-0.000 000 000 742 147 676 646 687 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 687 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
-0.000 000 000 742 147 676 646 687 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:

|-0.000 000 000 742 147 676 646 687 6| = 0.000 000 000 742 147 676 646 687 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: 0.000 000 000 742 147 676 646 687 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) 0.000 000 000 742 147 676 646 687 6 × 2 = 0 + 0.000 000 001 484 295 353 293 375 2;
  • 2) 0.000 000 001 484 295 353 293 375 2 × 2 = 0 + 0.000 000 002 968 590 706 586 750 4;
  • 3) 0.000 000 002 968 590 706 586 750 4 × 2 = 0 + 0.000 000 005 937 181 413 173 500 8;
  • 4) 0.000 000 005 937 181 413 173 500 8 × 2 = 0 + 0.000 000 011 874 362 826 347 001 6;
  • 5) 0.000 000 011 874 362 826 347 001 6 × 2 = 0 + 0.000 000 023 748 725 652 694 003 2;
  • 6) 0.000 000 023 748 725 652 694 003 2 × 2 = 0 + 0.000 000 047 497 451 305 388 006 4;
  • 7) 0.000 000 047 497 451 305 388 006 4 × 2 = 0 + 0.000 000 094 994 902 610 776 012 8;
  • 8) 0.000 000 094 994 902 610 776 012 8 × 2 = 0 + 0.000 000 189 989 805 221 552 025 6;
  • 9) 0.000 000 189 989 805 221 552 025 6 × 2 = 0 + 0.000 000 379 979 610 443 104 051 2;
  • 10) 0.000 000 379 979 610 443 104 051 2 × 2 = 0 + 0.000 000 759 959 220 886 208 102 4;
  • 11) 0.000 000 759 959 220 886 208 102 4 × 2 = 0 + 0.000 001 519 918 441 772 416 204 8;
  • 12) 0.000 001 519 918 441 772 416 204 8 × 2 = 0 + 0.000 003 039 836 883 544 832 409 6;
  • 13) 0.000 003 039 836 883 544 832 409 6 × 2 = 0 + 0.000 006 079 673 767 089 664 819 2;
  • 14) 0.000 006 079 673 767 089 664 819 2 × 2 = 0 + 0.000 012 159 347 534 179 329 638 4;
  • 15) 0.000 012 159 347 534 179 329 638 4 × 2 = 0 + 0.000 024 318 695 068 358 659 276 8;
  • 16) 0.000 024 318 695 068 358 659 276 8 × 2 = 0 + 0.000 048 637 390 136 717 318 553 6;
  • 17) 0.000 048 637 390 136 717 318 553 6 × 2 = 0 + 0.000 097 274 780 273 434 637 107 2;
  • 18) 0.000 097 274 780 273 434 637 107 2 × 2 = 0 + 0.000 194 549 560 546 869 274 214 4;
  • 19) 0.000 194 549 560 546 869 274 214 4 × 2 = 0 + 0.000 389 099 121 093 738 548 428 8;
  • 20) 0.000 389 099 121 093 738 548 428 8 × 2 = 0 + 0.000 778 198 242 187 477 096 857 6;
  • 21) 0.000 778 198 242 187 477 096 857 6 × 2 = 0 + 0.001 556 396 484 374 954 193 715 2;
  • 22) 0.001 556 396 484 374 954 193 715 2 × 2 = 0 + 0.003 112 792 968 749 908 387 430 4;
  • 23) 0.003 112 792 968 749 908 387 430 4 × 2 = 0 + 0.006 225 585 937 499 816 774 860 8;
  • 24) 0.006 225 585 937 499 816 774 860 8 × 2 = 0 + 0.012 451 171 874 999 633 549 721 6;
  • 25) 0.012 451 171 874 999 633 549 721 6 × 2 = 0 + 0.024 902 343 749 999 267 099 443 2;
  • 26) 0.024 902 343 749 999 267 099 443 2 × 2 = 0 + 0.049 804 687 499 998 534 198 886 4;
  • 27) 0.049 804 687 499 998 534 198 886 4 × 2 = 0 + 0.099 609 374 999 997 068 397 772 8;
  • 28) 0.099 609 374 999 997 068 397 772 8 × 2 = 0 + 0.199 218 749 999 994 136 795 545 6;
  • 29) 0.199 218 749 999 994 136 795 545 6 × 2 = 0 + 0.398 437 499 999 988 273 591 091 2;
  • 30) 0.398 437 499 999 988 273 591 091 2 × 2 = 0 + 0.796 874 999 999 976 547 182 182 4;
  • 31) 0.796 874 999 999 976 547 182 182 4 × 2 = 1 + 0.593 749 999 999 953 094 364 364 8;
  • 32) 0.593 749 999 999 953 094 364 364 8 × 2 = 1 + 0.187 499 999 999 906 188 728 729 6;
  • 33) 0.187 499 999 999 906 188 728 729 6 × 2 = 0 + 0.374 999 999 999 812 377 457 459 2;
  • 34) 0.374 999 999 999 812 377 457 459 2 × 2 = 0 + 0.749 999 999 999 624 754 914 918 4;
  • 35) 0.749 999 999 999 624 754 914 918 4 × 2 = 1 + 0.499 999 999 999 249 509 829 836 8;
  • 36) 0.499 999 999 999 249 509 829 836 8 × 2 = 0 + 0.999 999 999 998 499 019 659 673 6;
  • 37) 0.999 999 999 998 499 019 659 673 6 × 2 = 1 + 0.999 999 999 996 998 039 319 347 2;
  • 38) 0.999 999 999 996 998 039 319 347 2 × 2 = 1 + 0.999 999 999 993 996 078 638 694 4;
  • 39) 0.999 999 999 993 996 078 638 694 4 × 2 = 1 + 0.999 999 999 987 992 157 277 388 8;
  • 40) 0.999 999 999 987 992 157 277 388 8 × 2 = 1 + 0.999 999 999 975 984 314 554 777 6;
  • 41) 0.999 999 999 975 984 314 554 777 6 × 2 = 1 + 0.999 999 999 951 968 629 109 555 2;
  • 42) 0.999 999 999 951 968 629 109 555 2 × 2 = 1 + 0.999 999 999 903 937 258 219 110 4;
  • 43) 0.999 999 999 903 937 258 219 110 4 × 2 = 1 + 0.999 999 999 807 874 516 438 220 8;
  • 44) 0.999 999 999 807 874 516 438 220 8 × 2 = 1 + 0.999 999 999 615 749 032 876 441 6;
  • 45) 0.999 999 999 615 749 032 876 441 6 × 2 = 1 + 0.999 999 999 231 498 065 752 883 2;
  • 46) 0.999 999 999 231 498 065 752 883 2 × 2 = 1 + 0.999 999 998 462 996 131 505 766 4;
  • 47) 0.999 999 998 462 996 131 505 766 4 × 2 = 1 + 0.999 999 996 925 992 263 011 532 8;
  • 48) 0.999 999 996 925 992 263 011 532 8 × 2 = 1 + 0.999 999 993 851 984 526 023 065 6;
  • 49) 0.999 999 993 851 984 526 023 065 6 × 2 = 1 + 0.999 999 987 703 969 052 046 131 2;
  • 50) 0.999 999 987 703 969 052 046 131 2 × 2 = 1 + 0.999 999 975 407 938 104 092 262 4;
  • 51) 0.999 999 975 407 938 104 092 262 4 × 2 = 1 + 0.999 999 950 815 876 208 184 524 8;
  • 52) 0.999 999 950 815 876 208 184 524 8 × 2 = 1 + 0.999 999 901 631 752 416 369 049 6;
  • 53) 0.999 999 901 631 752 416 369 049 6 × 2 = 1 + 0.999 999 803 263 504 832 738 099 2;
  • 54) 0.999 999 803 263 504 832 738 099 2 × 2 = 1 + 0.999 999 606 527 009 665 476 198 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:


0.000 000 000 742 147 676 646 687 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 687 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:


0.000 000 000 742 147 676 646 687 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 -0.000 000 000 742 147 676 646 687 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