-0.000 000 000 742 147 676 618 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 618 4(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
-0.000 000 000 742 147 676 618 4(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:

|-0.000 000 000 742 147 676 618 4| = 0.000 000 000 742 147 676 618 4


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: 0.000 000 000 742 147 676 618 4.

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) 0.000 000 000 742 147 676 618 4 × 2 = 0 + 0.000 000 001 484 295 353 236 8;
  • 2) 0.000 000 001 484 295 353 236 8 × 2 = 0 + 0.000 000 002 968 590 706 473 6;
  • 3) 0.000 000 002 968 590 706 473 6 × 2 = 0 + 0.000 000 005 937 181 412 947 2;
  • 4) 0.000 000 005 937 181 412 947 2 × 2 = 0 + 0.000 000 011 874 362 825 894 4;
  • 5) 0.000 000 011 874 362 825 894 4 × 2 = 0 + 0.000 000 023 748 725 651 788 8;
  • 6) 0.000 000 023 748 725 651 788 8 × 2 = 0 + 0.000 000 047 497 451 303 577 6;
  • 7) 0.000 000 047 497 451 303 577 6 × 2 = 0 + 0.000 000 094 994 902 607 155 2;
  • 8) 0.000 000 094 994 902 607 155 2 × 2 = 0 + 0.000 000 189 989 805 214 310 4;
  • 9) 0.000 000 189 989 805 214 310 4 × 2 = 0 + 0.000 000 379 979 610 428 620 8;
  • 10) 0.000 000 379 979 610 428 620 8 × 2 = 0 + 0.000 000 759 959 220 857 241 6;
  • 11) 0.000 000 759 959 220 857 241 6 × 2 = 0 + 0.000 001 519 918 441 714 483 2;
  • 12) 0.000 001 519 918 441 714 483 2 × 2 = 0 + 0.000 003 039 836 883 428 966 4;
  • 13) 0.000 003 039 836 883 428 966 4 × 2 = 0 + 0.000 006 079 673 766 857 932 8;
  • 14) 0.000 006 079 673 766 857 932 8 × 2 = 0 + 0.000 012 159 347 533 715 865 6;
  • 15) 0.000 012 159 347 533 715 865 6 × 2 = 0 + 0.000 024 318 695 067 431 731 2;
  • 16) 0.000 024 318 695 067 431 731 2 × 2 = 0 + 0.000 048 637 390 134 863 462 4;
  • 17) 0.000 048 637 390 134 863 462 4 × 2 = 0 + 0.000 097 274 780 269 726 924 8;
  • 18) 0.000 097 274 780 269 726 924 8 × 2 = 0 + 0.000 194 549 560 539 453 849 6;
  • 19) 0.000 194 549 560 539 453 849 6 × 2 = 0 + 0.000 389 099 121 078 907 699 2;
  • 20) 0.000 389 099 121 078 907 699 2 × 2 = 0 + 0.000 778 198 242 157 815 398 4;
  • 21) 0.000 778 198 242 157 815 398 4 × 2 = 0 + 0.001 556 396 484 315 630 796 8;
  • 22) 0.001 556 396 484 315 630 796 8 × 2 = 0 + 0.003 112 792 968 631 261 593 6;
  • 23) 0.003 112 792 968 631 261 593 6 × 2 = 0 + 0.006 225 585 937 262 523 187 2;
  • 24) 0.006 225 585 937 262 523 187 2 × 2 = 0 + 0.012 451 171 874 525 046 374 4;
  • 25) 0.012 451 171 874 525 046 374 4 × 2 = 0 + 0.024 902 343 749 050 092 748 8;
  • 26) 0.024 902 343 749 050 092 748 8 × 2 = 0 + 0.049 804 687 498 100 185 497 6;
  • 27) 0.049 804 687 498 100 185 497 6 × 2 = 0 + 0.099 609 374 996 200 370 995 2;
  • 28) 0.099 609 374 996 200 370 995 2 × 2 = 0 + 0.199 218 749 992 400 741 990 4;
  • 29) 0.199 218 749 992 400 741 990 4 × 2 = 0 + 0.398 437 499 984 801 483 980 8;
  • 30) 0.398 437 499 984 801 483 980 8 × 2 = 0 + 0.796 874 999 969 602 967 961 6;
  • 31) 0.796 874 999 969 602 967 961 6 × 2 = 1 + 0.593 749 999 939 205 935 923 2;
  • 32) 0.593 749 999 939 205 935 923 2 × 2 = 1 + 0.187 499 999 878 411 871 846 4;
  • 33) 0.187 499 999 878 411 871 846 4 × 2 = 0 + 0.374 999 999 756 823 743 692 8;
  • 34) 0.374 999 999 756 823 743 692 8 × 2 = 0 + 0.749 999 999 513 647 487 385 6;
  • 35) 0.749 999 999 513 647 487 385 6 × 2 = 1 + 0.499 999 999 027 294 974 771 2;
  • 36) 0.499 999 999 027 294 974 771 2 × 2 = 0 + 0.999 999 998 054 589 949 542 4;
  • 37) 0.999 999 998 054 589 949 542 4 × 2 = 1 + 0.999 999 996 109 179 899 084 8;
  • 38) 0.999 999 996 109 179 899 084 8 × 2 = 1 + 0.999 999 992 218 359 798 169 6;
  • 39) 0.999 999 992 218 359 798 169 6 × 2 = 1 + 0.999 999 984 436 719 596 339 2;
  • 40) 0.999 999 984 436 719 596 339 2 × 2 = 1 + 0.999 999 968 873 439 192 678 4;
  • 41) 0.999 999 968 873 439 192 678 4 × 2 = 1 + 0.999 999 937 746 878 385 356 8;
  • 42) 0.999 999 937 746 878 385 356 8 × 2 = 1 + 0.999 999 875 493 756 770 713 6;
  • 43) 0.999 999 875 493 756 770 713 6 × 2 = 1 + 0.999 999 750 987 513 541 427 2;
  • 44) 0.999 999 750 987 513 541 427 2 × 2 = 1 + 0.999 999 501 975 027 082 854 4;
  • 45) 0.999 999 501 975 027 082 854 4 × 2 = 1 + 0.999 999 003 950 054 165 708 8;
  • 46) 0.999 999 003 950 054 165 708 8 × 2 = 1 + 0.999 998 007 900 108 331 417 6;
  • 47) 0.999 998 007 900 108 331 417 6 × 2 = 1 + 0.999 996 015 800 216 662 835 2;
  • 48) 0.999 996 015 800 216 662 835 2 × 2 = 1 + 0.999 992 031 600 433 325 670 4;
  • 49) 0.999 992 031 600 433 325 670 4 × 2 = 1 + 0.999 984 063 200 866 651 340 8;
  • 50) 0.999 984 063 200 866 651 340 8 × 2 = 1 + 0.999 968 126 401 733 302 681 6;
  • 51) 0.999 968 126 401 733 302 681 6 × 2 = 1 + 0.999 936 252 803 466 605 363 2;
  • 52) 0.999 936 252 803 466 605 363 2 × 2 = 1 + 0.999 872 505 606 933 210 726 4;
  • 53) 0.999 872 505 606 933 210 726 4 × 2 = 1 + 0.999 745 011 213 866 421 452 8;
  • 54) 0.999 745 011 213 866 421 452 8 × 2 = 1 + 0.999 490 022 427 732 842 905 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:


0.000 000 000 742 147 676 618 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 618 4(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:


0.000 000 000 742 147 676 618 4(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 -0.000 000 000 742 147 676 618 4 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