-0.000 000 000 742 148 38 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 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
-0.000 000 000 742 148 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:

|-0.000 000 000 742 148 38| = 0.000 000 000 742 148 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: 0.000 000 000 742 148 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) 0.000 000 000 742 148 38 × 2 = 0 + 0.000 000 001 484 296 76;
  • 2) 0.000 000 001 484 296 76 × 2 = 0 + 0.000 000 002 968 593 52;
  • 3) 0.000 000 002 968 593 52 × 2 = 0 + 0.000 000 005 937 187 04;
  • 4) 0.000 000 005 937 187 04 × 2 = 0 + 0.000 000 011 874 374 08;
  • 5) 0.000 000 011 874 374 08 × 2 = 0 + 0.000 000 023 748 748 16;
  • 6) 0.000 000 023 748 748 16 × 2 = 0 + 0.000 000 047 497 496 32;
  • 7) 0.000 000 047 497 496 32 × 2 = 0 + 0.000 000 094 994 992 64;
  • 8) 0.000 000 094 994 992 64 × 2 = 0 + 0.000 000 189 989 985 28;
  • 9) 0.000 000 189 989 985 28 × 2 = 0 + 0.000 000 379 979 970 56;
  • 10) 0.000 000 379 979 970 56 × 2 = 0 + 0.000 000 759 959 941 12;
  • 11) 0.000 000 759 959 941 12 × 2 = 0 + 0.000 001 519 919 882 24;
  • 12) 0.000 001 519 919 882 24 × 2 = 0 + 0.000 003 039 839 764 48;
  • 13) 0.000 003 039 839 764 48 × 2 = 0 + 0.000 006 079 679 528 96;
  • 14) 0.000 006 079 679 528 96 × 2 = 0 + 0.000 012 159 359 057 92;
  • 15) 0.000 012 159 359 057 92 × 2 = 0 + 0.000 024 318 718 115 84;
  • 16) 0.000 024 318 718 115 84 × 2 = 0 + 0.000 048 637 436 231 68;
  • 17) 0.000 048 637 436 231 68 × 2 = 0 + 0.000 097 274 872 463 36;
  • 18) 0.000 097 274 872 463 36 × 2 = 0 + 0.000 194 549 744 926 72;
  • 19) 0.000 194 549 744 926 72 × 2 = 0 + 0.000 389 099 489 853 44;
  • 20) 0.000 389 099 489 853 44 × 2 = 0 + 0.000 778 198 979 706 88;
  • 21) 0.000 778 198 979 706 88 × 2 = 0 + 0.001 556 397 959 413 76;
  • 22) 0.001 556 397 959 413 76 × 2 = 0 + 0.003 112 795 918 827 52;
  • 23) 0.003 112 795 918 827 52 × 2 = 0 + 0.006 225 591 837 655 04;
  • 24) 0.006 225 591 837 655 04 × 2 = 0 + 0.012 451 183 675 310 08;
  • 25) 0.012 451 183 675 310 08 × 2 = 0 + 0.024 902 367 350 620 16;
  • 26) 0.024 902 367 350 620 16 × 2 = 0 + 0.049 804 734 701 240 32;
  • 27) 0.049 804 734 701 240 32 × 2 = 0 + 0.099 609 469 402 480 64;
  • 28) 0.099 609 469 402 480 64 × 2 = 0 + 0.199 218 938 804 961 28;
  • 29) 0.199 218 938 804 961 28 × 2 = 0 + 0.398 437 877 609 922 56;
  • 30) 0.398 437 877 609 922 56 × 2 = 0 + 0.796 875 755 219 845 12;
  • 31) 0.796 875 755 219 845 12 × 2 = 1 + 0.593 751 510 439 690 24;
  • 32) 0.593 751 510 439 690 24 × 2 = 1 + 0.187 503 020 879 380 48;
  • 33) 0.187 503 020 879 380 48 × 2 = 0 + 0.375 006 041 758 760 96;
  • 34) 0.375 006 041 758 760 96 × 2 = 0 + 0.750 012 083 517 521 92;
  • 35) 0.750 012 083 517 521 92 × 2 = 1 + 0.500 024 167 035 043 84;
  • 36) 0.500 024 167 035 043 84 × 2 = 1 + 0.000 048 334 070 087 68;
  • 37) 0.000 048 334 070 087 68 × 2 = 0 + 0.000 096 668 140 175 36;
  • 38) 0.000 096 668 140 175 36 × 2 = 0 + 0.000 193 336 280 350 72;
  • 39) 0.000 193 336 280 350 72 × 2 = 0 + 0.000 386 672 560 701 44;
  • 40) 0.000 386 672 560 701 44 × 2 = 0 + 0.000 773 345 121 402 88;
  • 41) 0.000 773 345 121 402 88 × 2 = 0 + 0.001 546 690 242 805 76;
  • 42) 0.001 546 690 242 805 76 × 2 = 0 + 0.003 093 380 485 611 52;
  • 43) 0.003 093 380 485 611 52 × 2 = 0 + 0.006 186 760 971 223 04;
  • 44) 0.006 186 760 971 223 04 × 2 = 0 + 0.012 373 521 942 446 08;
  • 45) 0.012 373 521 942 446 08 × 2 = 0 + 0.024 747 043 884 892 16;
  • 46) 0.024 747 043 884 892 16 × 2 = 0 + 0.049 494 087 769 784 32;
  • 47) 0.049 494 087 769 784 32 × 2 = 0 + 0.098 988 175 539 568 64;
  • 48) 0.098 988 175 539 568 64 × 2 = 0 + 0.197 976 351 079 137 28;
  • 49) 0.197 976 351 079 137 28 × 2 = 0 + 0.395 952 702 158 274 56;
  • 50) 0.395 952 702 158 274 56 × 2 = 0 + 0.791 905 404 316 549 12;
  • 51) 0.791 905 404 316 549 12 × 2 = 1 + 0.583 810 808 633 098 24;
  • 52) 0.583 810 808 633 098 24 × 2 = 1 + 0.167 621 617 266 196 48;
  • 53) 0.167 621 617 266 196 48 × 2 = 0 + 0.335 243 234 532 392 96;
  • 54) 0.335 243 234 532 392 96 × 2 = 0 + 0.670 486 469 064 785 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:


0.000 000 000 742 148 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0011 00(2)

6. Positive number before normalization:

0.000 000 000 742 148 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0011 00(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:


0.000 000 000 742 148 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0011 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0011 00(2) × 20 =

1.1001 1000 0000 0000 0001 100(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 1000 0000 0000 0001 100


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 1100 0000 0000 0000 1100 =

100 1100 0000 0000 0000 1100


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 1100 0000 0000 0000 1100


Decimal number -0.000 000 000 742 148 38 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 1100

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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