-0.000 000 000 742 147 674 35 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 35(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
-0.000 000 000 742 147 674 35(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:

|-0.000 000 000 742 147 674 35| = 0.000 000 000 742 147 674 35


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: 0.000 000 000 742 147 674 35.

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) 0.000 000 000 742 147 674 35 × 2 = 0 + 0.000 000 001 484 295 348 7;
  • 2) 0.000 000 001 484 295 348 7 × 2 = 0 + 0.000 000 002 968 590 697 4;
  • 3) 0.000 000 002 968 590 697 4 × 2 = 0 + 0.000 000 005 937 181 394 8;
  • 4) 0.000 000 005 937 181 394 8 × 2 = 0 + 0.000 000 011 874 362 789 6;
  • 5) 0.000 000 011 874 362 789 6 × 2 = 0 + 0.000 000 023 748 725 579 2;
  • 6) 0.000 000 023 748 725 579 2 × 2 = 0 + 0.000 000 047 497 451 158 4;
  • 7) 0.000 000 047 497 451 158 4 × 2 = 0 + 0.000 000 094 994 902 316 8;
  • 8) 0.000 000 094 994 902 316 8 × 2 = 0 + 0.000 000 189 989 804 633 6;
  • 9) 0.000 000 189 989 804 633 6 × 2 = 0 + 0.000 000 379 979 609 267 2;
  • 10) 0.000 000 379 979 609 267 2 × 2 = 0 + 0.000 000 759 959 218 534 4;
  • 11) 0.000 000 759 959 218 534 4 × 2 = 0 + 0.000 001 519 918 437 068 8;
  • 12) 0.000 001 519 918 437 068 8 × 2 = 0 + 0.000 003 039 836 874 137 6;
  • 13) 0.000 003 039 836 874 137 6 × 2 = 0 + 0.000 006 079 673 748 275 2;
  • 14) 0.000 006 079 673 748 275 2 × 2 = 0 + 0.000 012 159 347 496 550 4;
  • 15) 0.000 012 159 347 496 550 4 × 2 = 0 + 0.000 024 318 694 993 100 8;
  • 16) 0.000 024 318 694 993 100 8 × 2 = 0 + 0.000 048 637 389 986 201 6;
  • 17) 0.000 048 637 389 986 201 6 × 2 = 0 + 0.000 097 274 779 972 403 2;
  • 18) 0.000 097 274 779 972 403 2 × 2 = 0 + 0.000 194 549 559 944 806 4;
  • 19) 0.000 194 549 559 944 806 4 × 2 = 0 + 0.000 389 099 119 889 612 8;
  • 20) 0.000 389 099 119 889 612 8 × 2 = 0 + 0.000 778 198 239 779 225 6;
  • 21) 0.000 778 198 239 779 225 6 × 2 = 0 + 0.001 556 396 479 558 451 2;
  • 22) 0.001 556 396 479 558 451 2 × 2 = 0 + 0.003 112 792 959 116 902 4;
  • 23) 0.003 112 792 959 116 902 4 × 2 = 0 + 0.006 225 585 918 233 804 8;
  • 24) 0.006 225 585 918 233 804 8 × 2 = 0 + 0.012 451 171 836 467 609 6;
  • 25) 0.012 451 171 836 467 609 6 × 2 = 0 + 0.024 902 343 672 935 219 2;
  • 26) 0.024 902 343 672 935 219 2 × 2 = 0 + 0.049 804 687 345 870 438 4;
  • 27) 0.049 804 687 345 870 438 4 × 2 = 0 + 0.099 609 374 691 740 876 8;
  • 28) 0.099 609 374 691 740 876 8 × 2 = 0 + 0.199 218 749 383 481 753 6;
  • 29) 0.199 218 749 383 481 753 6 × 2 = 0 + 0.398 437 498 766 963 507 2;
  • 30) 0.398 437 498 766 963 507 2 × 2 = 0 + 0.796 874 997 533 927 014 4;
  • 31) 0.796 874 997 533 927 014 4 × 2 = 1 + 0.593 749 995 067 854 028 8;
  • 32) 0.593 749 995 067 854 028 8 × 2 = 1 + 0.187 499 990 135 708 057 6;
  • 33) 0.187 499 990 135 708 057 6 × 2 = 0 + 0.374 999 980 271 416 115 2;
  • 34) 0.374 999 980 271 416 115 2 × 2 = 0 + 0.749 999 960 542 832 230 4;
  • 35) 0.749 999 960 542 832 230 4 × 2 = 1 + 0.499 999 921 085 664 460 8;
  • 36) 0.499 999 921 085 664 460 8 × 2 = 0 + 0.999 999 842 171 328 921 6;
  • 37) 0.999 999 842 171 328 921 6 × 2 = 1 + 0.999 999 684 342 657 843 2;
  • 38) 0.999 999 684 342 657 843 2 × 2 = 1 + 0.999 999 368 685 315 686 4;
  • 39) 0.999 999 368 685 315 686 4 × 2 = 1 + 0.999 998 737 370 631 372 8;
  • 40) 0.999 998 737 370 631 372 8 × 2 = 1 + 0.999 997 474 741 262 745 6;
  • 41) 0.999 997 474 741 262 745 6 × 2 = 1 + 0.999 994 949 482 525 491 2;
  • 42) 0.999 994 949 482 525 491 2 × 2 = 1 + 0.999 989 898 965 050 982 4;
  • 43) 0.999 989 898 965 050 982 4 × 2 = 1 + 0.999 979 797 930 101 964 8;
  • 44) 0.999 979 797 930 101 964 8 × 2 = 1 + 0.999 959 595 860 203 929 6;
  • 45) 0.999 959 595 860 203 929 6 × 2 = 1 + 0.999 919 191 720 407 859 2;
  • 46) 0.999 919 191 720 407 859 2 × 2 = 1 + 0.999 838 383 440 815 718 4;
  • 47) 0.999 838 383 440 815 718 4 × 2 = 1 + 0.999 676 766 881 631 436 8;
  • 48) 0.999 676 766 881 631 436 8 × 2 = 1 + 0.999 353 533 763 262 873 6;
  • 49) 0.999 353 533 763 262 873 6 × 2 = 1 + 0.998 707 067 526 525 747 2;
  • 50) 0.998 707 067 526 525 747 2 × 2 = 1 + 0.997 414 135 053 051 494 4;
  • 51) 0.997 414 135 053 051 494 4 × 2 = 1 + 0.994 828 270 106 102 988 8;
  • 52) 0.994 828 270 106 102 988 8 × 2 = 1 + 0.989 656 540 212 205 977 6;
  • 53) 0.989 656 540 212 205 977 6 × 2 = 1 + 0.979 313 080 424 411 955 2;
  • 54) 0.979 313 080 424 411 955 2 × 2 = 1 + 0.958 626 160 848 823 910 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:


0.000 000 000 742 147 674 35(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 674 35(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:


0.000 000 000 742 147 674 35(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 -0.000 000 000 742 147 674 35 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