-0.000 000 000 742 147 676 646 46 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 46(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
-0.000 000 000 742 147 676 646 46(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:

|-0.000 000 000 742 147 676 646 46| = 0.000 000 000 742 147 676 646 46


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: 0.000 000 000 742 147 676 646 46.

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) 0.000 000 000 742 147 676 646 46 × 2 = 0 + 0.000 000 001 484 295 353 292 92;
  • 2) 0.000 000 001 484 295 353 292 92 × 2 = 0 + 0.000 000 002 968 590 706 585 84;
  • 3) 0.000 000 002 968 590 706 585 84 × 2 = 0 + 0.000 000 005 937 181 413 171 68;
  • 4) 0.000 000 005 937 181 413 171 68 × 2 = 0 + 0.000 000 011 874 362 826 343 36;
  • 5) 0.000 000 011 874 362 826 343 36 × 2 = 0 + 0.000 000 023 748 725 652 686 72;
  • 6) 0.000 000 023 748 725 652 686 72 × 2 = 0 + 0.000 000 047 497 451 305 373 44;
  • 7) 0.000 000 047 497 451 305 373 44 × 2 = 0 + 0.000 000 094 994 902 610 746 88;
  • 8) 0.000 000 094 994 902 610 746 88 × 2 = 0 + 0.000 000 189 989 805 221 493 76;
  • 9) 0.000 000 189 989 805 221 493 76 × 2 = 0 + 0.000 000 379 979 610 442 987 52;
  • 10) 0.000 000 379 979 610 442 987 52 × 2 = 0 + 0.000 000 759 959 220 885 975 04;
  • 11) 0.000 000 759 959 220 885 975 04 × 2 = 0 + 0.000 001 519 918 441 771 950 08;
  • 12) 0.000 001 519 918 441 771 950 08 × 2 = 0 + 0.000 003 039 836 883 543 900 16;
  • 13) 0.000 003 039 836 883 543 900 16 × 2 = 0 + 0.000 006 079 673 767 087 800 32;
  • 14) 0.000 006 079 673 767 087 800 32 × 2 = 0 + 0.000 012 159 347 534 175 600 64;
  • 15) 0.000 012 159 347 534 175 600 64 × 2 = 0 + 0.000 024 318 695 068 351 201 28;
  • 16) 0.000 024 318 695 068 351 201 28 × 2 = 0 + 0.000 048 637 390 136 702 402 56;
  • 17) 0.000 048 637 390 136 702 402 56 × 2 = 0 + 0.000 097 274 780 273 404 805 12;
  • 18) 0.000 097 274 780 273 404 805 12 × 2 = 0 + 0.000 194 549 560 546 809 610 24;
  • 19) 0.000 194 549 560 546 809 610 24 × 2 = 0 + 0.000 389 099 121 093 619 220 48;
  • 20) 0.000 389 099 121 093 619 220 48 × 2 = 0 + 0.000 778 198 242 187 238 440 96;
  • 21) 0.000 778 198 242 187 238 440 96 × 2 = 0 + 0.001 556 396 484 374 476 881 92;
  • 22) 0.001 556 396 484 374 476 881 92 × 2 = 0 + 0.003 112 792 968 748 953 763 84;
  • 23) 0.003 112 792 968 748 953 763 84 × 2 = 0 + 0.006 225 585 937 497 907 527 68;
  • 24) 0.006 225 585 937 497 907 527 68 × 2 = 0 + 0.012 451 171 874 995 815 055 36;
  • 25) 0.012 451 171 874 995 815 055 36 × 2 = 0 + 0.024 902 343 749 991 630 110 72;
  • 26) 0.024 902 343 749 991 630 110 72 × 2 = 0 + 0.049 804 687 499 983 260 221 44;
  • 27) 0.049 804 687 499 983 260 221 44 × 2 = 0 + 0.099 609 374 999 966 520 442 88;
  • 28) 0.099 609 374 999 966 520 442 88 × 2 = 0 + 0.199 218 749 999 933 040 885 76;
  • 29) 0.199 218 749 999 933 040 885 76 × 2 = 0 + 0.398 437 499 999 866 081 771 52;
  • 30) 0.398 437 499 999 866 081 771 52 × 2 = 0 + 0.796 874 999 999 732 163 543 04;
  • 31) 0.796 874 999 999 732 163 543 04 × 2 = 1 + 0.593 749 999 999 464 327 086 08;
  • 32) 0.593 749 999 999 464 327 086 08 × 2 = 1 + 0.187 499 999 998 928 654 172 16;
  • 33) 0.187 499 999 998 928 654 172 16 × 2 = 0 + 0.374 999 999 997 857 308 344 32;
  • 34) 0.374 999 999 997 857 308 344 32 × 2 = 0 + 0.749 999 999 995 714 616 688 64;
  • 35) 0.749 999 999 995 714 616 688 64 × 2 = 1 + 0.499 999 999 991 429 233 377 28;
  • 36) 0.499 999 999 991 429 233 377 28 × 2 = 0 + 0.999 999 999 982 858 466 754 56;
  • 37) 0.999 999 999 982 858 466 754 56 × 2 = 1 + 0.999 999 999 965 716 933 509 12;
  • 38) 0.999 999 999 965 716 933 509 12 × 2 = 1 + 0.999 999 999 931 433 867 018 24;
  • 39) 0.999 999 999 931 433 867 018 24 × 2 = 1 + 0.999 999 999 862 867 734 036 48;
  • 40) 0.999 999 999 862 867 734 036 48 × 2 = 1 + 0.999 999 999 725 735 468 072 96;
  • 41) 0.999 999 999 725 735 468 072 96 × 2 = 1 + 0.999 999 999 451 470 936 145 92;
  • 42) 0.999 999 999 451 470 936 145 92 × 2 = 1 + 0.999 999 998 902 941 872 291 84;
  • 43) 0.999 999 998 902 941 872 291 84 × 2 = 1 + 0.999 999 997 805 883 744 583 68;
  • 44) 0.999 999 997 805 883 744 583 68 × 2 = 1 + 0.999 999 995 611 767 489 167 36;
  • 45) 0.999 999 995 611 767 489 167 36 × 2 = 1 + 0.999 999 991 223 534 978 334 72;
  • 46) 0.999 999 991 223 534 978 334 72 × 2 = 1 + 0.999 999 982 447 069 956 669 44;
  • 47) 0.999 999 982 447 069 956 669 44 × 2 = 1 + 0.999 999 964 894 139 913 338 88;
  • 48) 0.999 999 964 894 139 913 338 88 × 2 = 1 + 0.999 999 929 788 279 826 677 76;
  • 49) 0.999 999 929 788 279 826 677 76 × 2 = 1 + 0.999 999 859 576 559 653 355 52;
  • 50) 0.999 999 859 576 559 653 355 52 × 2 = 1 + 0.999 999 719 153 119 306 711 04;
  • 51) 0.999 999 719 153 119 306 711 04 × 2 = 1 + 0.999 999 438 306 238 613 422 08;
  • 52) 0.999 999 438 306 238 613 422 08 × 2 = 1 + 0.999 998 876 612 477 226 844 16;
  • 53) 0.999 998 876 612 477 226 844 16 × 2 = 1 + 0.999 997 753 224 954 453 688 32;
  • 54) 0.999 997 753 224 954 453 688 32 × 2 = 1 + 0.999 995 506 449 908 907 376 64;

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:


0.000 000 000 742 147 676 646 46(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 46(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:


0.000 000 000 742 147 676 646 46(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 -0.000 000 000 742 147 676 646 46 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