-0.000 000 000 742 147 676 637 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 637 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
-0.000 000 000 742 147 676 637 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:

|-0.000 000 000 742 147 676 637 8| = 0.000 000 000 742 147 676 637 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: 0.000 000 000 742 147 676 637 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) 0.000 000 000 742 147 676 637 8 × 2 = 0 + 0.000 000 001 484 295 353 275 6;
  • 2) 0.000 000 001 484 295 353 275 6 × 2 = 0 + 0.000 000 002 968 590 706 551 2;
  • 3) 0.000 000 002 968 590 706 551 2 × 2 = 0 + 0.000 000 005 937 181 413 102 4;
  • 4) 0.000 000 005 937 181 413 102 4 × 2 = 0 + 0.000 000 011 874 362 826 204 8;
  • 5) 0.000 000 011 874 362 826 204 8 × 2 = 0 + 0.000 000 023 748 725 652 409 6;
  • 6) 0.000 000 023 748 725 652 409 6 × 2 = 0 + 0.000 000 047 497 451 304 819 2;
  • 7) 0.000 000 047 497 451 304 819 2 × 2 = 0 + 0.000 000 094 994 902 609 638 4;
  • 8) 0.000 000 094 994 902 609 638 4 × 2 = 0 + 0.000 000 189 989 805 219 276 8;
  • 9) 0.000 000 189 989 805 219 276 8 × 2 = 0 + 0.000 000 379 979 610 438 553 6;
  • 10) 0.000 000 379 979 610 438 553 6 × 2 = 0 + 0.000 000 759 959 220 877 107 2;
  • 11) 0.000 000 759 959 220 877 107 2 × 2 = 0 + 0.000 001 519 918 441 754 214 4;
  • 12) 0.000 001 519 918 441 754 214 4 × 2 = 0 + 0.000 003 039 836 883 508 428 8;
  • 13) 0.000 003 039 836 883 508 428 8 × 2 = 0 + 0.000 006 079 673 767 016 857 6;
  • 14) 0.000 006 079 673 767 016 857 6 × 2 = 0 + 0.000 012 159 347 534 033 715 2;
  • 15) 0.000 012 159 347 534 033 715 2 × 2 = 0 + 0.000 024 318 695 068 067 430 4;
  • 16) 0.000 024 318 695 068 067 430 4 × 2 = 0 + 0.000 048 637 390 136 134 860 8;
  • 17) 0.000 048 637 390 136 134 860 8 × 2 = 0 + 0.000 097 274 780 272 269 721 6;
  • 18) 0.000 097 274 780 272 269 721 6 × 2 = 0 + 0.000 194 549 560 544 539 443 2;
  • 19) 0.000 194 549 560 544 539 443 2 × 2 = 0 + 0.000 389 099 121 089 078 886 4;
  • 20) 0.000 389 099 121 089 078 886 4 × 2 = 0 + 0.000 778 198 242 178 157 772 8;
  • 21) 0.000 778 198 242 178 157 772 8 × 2 = 0 + 0.001 556 396 484 356 315 545 6;
  • 22) 0.001 556 396 484 356 315 545 6 × 2 = 0 + 0.003 112 792 968 712 631 091 2;
  • 23) 0.003 112 792 968 712 631 091 2 × 2 = 0 + 0.006 225 585 937 425 262 182 4;
  • 24) 0.006 225 585 937 425 262 182 4 × 2 = 0 + 0.012 451 171 874 850 524 364 8;
  • 25) 0.012 451 171 874 850 524 364 8 × 2 = 0 + 0.024 902 343 749 701 048 729 6;
  • 26) 0.024 902 343 749 701 048 729 6 × 2 = 0 + 0.049 804 687 499 402 097 459 2;
  • 27) 0.049 804 687 499 402 097 459 2 × 2 = 0 + 0.099 609 374 998 804 194 918 4;
  • 28) 0.099 609 374 998 804 194 918 4 × 2 = 0 + 0.199 218 749 997 608 389 836 8;
  • 29) 0.199 218 749 997 608 389 836 8 × 2 = 0 + 0.398 437 499 995 216 779 673 6;
  • 30) 0.398 437 499 995 216 779 673 6 × 2 = 0 + 0.796 874 999 990 433 559 347 2;
  • 31) 0.796 874 999 990 433 559 347 2 × 2 = 1 + 0.593 749 999 980 867 118 694 4;
  • 32) 0.593 749 999 980 867 118 694 4 × 2 = 1 + 0.187 499 999 961 734 237 388 8;
  • 33) 0.187 499 999 961 734 237 388 8 × 2 = 0 + 0.374 999 999 923 468 474 777 6;
  • 34) 0.374 999 999 923 468 474 777 6 × 2 = 0 + 0.749 999 999 846 936 949 555 2;
  • 35) 0.749 999 999 846 936 949 555 2 × 2 = 1 + 0.499 999 999 693 873 899 110 4;
  • 36) 0.499 999 999 693 873 899 110 4 × 2 = 0 + 0.999 999 999 387 747 798 220 8;
  • 37) 0.999 999 999 387 747 798 220 8 × 2 = 1 + 0.999 999 998 775 495 596 441 6;
  • 38) 0.999 999 998 775 495 596 441 6 × 2 = 1 + 0.999 999 997 550 991 192 883 2;
  • 39) 0.999 999 997 550 991 192 883 2 × 2 = 1 + 0.999 999 995 101 982 385 766 4;
  • 40) 0.999 999 995 101 982 385 766 4 × 2 = 1 + 0.999 999 990 203 964 771 532 8;
  • 41) 0.999 999 990 203 964 771 532 8 × 2 = 1 + 0.999 999 980 407 929 543 065 6;
  • 42) 0.999 999 980 407 929 543 065 6 × 2 = 1 + 0.999 999 960 815 859 086 131 2;
  • 43) 0.999 999 960 815 859 086 131 2 × 2 = 1 + 0.999 999 921 631 718 172 262 4;
  • 44) 0.999 999 921 631 718 172 262 4 × 2 = 1 + 0.999 999 843 263 436 344 524 8;
  • 45) 0.999 999 843 263 436 344 524 8 × 2 = 1 + 0.999 999 686 526 872 689 049 6;
  • 46) 0.999 999 686 526 872 689 049 6 × 2 = 1 + 0.999 999 373 053 745 378 099 2;
  • 47) 0.999 999 373 053 745 378 099 2 × 2 = 1 + 0.999 998 746 107 490 756 198 4;
  • 48) 0.999 998 746 107 490 756 198 4 × 2 = 1 + 0.999 997 492 214 981 512 396 8;
  • 49) 0.999 997 492 214 981 512 396 8 × 2 = 1 + 0.999 994 984 429 963 024 793 6;
  • 50) 0.999 994 984 429 963 024 793 6 × 2 = 1 + 0.999 989 968 859 926 049 587 2;
  • 51) 0.999 989 968 859 926 049 587 2 × 2 = 1 + 0.999 979 937 719 852 099 174 4;
  • 52) 0.999 979 937 719 852 099 174 4 × 2 = 1 + 0.999 959 875 439 704 198 348 8;
  • 53) 0.999 959 875 439 704 198 348 8 × 2 = 1 + 0.999 919 750 879 408 396 697 6;
  • 54) 0.999 919 750 879 408 396 697 6 × 2 = 1 + 0.999 839 501 758 816 793 395 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:


0.000 000 000 742 147 676 637 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 637 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:


0.000 000 000 742 147 676 637 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 -0.000 000 000 742 147 676 637 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