-0.000 000 000 742 147 676 646 688 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 688 3(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
-0.000 000 000 742 147 676 646 688 3(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:

|-0.000 000 000 742 147 676 646 688 3| = 0.000 000 000 742 147 676 646 688 3


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: 0.000 000 000 742 147 676 646 688 3.

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) 0.000 000 000 742 147 676 646 688 3 × 2 = 0 + 0.000 000 001 484 295 353 293 376 6;
  • 2) 0.000 000 001 484 295 353 293 376 6 × 2 = 0 + 0.000 000 002 968 590 706 586 753 2;
  • 3) 0.000 000 002 968 590 706 586 753 2 × 2 = 0 + 0.000 000 005 937 181 413 173 506 4;
  • 4) 0.000 000 005 937 181 413 173 506 4 × 2 = 0 + 0.000 000 011 874 362 826 347 012 8;
  • 5) 0.000 000 011 874 362 826 347 012 8 × 2 = 0 + 0.000 000 023 748 725 652 694 025 6;
  • 6) 0.000 000 023 748 725 652 694 025 6 × 2 = 0 + 0.000 000 047 497 451 305 388 051 2;
  • 7) 0.000 000 047 497 451 305 388 051 2 × 2 = 0 + 0.000 000 094 994 902 610 776 102 4;
  • 8) 0.000 000 094 994 902 610 776 102 4 × 2 = 0 + 0.000 000 189 989 805 221 552 204 8;
  • 9) 0.000 000 189 989 805 221 552 204 8 × 2 = 0 + 0.000 000 379 979 610 443 104 409 6;
  • 10) 0.000 000 379 979 610 443 104 409 6 × 2 = 0 + 0.000 000 759 959 220 886 208 819 2;
  • 11) 0.000 000 759 959 220 886 208 819 2 × 2 = 0 + 0.000 001 519 918 441 772 417 638 4;
  • 12) 0.000 001 519 918 441 772 417 638 4 × 2 = 0 + 0.000 003 039 836 883 544 835 276 8;
  • 13) 0.000 003 039 836 883 544 835 276 8 × 2 = 0 + 0.000 006 079 673 767 089 670 553 6;
  • 14) 0.000 006 079 673 767 089 670 553 6 × 2 = 0 + 0.000 012 159 347 534 179 341 107 2;
  • 15) 0.000 012 159 347 534 179 341 107 2 × 2 = 0 + 0.000 024 318 695 068 358 682 214 4;
  • 16) 0.000 024 318 695 068 358 682 214 4 × 2 = 0 + 0.000 048 637 390 136 717 364 428 8;
  • 17) 0.000 048 637 390 136 717 364 428 8 × 2 = 0 + 0.000 097 274 780 273 434 728 857 6;
  • 18) 0.000 097 274 780 273 434 728 857 6 × 2 = 0 + 0.000 194 549 560 546 869 457 715 2;
  • 19) 0.000 194 549 560 546 869 457 715 2 × 2 = 0 + 0.000 389 099 121 093 738 915 430 4;
  • 20) 0.000 389 099 121 093 738 915 430 4 × 2 = 0 + 0.000 778 198 242 187 477 830 860 8;
  • 21) 0.000 778 198 242 187 477 830 860 8 × 2 = 0 + 0.001 556 396 484 374 955 661 721 6;
  • 22) 0.001 556 396 484 374 955 661 721 6 × 2 = 0 + 0.003 112 792 968 749 911 323 443 2;
  • 23) 0.003 112 792 968 749 911 323 443 2 × 2 = 0 + 0.006 225 585 937 499 822 646 886 4;
  • 24) 0.006 225 585 937 499 822 646 886 4 × 2 = 0 + 0.012 451 171 874 999 645 293 772 8;
  • 25) 0.012 451 171 874 999 645 293 772 8 × 2 = 0 + 0.024 902 343 749 999 290 587 545 6;
  • 26) 0.024 902 343 749 999 290 587 545 6 × 2 = 0 + 0.049 804 687 499 998 581 175 091 2;
  • 27) 0.049 804 687 499 998 581 175 091 2 × 2 = 0 + 0.099 609 374 999 997 162 350 182 4;
  • 28) 0.099 609 374 999 997 162 350 182 4 × 2 = 0 + 0.199 218 749 999 994 324 700 364 8;
  • 29) 0.199 218 749 999 994 324 700 364 8 × 2 = 0 + 0.398 437 499 999 988 649 400 729 6;
  • 30) 0.398 437 499 999 988 649 400 729 6 × 2 = 0 + 0.796 874 999 999 977 298 801 459 2;
  • 31) 0.796 874 999 999 977 298 801 459 2 × 2 = 1 + 0.593 749 999 999 954 597 602 918 4;
  • 32) 0.593 749 999 999 954 597 602 918 4 × 2 = 1 + 0.187 499 999 999 909 195 205 836 8;
  • 33) 0.187 499 999 999 909 195 205 836 8 × 2 = 0 + 0.374 999 999 999 818 390 411 673 6;
  • 34) 0.374 999 999 999 818 390 411 673 6 × 2 = 0 + 0.749 999 999 999 636 780 823 347 2;
  • 35) 0.749 999 999 999 636 780 823 347 2 × 2 = 1 + 0.499 999 999 999 273 561 646 694 4;
  • 36) 0.499 999 999 999 273 561 646 694 4 × 2 = 0 + 0.999 999 999 998 547 123 293 388 8;
  • 37) 0.999 999 999 998 547 123 293 388 8 × 2 = 1 + 0.999 999 999 997 094 246 586 777 6;
  • 38) 0.999 999 999 997 094 246 586 777 6 × 2 = 1 + 0.999 999 999 994 188 493 173 555 2;
  • 39) 0.999 999 999 994 188 493 173 555 2 × 2 = 1 + 0.999 999 999 988 376 986 347 110 4;
  • 40) 0.999 999 999 988 376 986 347 110 4 × 2 = 1 + 0.999 999 999 976 753 972 694 220 8;
  • 41) 0.999 999 999 976 753 972 694 220 8 × 2 = 1 + 0.999 999 999 953 507 945 388 441 6;
  • 42) 0.999 999 999 953 507 945 388 441 6 × 2 = 1 + 0.999 999 999 907 015 890 776 883 2;
  • 43) 0.999 999 999 907 015 890 776 883 2 × 2 = 1 + 0.999 999 999 814 031 781 553 766 4;
  • 44) 0.999 999 999 814 031 781 553 766 4 × 2 = 1 + 0.999 999 999 628 063 563 107 532 8;
  • 45) 0.999 999 999 628 063 563 107 532 8 × 2 = 1 + 0.999 999 999 256 127 126 215 065 6;
  • 46) 0.999 999 999 256 127 126 215 065 6 × 2 = 1 + 0.999 999 998 512 254 252 430 131 2;
  • 47) 0.999 999 998 512 254 252 430 131 2 × 2 = 1 + 0.999 999 997 024 508 504 860 262 4;
  • 48) 0.999 999 997 024 508 504 860 262 4 × 2 = 1 + 0.999 999 994 049 017 009 720 524 8;
  • 49) 0.999 999 994 049 017 009 720 524 8 × 2 = 1 + 0.999 999 988 098 034 019 441 049 6;
  • 50) 0.999 999 988 098 034 019 441 049 6 × 2 = 1 + 0.999 999 976 196 068 038 882 099 2;
  • 51) 0.999 999 976 196 068 038 882 099 2 × 2 = 1 + 0.999 999 952 392 136 077 764 198 4;
  • 52) 0.999 999 952 392 136 077 764 198 4 × 2 = 1 + 0.999 999 904 784 272 155 528 396 8;
  • 53) 0.999 999 904 784 272 155 528 396 8 × 2 = 1 + 0.999 999 809 568 544 311 056 793 6;
  • 54) 0.999 999 809 568 544 311 056 793 6 × 2 = 1 + 0.999 999 619 137 088 622 113 587 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:


0.000 000 000 742 147 676 646 688 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 688 3(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:


0.000 000 000 742 147 676 646 688 3(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 -0.000 000 000 742 147 676 646 688 3 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