-0.000 000 000 742 147 676 644 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 644 8(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
-0.000 000 000 742 147 676 644 8(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:

|-0.000 000 000 742 147 676 644 8| = 0.000 000 000 742 147 676 644 8


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: 0.000 000 000 742 147 676 644 8.

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) 0.000 000 000 742 147 676 644 8 × 2 = 0 + 0.000 000 001 484 295 353 289 6;
  • 2) 0.000 000 001 484 295 353 289 6 × 2 = 0 + 0.000 000 002 968 590 706 579 2;
  • 3) 0.000 000 002 968 590 706 579 2 × 2 = 0 + 0.000 000 005 937 181 413 158 4;
  • 4) 0.000 000 005 937 181 413 158 4 × 2 = 0 + 0.000 000 011 874 362 826 316 8;
  • 5) 0.000 000 011 874 362 826 316 8 × 2 = 0 + 0.000 000 023 748 725 652 633 6;
  • 6) 0.000 000 023 748 725 652 633 6 × 2 = 0 + 0.000 000 047 497 451 305 267 2;
  • 7) 0.000 000 047 497 451 305 267 2 × 2 = 0 + 0.000 000 094 994 902 610 534 4;
  • 8) 0.000 000 094 994 902 610 534 4 × 2 = 0 + 0.000 000 189 989 805 221 068 8;
  • 9) 0.000 000 189 989 805 221 068 8 × 2 = 0 + 0.000 000 379 979 610 442 137 6;
  • 10) 0.000 000 379 979 610 442 137 6 × 2 = 0 + 0.000 000 759 959 220 884 275 2;
  • 11) 0.000 000 759 959 220 884 275 2 × 2 = 0 + 0.000 001 519 918 441 768 550 4;
  • 12) 0.000 001 519 918 441 768 550 4 × 2 = 0 + 0.000 003 039 836 883 537 100 8;
  • 13) 0.000 003 039 836 883 537 100 8 × 2 = 0 + 0.000 006 079 673 767 074 201 6;
  • 14) 0.000 006 079 673 767 074 201 6 × 2 = 0 + 0.000 012 159 347 534 148 403 2;
  • 15) 0.000 012 159 347 534 148 403 2 × 2 = 0 + 0.000 024 318 695 068 296 806 4;
  • 16) 0.000 024 318 695 068 296 806 4 × 2 = 0 + 0.000 048 637 390 136 593 612 8;
  • 17) 0.000 048 637 390 136 593 612 8 × 2 = 0 + 0.000 097 274 780 273 187 225 6;
  • 18) 0.000 097 274 780 273 187 225 6 × 2 = 0 + 0.000 194 549 560 546 374 451 2;
  • 19) 0.000 194 549 560 546 374 451 2 × 2 = 0 + 0.000 389 099 121 092 748 902 4;
  • 20) 0.000 389 099 121 092 748 902 4 × 2 = 0 + 0.000 778 198 242 185 497 804 8;
  • 21) 0.000 778 198 242 185 497 804 8 × 2 = 0 + 0.001 556 396 484 370 995 609 6;
  • 22) 0.001 556 396 484 370 995 609 6 × 2 = 0 + 0.003 112 792 968 741 991 219 2;
  • 23) 0.003 112 792 968 741 991 219 2 × 2 = 0 + 0.006 225 585 937 483 982 438 4;
  • 24) 0.006 225 585 937 483 982 438 4 × 2 = 0 + 0.012 451 171 874 967 964 876 8;
  • 25) 0.012 451 171 874 967 964 876 8 × 2 = 0 + 0.024 902 343 749 935 929 753 6;
  • 26) 0.024 902 343 749 935 929 753 6 × 2 = 0 + 0.049 804 687 499 871 859 507 2;
  • 27) 0.049 804 687 499 871 859 507 2 × 2 = 0 + 0.099 609 374 999 743 719 014 4;
  • 28) 0.099 609 374 999 743 719 014 4 × 2 = 0 + 0.199 218 749 999 487 438 028 8;
  • 29) 0.199 218 749 999 487 438 028 8 × 2 = 0 + 0.398 437 499 998 974 876 057 6;
  • 30) 0.398 437 499 998 974 876 057 6 × 2 = 0 + 0.796 874 999 997 949 752 115 2;
  • 31) 0.796 874 999 997 949 752 115 2 × 2 = 1 + 0.593 749 999 995 899 504 230 4;
  • 32) 0.593 749 999 995 899 504 230 4 × 2 = 1 + 0.187 499 999 991 799 008 460 8;
  • 33) 0.187 499 999 991 799 008 460 8 × 2 = 0 + 0.374 999 999 983 598 016 921 6;
  • 34) 0.374 999 999 983 598 016 921 6 × 2 = 0 + 0.749 999 999 967 196 033 843 2;
  • 35) 0.749 999 999 967 196 033 843 2 × 2 = 1 + 0.499 999 999 934 392 067 686 4;
  • 36) 0.499 999 999 934 392 067 686 4 × 2 = 0 + 0.999 999 999 868 784 135 372 8;
  • 37) 0.999 999 999 868 784 135 372 8 × 2 = 1 + 0.999 999 999 737 568 270 745 6;
  • 38) 0.999 999 999 737 568 270 745 6 × 2 = 1 + 0.999 999 999 475 136 541 491 2;
  • 39) 0.999 999 999 475 136 541 491 2 × 2 = 1 + 0.999 999 998 950 273 082 982 4;
  • 40) 0.999 999 998 950 273 082 982 4 × 2 = 1 + 0.999 999 997 900 546 165 964 8;
  • 41) 0.999 999 997 900 546 165 964 8 × 2 = 1 + 0.999 999 995 801 092 331 929 6;
  • 42) 0.999 999 995 801 092 331 929 6 × 2 = 1 + 0.999 999 991 602 184 663 859 2;
  • 43) 0.999 999 991 602 184 663 859 2 × 2 = 1 + 0.999 999 983 204 369 327 718 4;
  • 44) 0.999 999 983 204 369 327 718 4 × 2 = 1 + 0.999 999 966 408 738 655 436 8;
  • 45) 0.999 999 966 408 738 655 436 8 × 2 = 1 + 0.999 999 932 817 477 310 873 6;
  • 46) 0.999 999 932 817 477 310 873 6 × 2 = 1 + 0.999 999 865 634 954 621 747 2;
  • 47) 0.999 999 865 634 954 621 747 2 × 2 = 1 + 0.999 999 731 269 909 243 494 4;
  • 48) 0.999 999 731 269 909 243 494 4 × 2 = 1 + 0.999 999 462 539 818 486 988 8;
  • 49) 0.999 999 462 539 818 486 988 8 × 2 = 1 + 0.999 998 925 079 636 973 977 6;
  • 50) 0.999 998 925 079 636 973 977 6 × 2 = 1 + 0.999 997 850 159 273 947 955 2;
  • 51) 0.999 997 850 159 273 947 955 2 × 2 = 1 + 0.999 995 700 318 547 895 910 4;
  • 52) 0.999 995 700 318 547 895 910 4 × 2 = 1 + 0.999 991 400 637 095 791 820 8;
  • 53) 0.999 991 400 637 095 791 820 8 × 2 = 1 + 0.999 982 801 274 191 583 641 6;
  • 54) 0.999 982 801 274 191 583 641 6 × 2 = 1 + 0.999 965 602 548 383 167 283 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:


0.000 000 000 742 147 676 644 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 644 8(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:


0.000 000 000 742 147 676 644 8(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 -0.000 000 000 742 147 676 644 8 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