-0.000 000 000 742 147 676 39 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 39(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 39(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 39| = 0.000 000 000 742 147 676 39


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 39.

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 39 × 2 = 0 + 0.000 000 001 484 295 352 78;
  • 2) 0.000 000 001 484 295 352 78 × 2 = 0 + 0.000 000 002 968 590 705 56;
  • 3) 0.000 000 002 968 590 705 56 × 2 = 0 + 0.000 000 005 937 181 411 12;
  • 4) 0.000 000 005 937 181 411 12 × 2 = 0 + 0.000 000 011 874 362 822 24;
  • 5) 0.000 000 011 874 362 822 24 × 2 = 0 + 0.000 000 023 748 725 644 48;
  • 6) 0.000 000 023 748 725 644 48 × 2 = 0 + 0.000 000 047 497 451 288 96;
  • 7) 0.000 000 047 497 451 288 96 × 2 = 0 + 0.000 000 094 994 902 577 92;
  • 8) 0.000 000 094 994 902 577 92 × 2 = 0 + 0.000 000 189 989 805 155 84;
  • 9) 0.000 000 189 989 805 155 84 × 2 = 0 + 0.000 000 379 979 610 311 68;
  • 10) 0.000 000 379 979 610 311 68 × 2 = 0 + 0.000 000 759 959 220 623 36;
  • 11) 0.000 000 759 959 220 623 36 × 2 = 0 + 0.000 001 519 918 441 246 72;
  • 12) 0.000 001 519 918 441 246 72 × 2 = 0 + 0.000 003 039 836 882 493 44;
  • 13) 0.000 003 039 836 882 493 44 × 2 = 0 + 0.000 006 079 673 764 986 88;
  • 14) 0.000 006 079 673 764 986 88 × 2 = 0 + 0.000 012 159 347 529 973 76;
  • 15) 0.000 012 159 347 529 973 76 × 2 = 0 + 0.000 024 318 695 059 947 52;
  • 16) 0.000 024 318 695 059 947 52 × 2 = 0 + 0.000 048 637 390 119 895 04;
  • 17) 0.000 048 637 390 119 895 04 × 2 = 0 + 0.000 097 274 780 239 790 08;
  • 18) 0.000 097 274 780 239 790 08 × 2 = 0 + 0.000 194 549 560 479 580 16;
  • 19) 0.000 194 549 560 479 580 16 × 2 = 0 + 0.000 389 099 120 959 160 32;
  • 20) 0.000 389 099 120 959 160 32 × 2 = 0 + 0.000 778 198 241 918 320 64;
  • 21) 0.000 778 198 241 918 320 64 × 2 = 0 + 0.001 556 396 483 836 641 28;
  • 22) 0.001 556 396 483 836 641 28 × 2 = 0 + 0.003 112 792 967 673 282 56;
  • 23) 0.003 112 792 967 673 282 56 × 2 = 0 + 0.006 225 585 935 346 565 12;
  • 24) 0.006 225 585 935 346 565 12 × 2 = 0 + 0.012 451 171 870 693 130 24;
  • 25) 0.012 451 171 870 693 130 24 × 2 = 0 + 0.024 902 343 741 386 260 48;
  • 26) 0.024 902 343 741 386 260 48 × 2 = 0 + 0.049 804 687 482 772 520 96;
  • 27) 0.049 804 687 482 772 520 96 × 2 = 0 + 0.099 609 374 965 545 041 92;
  • 28) 0.099 609 374 965 545 041 92 × 2 = 0 + 0.199 218 749 931 090 083 84;
  • 29) 0.199 218 749 931 090 083 84 × 2 = 0 + 0.398 437 499 862 180 167 68;
  • 30) 0.398 437 499 862 180 167 68 × 2 = 0 + 0.796 874 999 724 360 335 36;
  • 31) 0.796 874 999 724 360 335 36 × 2 = 1 + 0.593 749 999 448 720 670 72;
  • 32) 0.593 749 999 448 720 670 72 × 2 = 1 + 0.187 499 998 897 441 341 44;
  • 33) 0.187 499 998 897 441 341 44 × 2 = 0 + 0.374 999 997 794 882 682 88;
  • 34) 0.374 999 997 794 882 682 88 × 2 = 0 + 0.749 999 995 589 765 365 76;
  • 35) 0.749 999 995 589 765 365 76 × 2 = 1 + 0.499 999 991 179 530 731 52;
  • 36) 0.499 999 991 179 530 731 52 × 2 = 0 + 0.999 999 982 359 061 463 04;
  • 37) 0.999 999 982 359 061 463 04 × 2 = 1 + 0.999 999 964 718 122 926 08;
  • 38) 0.999 999 964 718 122 926 08 × 2 = 1 + 0.999 999 929 436 245 852 16;
  • 39) 0.999 999 929 436 245 852 16 × 2 = 1 + 0.999 999 858 872 491 704 32;
  • 40) 0.999 999 858 872 491 704 32 × 2 = 1 + 0.999 999 717 744 983 408 64;
  • 41) 0.999 999 717 744 983 408 64 × 2 = 1 + 0.999 999 435 489 966 817 28;
  • 42) 0.999 999 435 489 966 817 28 × 2 = 1 + 0.999 998 870 979 933 634 56;
  • 43) 0.999 998 870 979 933 634 56 × 2 = 1 + 0.999 997 741 959 867 269 12;
  • 44) 0.999 997 741 959 867 269 12 × 2 = 1 + 0.999 995 483 919 734 538 24;
  • 45) 0.999 995 483 919 734 538 24 × 2 = 1 + 0.999 990 967 839 469 076 48;
  • 46) 0.999 990 967 839 469 076 48 × 2 = 1 + 0.999 981 935 678 938 152 96;
  • 47) 0.999 981 935 678 938 152 96 × 2 = 1 + 0.999 963 871 357 876 305 92;
  • 48) 0.999 963 871 357 876 305 92 × 2 = 1 + 0.999 927 742 715 752 611 84;
  • 49) 0.999 927 742 715 752 611 84 × 2 = 1 + 0.999 855 485 431 505 223 68;
  • 50) 0.999 855 485 431 505 223 68 × 2 = 1 + 0.999 710 970 863 010 447 36;
  • 51) 0.999 710 970 863 010 447 36 × 2 = 1 + 0.999 421 941 726 020 894 72;
  • 52) 0.999 421 941 726 020 894 72 × 2 = 1 + 0.998 843 883 452 041 789 44;
  • 53) 0.998 843 883 452 041 789 44 × 2 = 1 + 0.997 687 766 904 083 578 88;
  • 54) 0.997 687 766 904 083 578 88 × 2 = 1 + 0.995 375 533 808 167 157 76;

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 39(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 39(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 39(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 39 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