-0.000 000 000 742 147 675 38 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 38(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
-0.000 000 000 742 147 675 38(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:

|-0.000 000 000 742 147 675 38| = 0.000 000 000 742 147 675 38


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: 0.000 000 000 742 147 675 38.

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) 0.000 000 000 742 147 675 38 × 2 = 0 + 0.000 000 001 484 295 350 76;
  • 2) 0.000 000 001 484 295 350 76 × 2 = 0 + 0.000 000 002 968 590 701 52;
  • 3) 0.000 000 002 968 590 701 52 × 2 = 0 + 0.000 000 005 937 181 403 04;
  • 4) 0.000 000 005 937 181 403 04 × 2 = 0 + 0.000 000 011 874 362 806 08;
  • 5) 0.000 000 011 874 362 806 08 × 2 = 0 + 0.000 000 023 748 725 612 16;
  • 6) 0.000 000 023 748 725 612 16 × 2 = 0 + 0.000 000 047 497 451 224 32;
  • 7) 0.000 000 047 497 451 224 32 × 2 = 0 + 0.000 000 094 994 902 448 64;
  • 8) 0.000 000 094 994 902 448 64 × 2 = 0 + 0.000 000 189 989 804 897 28;
  • 9) 0.000 000 189 989 804 897 28 × 2 = 0 + 0.000 000 379 979 609 794 56;
  • 10) 0.000 000 379 979 609 794 56 × 2 = 0 + 0.000 000 759 959 219 589 12;
  • 11) 0.000 000 759 959 219 589 12 × 2 = 0 + 0.000 001 519 918 439 178 24;
  • 12) 0.000 001 519 918 439 178 24 × 2 = 0 + 0.000 003 039 836 878 356 48;
  • 13) 0.000 003 039 836 878 356 48 × 2 = 0 + 0.000 006 079 673 756 712 96;
  • 14) 0.000 006 079 673 756 712 96 × 2 = 0 + 0.000 012 159 347 513 425 92;
  • 15) 0.000 012 159 347 513 425 92 × 2 = 0 + 0.000 024 318 695 026 851 84;
  • 16) 0.000 024 318 695 026 851 84 × 2 = 0 + 0.000 048 637 390 053 703 68;
  • 17) 0.000 048 637 390 053 703 68 × 2 = 0 + 0.000 097 274 780 107 407 36;
  • 18) 0.000 097 274 780 107 407 36 × 2 = 0 + 0.000 194 549 560 214 814 72;
  • 19) 0.000 194 549 560 214 814 72 × 2 = 0 + 0.000 389 099 120 429 629 44;
  • 20) 0.000 389 099 120 429 629 44 × 2 = 0 + 0.000 778 198 240 859 258 88;
  • 21) 0.000 778 198 240 859 258 88 × 2 = 0 + 0.001 556 396 481 718 517 76;
  • 22) 0.001 556 396 481 718 517 76 × 2 = 0 + 0.003 112 792 963 437 035 52;
  • 23) 0.003 112 792 963 437 035 52 × 2 = 0 + 0.006 225 585 926 874 071 04;
  • 24) 0.006 225 585 926 874 071 04 × 2 = 0 + 0.012 451 171 853 748 142 08;
  • 25) 0.012 451 171 853 748 142 08 × 2 = 0 + 0.024 902 343 707 496 284 16;
  • 26) 0.024 902 343 707 496 284 16 × 2 = 0 + 0.049 804 687 414 992 568 32;
  • 27) 0.049 804 687 414 992 568 32 × 2 = 0 + 0.099 609 374 829 985 136 64;
  • 28) 0.099 609 374 829 985 136 64 × 2 = 0 + 0.199 218 749 659 970 273 28;
  • 29) 0.199 218 749 659 970 273 28 × 2 = 0 + 0.398 437 499 319 940 546 56;
  • 30) 0.398 437 499 319 940 546 56 × 2 = 0 + 0.796 874 998 639 881 093 12;
  • 31) 0.796 874 998 639 881 093 12 × 2 = 1 + 0.593 749 997 279 762 186 24;
  • 32) 0.593 749 997 279 762 186 24 × 2 = 1 + 0.187 499 994 559 524 372 48;
  • 33) 0.187 499 994 559 524 372 48 × 2 = 0 + 0.374 999 989 119 048 744 96;
  • 34) 0.374 999 989 119 048 744 96 × 2 = 0 + 0.749 999 978 238 097 489 92;
  • 35) 0.749 999 978 238 097 489 92 × 2 = 1 + 0.499 999 956 476 194 979 84;
  • 36) 0.499 999 956 476 194 979 84 × 2 = 0 + 0.999 999 912 952 389 959 68;
  • 37) 0.999 999 912 952 389 959 68 × 2 = 1 + 0.999 999 825 904 779 919 36;
  • 38) 0.999 999 825 904 779 919 36 × 2 = 1 + 0.999 999 651 809 559 838 72;
  • 39) 0.999 999 651 809 559 838 72 × 2 = 1 + 0.999 999 303 619 119 677 44;
  • 40) 0.999 999 303 619 119 677 44 × 2 = 1 + 0.999 998 607 238 239 354 88;
  • 41) 0.999 998 607 238 239 354 88 × 2 = 1 + 0.999 997 214 476 478 709 76;
  • 42) 0.999 997 214 476 478 709 76 × 2 = 1 + 0.999 994 428 952 957 419 52;
  • 43) 0.999 994 428 952 957 419 52 × 2 = 1 + 0.999 988 857 905 914 839 04;
  • 44) 0.999 988 857 905 914 839 04 × 2 = 1 + 0.999 977 715 811 829 678 08;
  • 45) 0.999 977 715 811 829 678 08 × 2 = 1 + 0.999 955 431 623 659 356 16;
  • 46) 0.999 955 431 623 659 356 16 × 2 = 1 + 0.999 910 863 247 318 712 32;
  • 47) 0.999 910 863 247 318 712 32 × 2 = 1 + 0.999 821 726 494 637 424 64;
  • 48) 0.999 821 726 494 637 424 64 × 2 = 1 + 0.999 643 452 989 274 849 28;
  • 49) 0.999 643 452 989 274 849 28 × 2 = 1 + 0.999 286 905 978 549 698 56;
  • 50) 0.999 286 905 978 549 698 56 × 2 = 1 + 0.998 573 811 957 099 397 12;
  • 51) 0.998 573 811 957 099 397 12 × 2 = 1 + 0.997 147 623 914 198 794 24;
  • 52) 0.997 147 623 914 198 794 24 × 2 = 1 + 0.994 295 247 828 397 588 48;
  • 53) 0.994 295 247 828 397 588 48 × 2 = 1 + 0.988 590 495 656 795 176 96;
  • 54) 0.988 590 495 656 795 176 96 × 2 = 1 + 0.977 180 991 313 590 353 92;

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:


0.000 000 000 742 147 675 38(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 38(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:


0.000 000 000 742 147 675 38(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 -0.000 000 000 742 147 675 38 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