 0.999 967 563 451 779 881 369 6;
  • 51) 0.999 967 563 451 779 881 369 6 × 2 = 1 + 0.999 935 126 903 559 762 739 2;
  • 52) 0.999 935 126 903 559 762 739 2 × 2 = 1 + 0.999 870 253 807 119 525 478 4;
  • 53) 0.999 870 253 807 119 525 478 4 × 2 = 1 + 0.999 740 507 614 239 050 956 8;
  • 54) 0.999 740 507 614 239 050 956 8 × 2 = 1 + 0.999 481 015 228 478 101 913 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:


0.000 000 000 742 147 676 617 9(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 617 9(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:


0.000 000 000 742 147 676 617 9(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 -0.000 000 000 742 147 676 617 9 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