-0.000 000 000 742 147 676 623 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 623 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
-0.000 000 000 742 147 676 623 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:

|-0.000 000 000 742 147 676 623 8| = 0.000 000 000 742 147 676 623 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: 0.000 000 000 742 147 676 623 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) 0.000 000 000 742 147 676 623 8 × 2 = 0 + 0.000 000 001 484 295 353 247 6;
  • 2) 0.000 000 001 484 295 353 247 6 × 2 = 0 + 0.000 000 002 968 590 706 495 2;
  • 3) 0.000 000 002 968 590 706 495 2 × 2 = 0 + 0.000 000 005 937 181 412 990 4;
  • 4) 0.000 000 005 937 181 412 990 4 × 2 = 0 + 0.000 000 011 874 362 825 980 8;
  • 5) 0.000 000 011 874 362 825 980 8 × 2 = 0 + 0.000 000 023 748 725 651 961 6;
  • 6) 0.000 000 023 748 725 651 961 6 × 2 = 0 + 0.000 000 047 497 451 303 923 2;
  • 7) 0.000 000 047 497 451 303 923 2 × 2 = 0 + 0.000 000 094 994 902 607 846 4;
  • 8) 0.000 000 094 994 902 607 846 4 × 2 = 0 + 0.000 000 189 989 805 215 692 8;
  • 9) 0.000 000 189 989 805 215 692 8 × 2 = 0 + 0.000 000 379 979 610 431 385 6;
  • 10) 0.000 000 379 979 610 431 385 6 × 2 = 0 + 0.000 000 759 959 220 862 771 2;
  • 11) 0.000 000 759 959 220 862 771 2 × 2 = 0 + 0.000 001 519 918 441 725 542 4;
  • 12) 0.000 001 519 918 441 725 542 4 × 2 = 0 + 0.000 003 039 836 883 451 084 8;
  • 13) 0.000 003 039 836 883 451 084 8 × 2 = 0 + 0.000 006 079 673 766 902 169 6;
  • 14) 0.000 006 079 673 766 902 169 6 × 2 = 0 + 0.000 012 159 347 533 804 339 2;
  • 15) 0.000 012 159 347 533 804 339 2 × 2 = 0 + 0.000 024 318 695 067 608 678 4;
  • 16) 0.000 024 318 695 067 608 678 4 × 2 = 0 + 0.000 048 637 390 135 217 356 8;
  • 17) 0.000 048 637 390 135 217 356 8 × 2 = 0 + 0.000 097 274 780 270 434 713 6;
  • 18) 0.000 097 274 780 270 434 713 6 × 2 = 0 + 0.000 194 549 560 540 869 427 2;
  • 19) 0.000 194 549 560 540 869 427 2 × 2 = 0 + 0.000 389 099 121 081 738 854 4;
  • 20) 0.000 389 099 121 081 738 854 4 × 2 = 0 + 0.000 778 198 242 163 477 708 8;
  • 21) 0.000 778 198 242 163 477 708 8 × 2 = 0 + 0.001 556 396 484 326 955 417 6;
  • 22) 0.001 556 396 484 326 955 417 6 × 2 = 0 + 0.003 112 792 968 653 910 835 2;
  • 23) 0.003 112 792 968 653 910 835 2 × 2 = 0 + 0.006 225 585 937 307 821 670 4;
  • 24) 0.006 225 585 937 307 821 670 4 × 2 = 0 + 0.012 451 171 874 615 643 340 8;
  • 25) 0.012 451 171 874 615 643 340 8 × 2 = 0 + 0.024 902 343 749 231 286 681 6;
  • 26) 0.024 902 343 749 231 286 681 6 × 2 = 0 + 0.049 804 687 498 462 573 363 2;
  • 27) 0.049 804 687 498 462 573 363 2 × 2 = 0 + 0.099 609 374 996 925 146 726 4;
  • 28) 0.099 609 374 996 925 146 726 4 × 2 = 0 + 0.199 218 749 993 850 293 452 8;
  • 29) 0.199 218 749 993 850 293 452 8 × 2 = 0 + 0.398 437 499 987 700 586 905 6;
  • 30) 0.398 437 499 987 700 586 905 6 × 2 = 0 + 0.796 874 999 975 401 173 811 2;
  • 31) 0.796 874 999 975 401 173 811 2 × 2 = 1 + 0.593 749 999 950 802 347 622 4;
  • 32) 0.593 749 999 950 802 347 622 4 × 2 = 1 + 0.187 499 999 901 604 695 244 8;
  • 33) 0.187 499 999 901 604 695 244 8 × 2 = 0 + 0.374 999 999 803 209 390 489 6;
  • 34) 0.374 999 999 803 209 390 489 6 × 2 = 0 + 0.749 999 999 606 418 780 979 2;
  • 35) 0.749 999 999 606 418 780 979 2 × 2 = 1 + 0.499 999 999 212 837 561 958 4;
  • 36) 0.499 999 999 212 837 561 958 4 × 2 = 0 + 0.999 999 998 425 675 123 916 8;
  • 37) 0.999 999 998 425 675 123 916 8 × 2 = 1 + 0.999 999 996 851 350 247 833 6;
  • 38) 0.999 999 996 851 350 247 833 6 × 2 = 1 + 0.999 999 993 702 700 495 667 2;
  • 39) 0.999 999 993 702 700 495 667 2 × 2 = 1 + 0.999 999 987 405 400 991 334 4;
  • 40) 0.999 999 987 405 400 991 334 4 × 2 = 1 + 0.999 999 974 810 801 982 668 8;
  • 41) 0.999 999 974 810 801 982 668 8 × 2 = 1 + 0.999 999 949 621 603 965 337 6;
  • 42) 0.999 999 949 621 603 965 337 6 × 2 = 1 + 0.999 999 899 243 207 930 675 2;
  • 43) 0.999 999 899 243 207 930 675 2 × 2 = 1 + 0.999 999 798 486 415 861 350 4;
  • 44) 0.999 999 798 486 415 861 350 4 × 2 = 1 + 0.999 999 596 972 831 722 700 8;
  • 45) 0.999 999 596 972 831 722 700 8 × 2 = 1 + 0.999 999 193 945 663 445 401 6;
  • 46) 0.999 999 193 945 663 445 401 6 × 2 = 1 + 0.999 998 387 891 326 890 803 2;
  • 47) 0.999 998 387 891 326 890 803 2 × 2 = 1 + 0.999 996 775 782 653 781 606 4;
  • 48) 0.999 996 775 782 653 781 606 4 × 2 = 1 + 0.999 993 551 565 307 563 212 8;
  • 49) 0.999 993 551 565 307 563 212 8 × 2 = 1 + 0.999 987 103 130 615 126 425 6;
  • 50) 0.999 987 103 130 615 126 425 6 × 2 = 1 + 0.999 974 206 261 230 252 851 2;
  • 51) 0.999 974 206 261 230 252 851 2 × 2 = 1 + 0.999 948 412 522 460 505 702 4;
  • 52) 0.999 948 412 522 460 505 702 4 × 2 = 1 + 0.999 896 825 044 921 011 404 8;
  • 53) 0.999 896 825 044 921 011 404 8 × 2 = 1 + 0.999 793 650 089 842 022 809 6;
  • 54) 0.999 793 650 089 842 022 809 6 × 2 = 1 + 0.999 587 300 179 684 045 619 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:


0.000 000 000 742 147 676 623 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 623 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:


0.000 000 000 742 147 676 623 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 -0.000 000 000 742 147 676 623 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