-0.000 000 000 742 147 676 653 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 653 2(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
-0.000 000 000 742 147 676 653 2(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:

|-0.000 000 000 742 147 676 653 2| = 0.000 000 000 742 147 676 653 2


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: 0.000 000 000 742 147 676 653 2.

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) 0.000 000 000 742 147 676 653 2 × 2 = 0 + 0.000 000 001 484 295 353 306 4;
  • 2) 0.000 000 001 484 295 353 306 4 × 2 = 0 + 0.000 000 002 968 590 706 612 8;
  • 3) 0.000 000 002 968 590 706 612 8 × 2 = 0 + 0.000 000 005 937 181 413 225 6;
  • 4) 0.000 000 005 937 181 413 225 6 × 2 = 0 + 0.000 000 011 874 362 826 451 2;
  • 5) 0.000 000 011 874 362 826 451 2 × 2 = 0 + 0.000 000 023 748 725 652 902 4;
  • 6) 0.000 000 023 748 725 652 902 4 × 2 = 0 + 0.000 000 047 497 451 305 804 8;
  • 7) 0.000 000 047 497 451 305 804 8 × 2 = 0 + 0.000 000 094 994 902 611 609 6;
  • 8) 0.000 000 094 994 902 611 609 6 × 2 = 0 + 0.000 000 189 989 805 223 219 2;
  • 9) 0.000 000 189 989 805 223 219 2 × 2 = 0 + 0.000 000 379 979 610 446 438 4;
  • 10) 0.000 000 379 979 610 446 438 4 × 2 = 0 + 0.000 000 759 959 220 892 876 8;
  • 11) 0.000 000 759 959 220 892 876 8 × 2 = 0 + 0.000 001 519 918 441 785 753 6;
  • 12) 0.000 001 519 918 441 785 753 6 × 2 = 0 + 0.000 003 039 836 883 571 507 2;
  • 13) 0.000 003 039 836 883 571 507 2 × 2 = 0 + 0.000 006 079 673 767 143 014 4;
  • 14) 0.000 006 079 673 767 143 014 4 × 2 = 0 + 0.000 012 159 347 534 286 028 8;
  • 15) 0.000 012 159 347 534 286 028 8 × 2 = 0 + 0.000 024 318 695 068 572 057 6;
  • 16) 0.000 024 318 695 068 572 057 6 × 2 = 0 + 0.000 048 637 390 137 144 115 2;
  • 17) 0.000 048 637 390 137 144 115 2 × 2 = 0 + 0.000 097 274 780 274 288 230 4;
  • 18) 0.000 097 274 780 274 288 230 4 × 2 = 0 + 0.000 194 549 560 548 576 460 8;
  • 19) 0.000 194 549 560 548 576 460 8 × 2 = 0 + 0.000 389 099 121 097 152 921 6;
  • 20) 0.000 389 099 121 097 152 921 6 × 2 = 0 + 0.000 778 198 242 194 305 843 2;
  • 21) 0.000 778 198 242 194 305 843 2 × 2 = 0 + 0.001 556 396 484 388 611 686 4;
  • 22) 0.001 556 396 484 388 611 686 4 × 2 = 0 + 0.003 112 792 968 777 223 372 8;
  • 23) 0.003 112 792 968 777 223 372 8 × 2 = 0 + 0.006 225 585 937 554 446 745 6;
  • 24) 0.006 225 585 937 554 446 745 6 × 2 = 0 + 0.012 451 171 875 108 893 491 2;
  • 25) 0.012 451 171 875 108 893 491 2 × 2 = 0 + 0.024 902 343 750 217 786 982 4;
  • 26) 0.024 902 343 750 217 786 982 4 × 2 = 0 + 0.049 804 687 500 435 573 964 8;
  • 27) 0.049 804 687 500 435 573 964 8 × 2 = 0 + 0.099 609 375 000 871 147 929 6;
  • 28) 0.099 609 375 000 871 147 929 6 × 2 = 0 + 0.199 218 750 001 742 295 859 2;
  • 29) 0.199 218 750 001 742 295 859 2 × 2 = 0 + 0.398 437 500 003 484 591 718 4;
  • 30) 0.398 437 500 003 484 591 718 4 × 2 = 0 + 0.796 875 000 006 969 183 436 8;
  • 31) 0.796 875 000 006 969 183 436 8 × 2 = 1 + 0.593 750 000 013 938 366 873 6;
  • 32) 0.593 750 000 013 938 366 873 6 × 2 = 1 + 0.187 500 000 027 876 733 747 2;
  • 33) 0.187 500 000 027 876 733 747 2 × 2 = 0 + 0.375 000 000 055 753 467 494 4;
  • 34) 0.375 000 000 055 753 467 494 4 × 2 = 0 + 0.750 000 000 111 506 934 988 8;
  • 35) 0.750 000 000 111 506 934 988 8 × 2 = 1 + 0.500 000 000 223 013 869 977 6;
  • 36) 0.500 000 000 223 013 869 977 6 × 2 = 1 + 0.000 000 000 446 027 739 955 2;
  • 37) 0.000 000 000 446 027 739 955 2 × 2 = 0 + 0.000 000 000 892 055 479 910 4;
  • 38) 0.000 000 000 892 055 479 910 4 × 2 = 0 + 0.000 000 001 784 110 959 820 8;
  • 39) 0.000 000 001 784 110 959 820 8 × 2 = 0 + 0.000 000 003 568 221 919 641 6;
  • 40) 0.000 000 003 568 221 919 641 6 × 2 = 0 + 0.000 000 007 136 443 839 283 2;
  • 41) 0.000 000 007 136 443 839 283 2 × 2 = 0 + 0.000 000 014 272 887 678 566 4;
  • 42) 0.000 000 014 272 887 678 566 4 × 2 = 0 + 0.000 000 028 545 775 357 132 8;
  • 43) 0.000 000 028 545 775 357 132 8 × 2 = 0 + 0.000 000 057 091 550 714 265 6;
  • 44) 0.000 000 057 091 550 714 265 6 × 2 = 0 + 0.000 000 114 183 101 428 531 2;
  • 45) 0.000 000 114 183 101 428 531 2 × 2 = 0 + 0.000 000 228 366 202 857 062 4;
  • 46) 0.000 000 228 366 202 857 062 4 × 2 = 0 + 0.000 000 456 732 405 714 124 8;
  • 47) 0.000 000 456 732 405 714 124 8 × 2 = 0 + 0.000 000 913 464 811 428 249 6;
  • 48) 0.000 000 913 464 811 428 249 6 × 2 = 0 + 0.000 001 826 929 622 856 499 2;
  • 49) 0.000 001 826 929 622 856 499 2 × 2 = 0 + 0.000 003 653 859 245 712 998 4;
  • 50) 0.000 003 653 859 245 712 998 4 × 2 = 0 + 0.000 007 307 718 491 425 996 8;
  • 51) 0.000 007 307 718 491 425 996 8 × 2 = 0 + 0.000 014 615 436 982 851 993 6;
  • 52) 0.000 014 615 436 982 851 993 6 × 2 = 0 + 0.000 029 230 873 965 703 987 2;
  • 53) 0.000 029 230 873 965 703 987 2 × 2 = 0 + 0.000 058 461 747 931 407 974 4;
  • 54) 0.000 058 461 747 931 407 974 4 × 2 = 0 + 0.000 116 923 495 862 815 948 8;

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:


0.000 000 000 742 147 676 653 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 653 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 653 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 653 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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