-0.000 000 000 742 147 656 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 656(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
-0.000 000 000 742 147 656(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:

|-0.000 000 000 742 147 656| = 0.000 000 000 742 147 656


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: 0.000 000 000 742 147 656.

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) 0.000 000 000 742 147 656 × 2 = 0 + 0.000 000 001 484 295 312;
  • 2) 0.000 000 001 484 295 312 × 2 = 0 + 0.000 000 002 968 590 624;
  • 3) 0.000 000 002 968 590 624 × 2 = 0 + 0.000 000 005 937 181 248;
  • 4) 0.000 000 005 937 181 248 × 2 = 0 + 0.000 000 011 874 362 496;
  • 5) 0.000 000 011 874 362 496 × 2 = 0 + 0.000 000 023 748 724 992;
  • 6) 0.000 000 023 748 724 992 × 2 = 0 + 0.000 000 047 497 449 984;
  • 7) 0.000 000 047 497 449 984 × 2 = 0 + 0.000 000 094 994 899 968;
  • 8) 0.000 000 094 994 899 968 × 2 = 0 + 0.000 000 189 989 799 936;
  • 9) 0.000 000 189 989 799 936 × 2 = 0 + 0.000 000 379 979 599 872;
  • 10) 0.000 000 379 979 599 872 × 2 = 0 + 0.000 000 759 959 199 744;
  • 11) 0.000 000 759 959 199 744 × 2 = 0 + 0.000 001 519 918 399 488;
  • 12) 0.000 001 519 918 399 488 × 2 = 0 + 0.000 003 039 836 798 976;
  • 13) 0.000 003 039 836 798 976 × 2 = 0 + 0.000 006 079 673 597 952;
  • 14) 0.000 006 079 673 597 952 × 2 = 0 + 0.000 012 159 347 195 904;
  • 15) 0.000 012 159 347 195 904 × 2 = 0 + 0.000 024 318 694 391 808;
  • 16) 0.000 024 318 694 391 808 × 2 = 0 + 0.000 048 637 388 783 616;
  • 17) 0.000 048 637 388 783 616 × 2 = 0 + 0.000 097 274 777 567 232;
  • 18) 0.000 097 274 777 567 232 × 2 = 0 + 0.000 194 549 555 134 464;
  • 19) 0.000 194 549 555 134 464 × 2 = 0 + 0.000 389 099 110 268 928;
  • 20) 0.000 389 099 110 268 928 × 2 = 0 + 0.000 778 198 220 537 856;
  • 21) 0.000 778 198 220 537 856 × 2 = 0 + 0.001 556 396 441 075 712;
  • 22) 0.001 556 396 441 075 712 × 2 = 0 + 0.003 112 792 882 151 424;
  • 23) 0.003 112 792 882 151 424 × 2 = 0 + 0.006 225 585 764 302 848;
  • 24) 0.006 225 585 764 302 848 × 2 = 0 + 0.012 451 171 528 605 696;
  • 25) 0.012 451 171 528 605 696 × 2 = 0 + 0.024 902 343 057 211 392;
  • 26) 0.024 902 343 057 211 392 × 2 = 0 + 0.049 804 686 114 422 784;
  • 27) 0.049 804 686 114 422 784 × 2 = 0 + 0.099 609 372 228 845 568;
  • 28) 0.099 609 372 228 845 568 × 2 = 0 + 0.199 218 744 457 691 136;
  • 29) 0.199 218 744 457 691 136 × 2 = 0 + 0.398 437 488 915 382 272;
  • 30) 0.398 437 488 915 382 272 × 2 = 0 + 0.796 874 977 830 764 544;
  • 31) 0.796 874 977 830 764 544 × 2 = 1 + 0.593 749 955 661 529 088;
  • 32) 0.593 749 955 661 529 088 × 2 = 1 + 0.187 499 911 323 058 176;
  • 33) 0.187 499 911 323 058 176 × 2 = 0 + 0.374 999 822 646 116 352;
  • 34) 0.374 999 822 646 116 352 × 2 = 0 + 0.749 999 645 292 232 704;
  • 35) 0.749 999 645 292 232 704 × 2 = 1 + 0.499 999 290 584 465 408;
  • 36) 0.499 999 290 584 465 408 × 2 = 0 + 0.999 998 581 168 930 816;
  • 37) 0.999 998 581 168 930 816 × 2 = 1 + 0.999 997 162 337 861 632;
  • 38) 0.999 997 162 337 861 632 × 2 = 1 + 0.999 994 324 675 723 264;
  • 39) 0.999 994 324 675 723 264 × 2 = 1 + 0.999 988 649 351 446 528;
  • 40) 0.999 988 649 351 446 528 × 2 = 1 + 0.999 977 298 702 893 056;
  • 41) 0.999 977 298 702 893 056 × 2 = 1 + 0.999 954 597 405 786 112;
  • 42) 0.999 954 597 405 786 112 × 2 = 1 + 0.999 909 194 811 572 224;
  • 43) 0.999 909 194 811 572 224 × 2 = 1 + 0.999 818 389 623 144 448;
  • 44) 0.999 818 389 623 144 448 × 2 = 1 + 0.999 636 779 246 288 896;
  • 45) 0.999 636 779 246 288 896 × 2 = 1 + 0.999 273 558 492 577 792;
  • 46) 0.999 273 558 492 577 792 × 2 = 1 + 0.998 547 116 985 155 584;
  • 47) 0.998 547 116 985 155 584 × 2 = 1 + 0.997 094 233 970 311 168;
  • 48) 0.997 094 233 970 311 168 × 2 = 1 + 0.994 188 467 940 622 336;
  • 49) 0.994 188 467 940 622 336 × 2 = 1 + 0.988 376 935 881 244 672;
  • 50) 0.988 376 935 881 244 672 × 2 = 1 + 0.976 753 871 762 489 344;
  • 51) 0.976 753 871 762 489 344 × 2 = 1 + 0.953 507 743 524 978 688;
  • 52) 0.953 507 743 524 978 688 × 2 = 1 + 0.907 015 487 049 957 376;
  • 53) 0.907 015 487 049 957 376 × 2 = 1 + 0.814 030 974 099 914 752;
  • 54) 0.814 030 974 099 914 752 × 2 = 1 + 0.628 061 948 199 829 504;

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:


0.000 000 000 742 147 656(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 656(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:


0.000 000 000 742 147 656(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 -0.000 000 000 742 147 656 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