-0.000 000 000 742 147 676 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 4(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 4(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 4| = 0.000 000 000 742 147 676 4


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 4.

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 4 × 2 = 0 + 0.000 000 001 484 295 352 8;
  • 2) 0.000 000 001 484 295 352 8 × 2 = 0 + 0.000 000 002 968 590 705 6;
  • 3) 0.000 000 002 968 590 705 6 × 2 = 0 + 0.000 000 005 937 181 411 2;
  • 4) 0.000 000 005 937 181 411 2 × 2 = 0 + 0.000 000 011 874 362 822 4;
  • 5) 0.000 000 011 874 362 822 4 × 2 = 0 + 0.000 000 023 748 725 644 8;
  • 6) 0.000 000 023 748 725 644 8 × 2 = 0 + 0.000 000 047 497 451 289 6;
  • 7) 0.000 000 047 497 451 289 6 × 2 = 0 + 0.000 000 094 994 902 579 2;
  • 8) 0.000 000 094 994 902 579 2 × 2 = 0 + 0.000 000 189 989 805 158 4;
  • 9) 0.000 000 189 989 805 158 4 × 2 = 0 + 0.000 000 379 979 610 316 8;
  • 10) 0.000 000 379 979 610 316 8 × 2 = 0 + 0.000 000 759 959 220 633 6;
  • 11) 0.000 000 759 959 220 633 6 × 2 = 0 + 0.000 001 519 918 441 267 2;
  • 12) 0.000 001 519 918 441 267 2 × 2 = 0 + 0.000 003 039 836 882 534 4;
  • 13) 0.000 003 039 836 882 534 4 × 2 = 0 + 0.000 006 079 673 765 068 8;
  • 14) 0.000 006 079 673 765 068 8 × 2 = 0 + 0.000 012 159 347 530 137 6;
  • 15) 0.000 012 159 347 530 137 6 × 2 = 0 + 0.000 024 318 695 060 275 2;
  • 16) 0.000 024 318 695 060 275 2 × 2 = 0 + 0.000 048 637 390 120 550 4;
  • 17) 0.000 048 637 390 120 550 4 × 2 = 0 + 0.000 097 274 780 241 100 8;
  • 18) 0.000 097 274 780 241 100 8 × 2 = 0 + 0.000 194 549 560 482 201 6;
  • 19) 0.000 194 549 560 482 201 6 × 2 = 0 + 0.000 389 099 120 964 403 2;
  • 20) 0.000 389 099 120 964 403 2 × 2 = 0 + 0.000 778 198 241 928 806 4;
  • 21) 0.000 778 198 241 928 806 4 × 2 = 0 + 0.001 556 396 483 857 612 8;
  • 22) 0.001 556 396 483 857 612 8 × 2 = 0 + 0.003 112 792 967 715 225 6;
  • 23) 0.003 112 792 967 715 225 6 × 2 = 0 + 0.006 225 585 935 430 451 2;
  • 24) 0.006 225 585 935 430 451 2 × 2 = 0 + 0.012 451 171 870 860 902 4;
  • 25) 0.012 451 171 870 860 902 4 × 2 = 0 + 0.024 902 343 741 721 804 8;
  • 26) 0.024 902 343 741 721 804 8 × 2 = 0 + 0.049 804 687 483 443 609 6;
  • 27) 0.049 804 687 483 443 609 6 × 2 = 0 + 0.099 609 374 966 887 219 2;
  • 28) 0.099 609 374 966 887 219 2 × 2 = 0 + 0.199 218 749 933 774 438 4;
  • 29) 0.199 218 749 933 774 438 4 × 2 = 0 + 0.398 437 499 867 548 876 8;
  • 30) 0.398 437 499 867 548 876 8 × 2 = 0 + 0.796 874 999 735 097 753 6;
  • 31) 0.796 874 999 735 097 753 6 × 2 = 1 + 0.593 749 999 470 195 507 2;
  • 32) 0.593 749 999 470 195 507 2 × 2 = 1 + 0.187 499 998 940 391 014 4;
  • 33) 0.187 499 998 940 391 014 4 × 2 = 0 + 0.374 999 997 880 782 028 8;
  • 34) 0.374 999 997 880 782 028 8 × 2 = 0 + 0.749 999 995 761 564 057 6;
  • 35) 0.749 999 995 761 564 057 6 × 2 = 1 + 0.499 999 991 523 128 115 2;
  • 36) 0.499 999 991 523 128 115 2 × 2 = 0 + 0.999 999 983 046 256 230 4;
  • 37) 0.999 999 983 046 256 230 4 × 2 = 1 + 0.999 999 966 092 512 460 8;
  • 38) 0.999 999 966 092 512 460 8 × 2 = 1 + 0.999 999 932 185 024 921 6;
  • 39) 0.999 999 932 185 024 921 6 × 2 = 1 + 0.999 999 864 370 049 843 2;
  • 40) 0.999 999 864 370 049 843 2 × 2 = 1 + 0.999 999 728 740 099 686 4;
  • 41) 0.999 999 728 740 099 686 4 × 2 = 1 + 0.999 999 457 480 199 372 8;
  • 42) 0.999 999 457 480 199 372 8 × 2 = 1 + 0.999 998 914 960 398 745 6;
  • 43) 0.999 998 914 960 398 745 6 × 2 = 1 + 0.999 997 829 920 797 491 2;
  • 44) 0.999 997 829 920 797 491 2 × 2 = 1 + 0.999 995 659 841 594 982 4;
  • 45) 0.999 995 659 841 594 982 4 × 2 = 1 + 0.999 991 319 683 189 964 8;
  • 46) 0.999 991 319 683 189 964 8 × 2 = 1 + 0.999 982 639 366 379 929 6;
  • 47) 0.999 982 639 366 379 929 6 × 2 = 1 + 0.999 965 278 732 759 859 2;
  • 48) 0.999 965 278 732 759 859 2 × 2 = 1 + 0.999 930 557 465 519 718 4;
  • 49) 0.999 930 557 465 519 718 4 × 2 = 1 + 0.999 861 114 931 039 436 8;
  • 50) 0.999 861 114 931 039 436 8 × 2 = 1 + 0.999 722 229 862 078 873 6;
  • 51) 0.999 722 229 862 078 873 6 × 2 = 1 + 0.999 444 459 724 157 747 2;
  • 52) 0.999 444 459 724 157 747 2 × 2 = 1 + 0.998 888 919 448 315 494 4;
  • 53) 0.998 888 919 448 315 494 4 × 2 = 1 + 0.997 777 838 896 630 988 8;
  • 54) 0.997 777 838 896 630 988 8 × 2 = 1 + 0.995 555 677 793 261 977 6;

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 4(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 4(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 4(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 4 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