-0.000 000 000 742 147 676 642 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 642 6(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
-0.000 000 000 742 147 676 642 6(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:

|-0.000 000 000 742 147 676 642 6| = 0.000 000 000 742 147 676 642 6


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: 0.000 000 000 742 147 676 642 6.

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) 0.000 000 000 742 147 676 642 6 × 2 = 0 + 0.000 000 001 484 295 353 285 2;
  • 2) 0.000 000 001 484 295 353 285 2 × 2 = 0 + 0.000 000 002 968 590 706 570 4;
  • 3) 0.000 000 002 968 590 706 570 4 × 2 = 0 + 0.000 000 005 937 181 413 140 8;
  • 4) 0.000 000 005 937 181 413 140 8 × 2 = 0 + 0.000 000 011 874 362 826 281 6;
  • 5) 0.000 000 011 874 362 826 281 6 × 2 = 0 + 0.000 000 023 748 725 652 563 2;
  • 6) 0.000 000 023 748 725 652 563 2 × 2 = 0 + 0.000 000 047 497 451 305 126 4;
  • 7) 0.000 000 047 497 451 305 126 4 × 2 = 0 + 0.000 000 094 994 902 610 252 8;
  • 8) 0.000 000 094 994 902 610 252 8 × 2 = 0 + 0.000 000 189 989 805 220 505 6;
  • 9) 0.000 000 189 989 805 220 505 6 × 2 = 0 + 0.000 000 379 979 610 441 011 2;
  • 10) 0.000 000 379 979 610 441 011 2 × 2 = 0 + 0.000 000 759 959 220 882 022 4;
  • 11) 0.000 000 759 959 220 882 022 4 × 2 = 0 + 0.000 001 519 918 441 764 044 8;
  • 12) 0.000 001 519 918 441 764 044 8 × 2 = 0 + 0.000 003 039 836 883 528 089 6;
  • 13) 0.000 003 039 836 883 528 089 6 × 2 = 0 + 0.000 006 079 673 767 056 179 2;
  • 14) 0.000 006 079 673 767 056 179 2 × 2 = 0 + 0.000 012 159 347 534 112 358 4;
  • 15) 0.000 012 159 347 534 112 358 4 × 2 = 0 + 0.000 024 318 695 068 224 716 8;
  • 16) 0.000 024 318 695 068 224 716 8 × 2 = 0 + 0.000 048 637 390 136 449 433 6;
  • 17) 0.000 048 637 390 136 449 433 6 × 2 = 0 + 0.000 097 274 780 272 898 867 2;
  • 18) 0.000 097 274 780 272 898 867 2 × 2 = 0 + 0.000 194 549 560 545 797 734 4;
  • 19) 0.000 194 549 560 545 797 734 4 × 2 = 0 + 0.000 389 099 121 091 595 468 8;
  • 20) 0.000 389 099 121 091 595 468 8 × 2 = 0 + 0.000 778 198 242 183 190 937 6;
  • 21) 0.000 778 198 242 183 190 937 6 × 2 = 0 + 0.001 556 396 484 366 381 875 2;
  • 22) 0.001 556 396 484 366 381 875 2 × 2 = 0 + 0.003 112 792 968 732 763 750 4;
  • 23) 0.003 112 792 968 732 763 750 4 × 2 = 0 + 0.006 225 585 937 465 527 500 8;
  • 24) 0.006 225 585 937 465 527 500 8 × 2 = 0 + 0.012 451 171 874 931 055 001 6;
  • 25) 0.012 451 171 874 931 055 001 6 × 2 = 0 + 0.024 902 343 749 862 110 003 2;
  • 26) 0.024 902 343 749 862 110 003 2 × 2 = 0 + 0.049 804 687 499 724 220 006 4;
  • 27) 0.049 804 687 499 724 220 006 4 × 2 = 0 + 0.099 609 374 999 448 440 012 8;
  • 28) 0.099 609 374 999 448 440 012 8 × 2 = 0 + 0.199 218 749 998 896 880 025 6;
  • 29) 0.199 218 749 998 896 880 025 6 × 2 = 0 + 0.398 437 499 997 793 760 051 2;
  • 30) 0.398 437 499 997 793 760 051 2 × 2 = 0 + 0.796 874 999 995 587 520 102 4;
  • 31) 0.796 874 999 995 587 520 102 4 × 2 = 1 + 0.593 749 999 991 175 040 204 8;
  • 32) 0.593 749 999 991 175 040 204 8 × 2 = 1 + 0.187 499 999 982 350 080 409 6;
  • 33) 0.187 499 999 982 350 080 409 6 × 2 = 0 + 0.374 999 999 964 700 160 819 2;
  • 34) 0.374 999 999 964 700 160 819 2 × 2 = 0 + 0.749 999 999 929 400 321 638 4;
  • 35) 0.749 999 999 929 400 321 638 4 × 2 = 1 + 0.499 999 999 858 800 643 276 8;
  • 36) 0.499 999 999 858 800 643 276 8 × 2 = 0 + 0.999 999 999 717 601 286 553 6;
  • 37) 0.999 999 999 717 601 286 553 6 × 2 = 1 + 0.999 999 999 435 202 573 107 2;
  • 38) 0.999 999 999 435 202 573 107 2 × 2 = 1 + 0.999 999 998 870 405 146 214 4;
  • 39) 0.999 999 998 870 405 146 214 4 × 2 = 1 + 0.999 999 997 740 810 292 428 8;
  • 40) 0.999 999 997 740 810 292 428 8 × 2 = 1 + 0.999 999 995 481 620 584 857 6;
  • 41) 0.999 999 995 481 620 584 857 6 × 2 = 1 + 0.999 999 990 963 241 169 715 2;
  • 42) 0.999 999 990 963 241 169 715 2 × 2 = 1 + 0.999 999 981 926 482 339 430 4;
  • 43) 0.999 999 981 926 482 339 430 4 × 2 = 1 + 0.999 999 963 852 964 678 860 8;
  • 44) 0.999 999 963 852 964 678 860 8 × 2 = 1 + 0.999 999 927 705 929 357 721 6;
  • 45) 0.999 999 927 705 929 357 721 6 × 2 = 1 + 0.999 999 855 411 858 715 443 2;
  • 46) 0.999 999 855 411 858 715 443 2 × 2 = 1 + 0.999 999 710 823 717 430 886 4;
  • 47) 0.999 999 710 823 717 430 886 4 × 2 = 1 + 0.999 999 421 647 434 861 772 8;
  • 48) 0.999 999 421 647 434 861 772 8 × 2 = 1 + 0.999 998 843 294 869 723 545 6;
  • 49) 0.999 998 843 294 869 723 545 6 × 2 = 1 + 0.999 997 686 589 739 447 091 2;
  • 50) 0.999 997 686 589 739 447 091 2 × 2 = 1 + 0.999 995 373 179 478 894 182 4;
  • 51) 0.999 995 373 179 478 894 182 4 × 2 = 1 + 0.999 990 746 358 957 788 364 8;
  • 52) 0.999 990 746 358 957 788 364 8 × 2 = 1 + 0.999 981 492 717 915 576 729 6;
  • 53) 0.999 981 492 717 915 576 729 6 × 2 = 1 + 0.999 962 985 435 831 153 459 2;
  • 54) 0.999 962 985 435 831 153 459 2 × 2 = 1 + 0.999 925 970 871 662 306 918 4;

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:


0.000 000 000 742 147 676 642 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 642 6(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:


0.000 000 000 742 147 676 642 6(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 -0.000 000 000 742 147 676 642 6 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