-0.000 000 000 742 147 676 646 05 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 05(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
-0.000 000 000 742 147 676 646 05(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:

|-0.000 000 000 742 147 676 646 05| = 0.000 000 000 742 147 676 646 05


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: 0.000 000 000 742 147 676 646 05.

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) 0.000 000 000 742 147 676 646 05 × 2 = 0 + 0.000 000 001 484 295 353 292 1;
  • 2) 0.000 000 001 484 295 353 292 1 × 2 = 0 + 0.000 000 002 968 590 706 584 2;
  • 3) 0.000 000 002 968 590 706 584 2 × 2 = 0 + 0.000 000 005 937 181 413 168 4;
  • 4) 0.000 000 005 937 181 413 168 4 × 2 = 0 + 0.000 000 011 874 362 826 336 8;
  • 5) 0.000 000 011 874 362 826 336 8 × 2 = 0 + 0.000 000 023 748 725 652 673 6;
  • 6) 0.000 000 023 748 725 652 673 6 × 2 = 0 + 0.000 000 047 497 451 305 347 2;
  • 7) 0.000 000 047 497 451 305 347 2 × 2 = 0 + 0.000 000 094 994 902 610 694 4;
  • 8) 0.000 000 094 994 902 610 694 4 × 2 = 0 + 0.000 000 189 989 805 221 388 8;
  • 9) 0.000 000 189 989 805 221 388 8 × 2 = 0 + 0.000 000 379 979 610 442 777 6;
  • 10) 0.000 000 379 979 610 442 777 6 × 2 = 0 + 0.000 000 759 959 220 885 555 2;
  • 11) 0.000 000 759 959 220 885 555 2 × 2 = 0 + 0.000 001 519 918 441 771 110 4;
  • 12) 0.000 001 519 918 441 771 110 4 × 2 = 0 + 0.000 003 039 836 883 542 220 8;
  • 13) 0.000 003 039 836 883 542 220 8 × 2 = 0 + 0.000 006 079 673 767 084 441 6;
  • 14) 0.000 006 079 673 767 084 441 6 × 2 = 0 + 0.000 012 159 347 534 168 883 2;
  • 15) 0.000 012 159 347 534 168 883 2 × 2 = 0 + 0.000 024 318 695 068 337 766 4;
  • 16) 0.000 024 318 695 068 337 766 4 × 2 = 0 + 0.000 048 637 390 136 675 532 8;
  • 17) 0.000 048 637 390 136 675 532 8 × 2 = 0 + 0.000 097 274 780 273 351 065 6;
  • 18) 0.000 097 274 780 273 351 065 6 × 2 = 0 + 0.000 194 549 560 546 702 131 2;
  • 19) 0.000 194 549 560 546 702 131 2 × 2 = 0 + 0.000 389 099 121 093 404 262 4;
  • 20) 0.000 389 099 121 093 404 262 4 × 2 = 0 + 0.000 778 198 242 186 808 524 8;
  • 21) 0.000 778 198 242 186 808 524 8 × 2 = 0 + 0.001 556 396 484 373 617 049 6;
  • 22) 0.001 556 396 484 373 617 049 6 × 2 = 0 + 0.003 112 792 968 747 234 099 2;
  • 23) 0.003 112 792 968 747 234 099 2 × 2 = 0 + 0.006 225 585 937 494 468 198 4;
  • 24) 0.006 225 585 937 494 468 198 4 × 2 = 0 + 0.012 451 171 874 988 936 396 8;
  • 25) 0.012 451 171 874 988 936 396 8 × 2 = 0 + 0.024 902 343 749 977 872 793 6;
  • 26) 0.024 902 343 749 977 872 793 6 × 2 = 0 + 0.049 804 687 499 955 745 587 2;
  • 27) 0.049 804 687 499 955 745 587 2 × 2 = 0 + 0.099 609 374 999 911 491 174 4;
  • 28) 0.099 609 374 999 911 491 174 4 × 2 = 0 + 0.199 218 749 999 822 982 348 8;
  • 29) 0.199 218 749 999 822 982 348 8 × 2 = 0 + 0.398 437 499 999 645 964 697 6;
  • 30) 0.398 437 499 999 645 964 697 6 × 2 = 0 + 0.796 874 999 999 291 929 395 2;
  • 31) 0.796 874 999 999 291 929 395 2 × 2 = 1 + 0.593 749 999 998 583 858 790 4;
  • 32) 0.593 749 999 998 583 858 790 4 × 2 = 1 + 0.187 499 999 997 167 717 580 8;
  • 33) 0.187 499 999 997 167 717 580 8 × 2 = 0 + 0.374 999 999 994 335 435 161 6;
  • 34) 0.374 999 999 994 335 435 161 6 × 2 = 0 + 0.749 999 999 988 670 870 323 2;
  • 35) 0.749 999 999 988 670 870 323 2 × 2 = 1 + 0.499 999 999 977 341 740 646 4;
  • 36) 0.499 999 999 977 341 740 646 4 × 2 = 0 + 0.999 999 999 954 683 481 292 8;
  • 37) 0.999 999 999 954 683 481 292 8 × 2 = 1 + 0.999 999 999 909 366 962 585 6;
  • 38) 0.999 999 999 909 366 962 585 6 × 2 = 1 + 0.999 999 999 818 733 925 171 2;
  • 39) 0.999 999 999 818 733 925 171 2 × 2 = 1 + 0.999 999 999 637 467 850 342 4;
  • 40) 0.999 999 999 637 467 850 342 4 × 2 = 1 + 0.999 999 999 274 935 700 684 8;
  • 41) 0.999 999 999 274 935 700 684 8 × 2 = 1 + 0.999 999 998 549 871 401 369 6;
  • 42) 0.999 999 998 549 871 401 369 6 × 2 = 1 + 0.999 999 997 099 742 802 739 2;
  • 43) 0.999 999 997 099 742 802 739 2 × 2 = 1 + 0.999 999 994 199 485 605 478 4;
  • 44) 0.999 999 994 199 485 605 478 4 × 2 = 1 + 0.999 999 988 398 971 210 956 8;
  • 45) 0.999 999 988 398 971 210 956 8 × 2 = 1 + 0.999 999 976 797 942 421 913 6;
  • 46) 0.999 999 976 797 942 421 913 6 × 2 = 1 + 0.999 999 953 595 884 843 827 2;
  • 47) 0.999 999 953 595 884 843 827 2 × 2 = 1 + 0.999 999 907 191 769 687 654 4;
  • 48) 0.999 999 907 191 769 687 654 4 × 2 = 1 + 0.999 999 814 383 539 375 308 8;
  • 49) 0.999 999 814 383 539 375 308 8 × 2 = 1 + 0.999 999 628 767 078 750 617 6;
  • 50) 0.999 999 628 767 078 750 617 6 × 2 = 1 + 0.999 999 257 534 157 501 235 2;
  • 51) 0.999 999 257 534 157 501 235 2 × 2 = 1 + 0.999 998 515 068 315 002 470 4;
  • 52) 0.999 998 515 068 315 002 470 4 × 2 = 1 + 0.999 997 030 136 630 004 940 8;
  • 53) 0.999 997 030 136 630 004 940 8 × 2 = 1 + 0.999 994 060 273 260 009 881 6;
  • 54) 0.999 994 060 273 260 009 881 6 × 2 = 1 + 0.999 988 120 546 520 019 763 2;

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:


0.000 000 000 742 147 676 646 05(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 05(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:


0.000 000 000 742 147 676 646 05(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 -0.000 000 000 742 147 676 646 05 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