-0.000 000 000 742 147 676 617 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

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

|-0.000 000 000 742 147 676 617 4| = 0.000 000 000 742 147 676 617 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: 0.000 000 000 742 147 676 617 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) 0.000 000 000 742 147 676 617 4 × 2 = 0 + 0.000 000 001 484 295 353 234 8;
  • 2) 0.000 000 001 484 295 353 234 8 × 2 = 0 + 0.000 000 002 968 590 706 469 6;
  • 3) 0.000 000 002 968 590 706 469 6 × 2 = 0 + 0.000 000 005 937 181 412 939 2;
  • 4) 0.000 000 005 937 181 412 939 2 × 2 = 0 + 0.000 000 011 874 362 825 878 4;
  • 5) 0.000 000 011 874 362 825 878 4 × 2 = 0 + 0.000 000 023 748 725 651 756 8;
  • 6) 0.000 000 023 748 725 651 756 8 × 2 = 0 + 0.000 000 047 497 451 303 513 6;
  • 7) 0.000 000 047 497 451 303 513 6 × 2 = 0 + 0.000 000 094 994 902 607 027 2;
  • 8) 0.000 000 094 994 902 607 027 2 × 2 = 0 + 0.000 000 189 989 805 214 054 4;
  • 9) 0.000 000 189 989 805 214 054 4 × 2 = 0 + 0.000 000 379 979 610 428 108 8;
  • 10) 0.000 000 379 979 610 428 108 8 × 2 = 0 + 0.000 000 759 959 220 856 217 6;
  • 11) 0.000 000 759 959 220 856 217 6 × 2 = 0 + 0.000 001 519 918 441 712 435 2;
  • 12) 0.000 001 519 918 441 712 435 2 × 2 = 0 + 0.000 003 039 836 883 424 870 4;
  • 13) 0.000 003 039 836 883 424 870 4 × 2 = 0 + 0.000 006 079 673 766 849 740 8;
  • 14) 0.000 006 079 673 766 849 740 8 × 2 = 0 + 0.000 012 159 347 533 699 481 6;
  • 15) 0.000 012 159 347 533 699 481 6 × 2 = 0 + 0.000 024 318 695 067 398 963 2;
  • 16) 0.000 024 318 695 067 398 963 2 × 2 = 0 + 0.000 048 637 390 134 797 926 4;
  • 17) 0.000 048 637 390 134 797 926 4 × 2 = 0 + 0.000 097 274 780 269 595 852 8;
  • 18) 0.000 097 274 780 269 595 852 8 × 2 = 0 + 0.000 194 549 560 539 191 705 6;
  • 19) 0.000 194 549 560 539 191 705 6 × 2 = 0 + 0.000 389 099 121 078 383 411 2;
  • 20) 0.000 389 099 121 078 383 411 2 × 2 = 0 + 0.000 778 198 242 156 766 822 4;
  • 21) 0.000 778 198 242 156 766 822 4 × 2 = 0 + 0.001 556 396 484 313 533 644 8;
  • 22) 0.001 556 396 484 313 533 644 8 × 2 = 0 + 0.003 112 792 968 627 067 289 6;
  • 23) 0.003 112 792 968 627 067 289 6 × 2 = 0 + 0.006 225 585 937 254 134 579 2;
  • 24) 0.006 225 585 937 254 134 579 2 × 2 = 0 + 0.012 451 171 874 508 269 158 4;
  • 25) 0.012 451 171 874 508 269 158 4 × 2 = 0 + 0.024 902 343 749 016 538 316 8;
  • 26) 0.024 902 343 749 016 538 316 8 × 2 = 0 + 0.049 804 687 498 033 076 633 6;
  • 27) 0.049 804 687 498 033 076 633 6 × 2 = 0 + 0.099 609 374 996 066 153 267 2;
  • 28) 0.099 609 374 996 066 153 267 2 × 2 = 0 + 0.199 218 749 992 132 306 534 4;
  • 29) 0.199 218 749 992 132 306 534 4 × 2 = 0 + 0.398 437 499 984 264 613 068 8;
  • 30) 0.398 437 499 984 264 613 068 8 × 2 = 0 + 0.796 874 999 968 529 226 137 6;
  • 31) 0.796 874 999 968 529 226 137 6 × 2 = 1 + 0.593 749 999 937 058 452 275 2;
  • 32) 0.593 749 999 937 058 452 275 2 × 2 = 1 + 0.187 499 999 874 116 904 550 4;
  • 33) 0.187 499 999 874 116 904 550 4 × 2 = 0 + 0.374 999 999 748 233 809 100 8;
  • 34) 0.374 999 999 748 233 809 100 8 × 2 = 0 + 0.749 999 999 496 467 618 201 6;
  • 35) 0.749 999 999 496 467 618 201 6 × 2 = 1 + 0.499 999 998 992 935 236 403 2;
  • 36) 0.499 999 998 992 935 236 403 2 × 2 = 0 + 0.999 999 997 985 870 472 806 4;
  • 37) 0.999 999 997 985 870 472 806 4 × 2 = 1 + 0.999 999 995 971 740 945 612 8;
  • 38) 0.999 999 995 971 740 945 612 8 × 2 = 1 + 0.999 999 991 943 481 891 225 6;
  • 39) 0.999 999 991 943 481 891 225 6 × 2 = 1 + 0.999 999 983 886 963 782 451 2;
  • 40) 0.999 999 983 886 963 782 451 2 × 2 = 1 + 0.999 999 967 773 927 564 902 4;
  • 41) 0.999 999 967 773 927 564 902 4 × 2 = 1 + 0.999 999 935 547 855 129 804 8;
  • 42) 0.999 999 935 547 855 129 804 8 × 2 = 1 + 0.999 999 871 095 710 259 609 6;
  • 43) 0.999 999 871 095 710 259 609 6 × 2 = 1 + 0.999 999 742 191 420 519 219 2;
  • 44) 0.999 999 742 191 420 519 219 2 × 2 = 1 + 0.999 999 484 382 841 038 438 4;
  • 45) 0.999 999 484 382 841 038 438 4 × 2 = 1 + 0.999 998 968 765 682 076 876 8;
  • 46) 0.999 998 968 765 682 076 876 8 × 2 = 1 + 0.999 997 937 531 364 153 753 6;
  • 47) 0.999 997 937 531 364 153 753 6 × 2 = 1 + 0.999 995 875 062 728 307 507 2;
  • 48) 0.999 995 875 062 728 307 507 2 × 2 = 1 + 0.999 991 750 125 456 615 014 4;
  • 49) 0.999 991 750 125 456 615 014 4 × 2 = 1 + 0.999 983 500 250 913 230 028 8;
  • 50) 0.999 983 500 250 913 230 028 8 × 2 = 1 + 0.999 967 000 501 826 460 057 6;
  • 51) 0.999 967 000 501 826 460 057 6 × 2 = 1 + 0.999 934 001 003 652 920 115 2;
  • 52) 0.999 934 001 003 652 920 115 2 × 2 = 1 + 0.999 868 002 007 305 840 230 4;
  • 53) 0.999 868 002 007 305 840 230 4 × 2 = 1 + 0.999 736 004 014 611 680 460 8;
  • 54) 0.999 736 004 014 611 680 460 8 × 2 = 1 + 0.999 472 008 029 223 360 921 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 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 617 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:


0.000 000 000 742 147 676 617 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 -0.000 000 000 742 147 676 617 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