-0.000 000 000 742 147 676 622 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 622 6(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
-0.000 000 000 742 147 676 622 6(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:

|-0.000 000 000 742 147 676 622 6| = 0.000 000 000 742 147 676 622 6


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: 0.000 000 000 742 147 676 622 6.

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) 0.000 000 000 742 147 676 622 6 × 2 = 0 + 0.000 000 001 484 295 353 245 2;
  • 2) 0.000 000 001 484 295 353 245 2 × 2 = 0 + 0.000 000 002 968 590 706 490 4;
  • 3) 0.000 000 002 968 590 706 490 4 × 2 = 0 + 0.000 000 005 937 181 412 980 8;
  • 4) 0.000 000 005 937 181 412 980 8 × 2 = 0 + 0.000 000 011 874 362 825 961 6;
  • 5) 0.000 000 011 874 362 825 961 6 × 2 = 0 + 0.000 000 023 748 725 651 923 2;
  • 6) 0.000 000 023 748 725 651 923 2 × 2 = 0 + 0.000 000 047 497 451 303 846 4;
  • 7) 0.000 000 047 497 451 303 846 4 × 2 = 0 + 0.000 000 094 994 902 607 692 8;
  • 8) 0.000 000 094 994 902 607 692 8 × 2 = 0 + 0.000 000 189 989 805 215 385 6;
  • 9) 0.000 000 189 989 805 215 385 6 × 2 = 0 + 0.000 000 379 979 610 430 771 2;
  • 10) 0.000 000 379 979 610 430 771 2 × 2 = 0 + 0.000 000 759 959 220 861 542 4;
  • 11) 0.000 000 759 959 220 861 542 4 × 2 = 0 + 0.000 001 519 918 441 723 084 8;
  • 12) 0.000 001 519 918 441 723 084 8 × 2 = 0 + 0.000 003 039 836 883 446 169 6;
  • 13) 0.000 003 039 836 883 446 169 6 × 2 = 0 + 0.000 006 079 673 766 892 339 2;
  • 14) 0.000 006 079 673 766 892 339 2 × 2 = 0 + 0.000 012 159 347 533 784 678 4;
  • 15) 0.000 012 159 347 533 784 678 4 × 2 = 0 + 0.000 024 318 695 067 569 356 8;
  • 16) 0.000 024 318 695 067 569 356 8 × 2 = 0 + 0.000 048 637 390 135 138 713 6;
  • 17) 0.000 048 637 390 135 138 713 6 × 2 = 0 + 0.000 097 274 780 270 277 427 2;
  • 18) 0.000 097 274 780 270 277 427 2 × 2 = 0 + 0.000 194 549 560 540 554 854 4;
  • 19) 0.000 194 549 560 540 554 854 4 × 2 = 0 + 0.000 389 099 121 081 109 708 8;
  • 20) 0.000 389 099 121 081 109 708 8 × 2 = 0 + 0.000 778 198 242 162 219 417 6;
  • 21) 0.000 778 198 242 162 219 417 6 × 2 = 0 + 0.001 556 396 484 324 438 835 2;
  • 22) 0.001 556 396 484 324 438 835 2 × 2 = 0 + 0.003 112 792 968 648 877 670 4;
  • 23) 0.003 112 792 968 648 877 670 4 × 2 = 0 + 0.006 225 585 937 297 755 340 8;
  • 24) 0.006 225 585 937 297 755 340 8 × 2 = 0 + 0.012 451 171 874 595 510 681 6;
  • 25) 0.012 451 171 874 595 510 681 6 × 2 = 0 + 0.024 902 343 749 191 021 363 2;
  • 26) 0.024 902 343 749 191 021 363 2 × 2 = 0 + 0.049 804 687 498 382 042 726 4;
  • 27) 0.049 804 687 498 382 042 726 4 × 2 = 0 + 0.099 609 374 996 764 085 452 8;
  • 28) 0.099 609 374 996 764 085 452 8 × 2 = 0 + 0.199 218 749 993 528 170 905 6;
  • 29) 0.199 218 749 993 528 170 905 6 × 2 = 0 + 0.398 437 499 987 056 341 811 2;
  • 30) 0.398 437 499 987 056 341 811 2 × 2 = 0 + 0.796 874 999 974 112 683 622 4;
  • 31) 0.796 874 999 974 112 683 622 4 × 2 = 1 + 0.593 749 999 948 225 367 244 8;
  • 32) 0.593 749 999 948 225 367 244 8 × 2 = 1 + 0.187 499 999 896 450 734 489 6;
  • 33) 0.187 499 999 896 450 734 489 6 × 2 = 0 + 0.374 999 999 792 901 468 979 2;
  • 34) 0.374 999 999 792 901 468 979 2 × 2 = 0 + 0.749 999 999 585 802 937 958 4;
  • 35) 0.749 999 999 585 802 937 958 4 × 2 = 1 + 0.499 999 999 171 605 875 916 8;
  • 36) 0.499 999 999 171 605 875 916 8 × 2 = 0 + 0.999 999 998 343 211 751 833 6;
  • 37) 0.999 999 998 343 211 751 833 6 × 2 = 1 + 0.999 999 996 686 423 503 667 2;
  • 38) 0.999 999 996 686 423 503 667 2 × 2 = 1 + 0.999 999 993 372 847 007 334 4;
  • 39) 0.999 999 993 372 847 007 334 4 × 2 = 1 + 0.999 999 986 745 694 014 668 8;
  • 40) 0.999 999 986 745 694 014 668 8 × 2 = 1 + 0.999 999 973 491 388 029 337 6;
  • 41) 0.999 999 973 491 388 029 337 6 × 2 = 1 + 0.999 999 946 982 776 058 675 2;
  • 42) 0.999 999 946 982 776 058 675 2 × 2 = 1 + 0.999 999 893 965 552 117 350 4;
  • 43) 0.999 999 893 965 552 117 350 4 × 2 = 1 + 0.999 999 787 931 104 234 700 8;
  • 44) 0.999 999 787 931 104 234 700 8 × 2 = 1 + 0.999 999 575 862 208 469 401 6;
  • 45) 0.999 999 575 862 208 469 401 6 × 2 = 1 + 0.999 999 151 724 416 938 803 2;
  • 46) 0.999 999 151 724 416 938 803 2 × 2 = 1 + 0.999 998 303 448 833 877 606 4;
  • 47) 0.999 998 303 448 833 877 606 4 × 2 = 1 + 0.999 996 606 897 667 755 212 8;
  • 48) 0.999 996 606 897 667 755 212 8 × 2 = 1 + 0.999 993 213 795 335 510 425 6;
  • 49) 0.999 993 213 795 335 510 425 6 × 2 = 1 + 0.999 986 427 590 671 020 851 2;
  • 50) 0.999 986 427 590 671 020 851 2 × 2 = 1 + 0.999 972 855 181 342 041 702 4;
  • 51) 0.999 972 855 181 342 041 702 4 × 2 = 1 + 0.999 945 710 362 684 083 404 8;
  • 52) 0.999 945 710 362 684 083 404 8 × 2 = 1 + 0.999 891 420 725 368 166 809 6;
  • 53) 0.999 891 420 725 368 166 809 6 × 2 = 1 + 0.999 782 841 450 736 333 619 2;
  • 54) 0.999 782 841 450 736 333 619 2 × 2 = 1 + 0.999 565 682 901 472 667 238 4;

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:


0.000 000 000 742 147 676 622 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 622 6(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:


0.000 000 000 742 147 676 622 6(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 -0.000 000 000 742 147 676 622 6 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