-0.000 000 000 742 147 676 639 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 639 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
-0.000 000 000 742 147 676 639 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:

|-0.000 000 000 742 147 676 639 8| = 0.000 000 000 742 147 676 639 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: 0.000 000 000 742 147 676 639 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) 0.000 000 000 742 147 676 639 8 × 2 = 0 + 0.000 000 001 484 295 353 279 6;
  • 2) 0.000 000 001 484 295 353 279 6 × 2 = 0 + 0.000 000 002 968 590 706 559 2;
  • 3) 0.000 000 002 968 590 706 559 2 × 2 = 0 + 0.000 000 005 937 181 413 118 4;
  • 4) 0.000 000 005 937 181 413 118 4 × 2 = 0 + 0.000 000 011 874 362 826 236 8;
  • 5) 0.000 000 011 874 362 826 236 8 × 2 = 0 + 0.000 000 023 748 725 652 473 6;
  • 6) 0.000 000 023 748 725 652 473 6 × 2 = 0 + 0.000 000 047 497 451 304 947 2;
  • 7) 0.000 000 047 497 451 304 947 2 × 2 = 0 + 0.000 000 094 994 902 609 894 4;
  • 8) 0.000 000 094 994 902 609 894 4 × 2 = 0 + 0.000 000 189 989 805 219 788 8;
  • 9) 0.000 000 189 989 805 219 788 8 × 2 = 0 + 0.000 000 379 979 610 439 577 6;
  • 10) 0.000 000 379 979 610 439 577 6 × 2 = 0 + 0.000 000 759 959 220 879 155 2;
  • 11) 0.000 000 759 959 220 879 155 2 × 2 = 0 + 0.000 001 519 918 441 758 310 4;
  • 12) 0.000 001 519 918 441 758 310 4 × 2 = 0 + 0.000 003 039 836 883 516 620 8;
  • 13) 0.000 003 039 836 883 516 620 8 × 2 = 0 + 0.000 006 079 673 767 033 241 6;
  • 14) 0.000 006 079 673 767 033 241 6 × 2 = 0 + 0.000 012 159 347 534 066 483 2;
  • 15) 0.000 012 159 347 534 066 483 2 × 2 = 0 + 0.000 024 318 695 068 132 966 4;
  • 16) 0.000 024 318 695 068 132 966 4 × 2 = 0 + 0.000 048 637 390 136 265 932 8;
  • 17) 0.000 048 637 390 136 265 932 8 × 2 = 0 + 0.000 097 274 780 272 531 865 6;
  • 18) 0.000 097 274 780 272 531 865 6 × 2 = 0 + 0.000 194 549 560 545 063 731 2;
  • 19) 0.000 194 549 560 545 063 731 2 × 2 = 0 + 0.000 389 099 121 090 127 462 4;
  • 20) 0.000 389 099 121 090 127 462 4 × 2 = 0 + 0.000 778 198 242 180 254 924 8;
  • 21) 0.000 778 198 242 180 254 924 8 × 2 = 0 + 0.001 556 396 484 360 509 849 6;
  • 22) 0.001 556 396 484 360 509 849 6 × 2 = 0 + 0.003 112 792 968 721 019 699 2;
  • 23) 0.003 112 792 968 721 019 699 2 × 2 = 0 + 0.006 225 585 937 442 039 398 4;
  • 24) 0.006 225 585 937 442 039 398 4 × 2 = 0 + 0.012 451 171 874 884 078 796 8;
  • 25) 0.012 451 171 874 884 078 796 8 × 2 = 0 + 0.024 902 343 749 768 157 593 6;
  • 26) 0.024 902 343 749 768 157 593 6 × 2 = 0 + 0.049 804 687 499 536 315 187 2;
  • 27) 0.049 804 687 499 536 315 187 2 × 2 = 0 + 0.099 609 374 999 072 630 374 4;
  • 28) 0.099 609 374 999 072 630 374 4 × 2 = 0 + 0.199 218 749 998 145 260 748 8;
  • 29) 0.199 218 749 998 145 260 748 8 × 2 = 0 + 0.398 437 499 996 290 521 497 6;
  • 30) 0.398 437 499 996 290 521 497 6 × 2 = 0 + 0.796 874 999 992 581 042 995 2;
  • 31) 0.796 874 999 992 581 042 995 2 × 2 = 1 + 0.593 749 999 985 162 085 990 4;
  • 32) 0.593 749 999 985 162 085 990 4 × 2 = 1 + 0.187 499 999 970 324 171 980 8;
  • 33) 0.187 499 999 970 324 171 980 8 × 2 = 0 + 0.374 999 999 940 648 343 961 6;
  • 34) 0.374 999 999 940 648 343 961 6 × 2 = 0 + 0.749 999 999 881 296 687 923 2;
  • 35) 0.749 999 999 881 296 687 923 2 × 2 = 1 + 0.499 999 999 762 593 375 846 4;
  • 36) 0.499 999 999 762 593 375 846 4 × 2 = 0 + 0.999 999 999 525 186 751 692 8;
  • 37) 0.999 999 999 525 186 751 692 8 × 2 = 1 + 0.999 999 999 050 373 503 385 6;
  • 38) 0.999 999 999 050 373 503 385 6 × 2 = 1 + 0.999 999 998 100 747 006 771 2;
  • 39) 0.999 999 998 100 747 006 771 2 × 2 = 1 + 0.999 999 996 201 494 013 542 4;
  • 40) 0.999 999 996 201 494 013 542 4 × 2 = 1 + 0.999 999 992 402 988 027 084 8;
  • 41) 0.999 999 992 402 988 027 084 8 × 2 = 1 + 0.999 999 984 805 976 054 169 6;
  • 42) 0.999 999 984 805 976 054 169 6 × 2 = 1 + 0.999 999 969 611 952 108 339 2;
  • 43) 0.999 999 969 611 952 108 339 2 × 2 = 1 + 0.999 999 939 223 904 216 678 4;
  • 44) 0.999 999 939 223 904 216 678 4 × 2 = 1 + 0.999 999 878 447 808 433 356 8;
  • 45) 0.999 999 878 447 808 433 356 8 × 2 = 1 + 0.999 999 756 895 616 866 713 6;
  • 46) 0.999 999 756 895 616 866 713 6 × 2 = 1 + 0.999 999 513 791 233 733 427 2;
  • 47) 0.999 999 513 791 233 733 427 2 × 2 = 1 + 0.999 999 027 582 467 466 854 4;
  • 48) 0.999 999 027 582 467 466 854 4 × 2 = 1 + 0.999 998 055 164 934 933 708 8;
  • 49) 0.999 998 055 164 934 933 708 8 × 2 = 1 + 0.999 996 110 329 869 867 417 6;
  • 50) 0.999 996 110 329 869 867 417 6 × 2 = 1 + 0.999 992 220 659 739 734 835 2;
  • 51) 0.999 992 220 659 739 734 835 2 × 2 = 1 + 0.999 984 441 319 479 469 670 4;
  • 52) 0.999 984 441 319 479 469 670 4 × 2 = 1 + 0.999 968 882 638 958 939 340 8;
  • 53) 0.999 968 882 638 958 939 340 8 × 2 = 1 + 0.999 937 765 277 917 878 681 6;
  • 54) 0.999 937 765 277 917 878 681 6 × 2 = 1 + 0.999 875 530 555 835 757 363 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:


0.000 000 000 742 147 676 639 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 639 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:


0.000 000 000 742 147 676 639 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 -0.000 000 000 742 147 676 639 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