-0.000 000 000 742 150 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 150 2(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
-0.000 000 000 742 150 2(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:

|-0.000 000 000 742 150 2| = 0.000 000 000 742 150 2


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: 0.000 000 000 742 150 2.

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) 0.000 000 000 742 150 2 × 2 = 0 + 0.000 000 001 484 300 4;
  • 2) 0.000 000 001 484 300 4 × 2 = 0 + 0.000 000 002 968 600 8;
  • 3) 0.000 000 002 968 600 8 × 2 = 0 + 0.000 000 005 937 201 6;
  • 4) 0.000 000 005 937 201 6 × 2 = 0 + 0.000 000 011 874 403 2;
  • 5) 0.000 000 011 874 403 2 × 2 = 0 + 0.000 000 023 748 806 4;
  • 6) 0.000 000 023 748 806 4 × 2 = 0 + 0.000 000 047 497 612 8;
  • 7) 0.000 000 047 497 612 8 × 2 = 0 + 0.000 000 094 995 225 6;
  • 8) 0.000 000 094 995 225 6 × 2 = 0 + 0.000 000 189 990 451 2;
  • 9) 0.000 000 189 990 451 2 × 2 = 0 + 0.000 000 379 980 902 4;
  • 10) 0.000 000 379 980 902 4 × 2 = 0 + 0.000 000 759 961 804 8;
  • 11) 0.000 000 759 961 804 8 × 2 = 0 + 0.000 001 519 923 609 6;
  • 12) 0.000 001 519 923 609 6 × 2 = 0 + 0.000 003 039 847 219 2;
  • 13) 0.000 003 039 847 219 2 × 2 = 0 + 0.000 006 079 694 438 4;
  • 14) 0.000 006 079 694 438 4 × 2 = 0 + 0.000 012 159 388 876 8;
  • 15) 0.000 012 159 388 876 8 × 2 = 0 + 0.000 024 318 777 753 6;
  • 16) 0.000 024 318 777 753 6 × 2 = 0 + 0.000 048 637 555 507 2;
  • 17) 0.000 048 637 555 507 2 × 2 = 0 + 0.000 097 275 111 014 4;
  • 18) 0.000 097 275 111 014 4 × 2 = 0 + 0.000 194 550 222 028 8;
  • 19) 0.000 194 550 222 028 8 × 2 = 0 + 0.000 389 100 444 057 6;
  • 20) 0.000 389 100 444 057 6 × 2 = 0 + 0.000 778 200 888 115 2;
  • 21) 0.000 778 200 888 115 2 × 2 = 0 + 0.001 556 401 776 230 4;
  • 22) 0.001 556 401 776 230 4 × 2 = 0 + 0.003 112 803 552 460 8;
  • 23) 0.003 112 803 552 460 8 × 2 = 0 + 0.006 225 607 104 921 6;
  • 24) 0.006 225 607 104 921 6 × 2 = 0 + 0.012 451 214 209 843 2;
  • 25) 0.012 451 214 209 843 2 × 2 = 0 + 0.024 902 428 419 686 4;
  • 26) 0.024 902 428 419 686 4 × 2 = 0 + 0.049 804 856 839 372 8;
  • 27) 0.049 804 856 839 372 8 × 2 = 0 + 0.099 609 713 678 745 6;
  • 28) 0.099 609 713 678 745 6 × 2 = 0 + 0.199 219 427 357 491 2;
  • 29) 0.199 219 427 357 491 2 × 2 = 0 + 0.398 438 854 714 982 4;
  • 30) 0.398 438 854 714 982 4 × 2 = 0 + 0.796 877 709 429 964 8;
  • 31) 0.796 877 709 429 964 8 × 2 = 1 + 0.593 755 418 859 929 6;
  • 32) 0.593 755 418 859 929 6 × 2 = 1 + 0.187 510 837 719 859 2;
  • 33) 0.187 510 837 719 859 2 × 2 = 0 + 0.375 021 675 439 718 4;
  • 34) 0.375 021 675 439 718 4 × 2 = 0 + 0.750 043 350 879 436 8;
  • 35) 0.750 043 350 879 436 8 × 2 = 1 + 0.500 086 701 758 873 6;
  • 36) 0.500 086 701 758 873 6 × 2 = 1 + 0.000 173 403 517 747 2;
  • 37) 0.000 173 403 517 747 2 × 2 = 0 + 0.000 346 807 035 494 4;
  • 38) 0.000 346 807 035 494 4 × 2 = 0 + 0.000 693 614 070 988 8;
  • 39) 0.000 693 614 070 988 8 × 2 = 0 + 0.001 387 228 141 977 6;
  • 40) 0.001 387 228 141 977 6 × 2 = 0 + 0.002 774 456 283 955 2;
  • 41) 0.002 774 456 283 955 2 × 2 = 0 + 0.005 548 912 567 910 4;
  • 42) 0.005 548 912 567 910 4 × 2 = 0 + 0.011 097 825 135 820 8;
  • 43) 0.011 097 825 135 820 8 × 2 = 0 + 0.022 195 650 271 641 6;
  • 44) 0.022 195 650 271 641 6 × 2 = 0 + 0.044 391 300 543 283 2;
  • 45) 0.044 391 300 543 283 2 × 2 = 0 + 0.088 782 601 086 566 4;
  • 46) 0.088 782 601 086 566 4 × 2 = 0 + 0.177 565 202 173 132 8;
  • 47) 0.177 565 202 173 132 8 × 2 = 0 + 0.355 130 404 346 265 6;
  • 48) 0.355 130 404 346 265 6 × 2 = 0 + 0.710 260 808 692 531 2;
  • 49) 0.710 260 808 692 531 2 × 2 = 1 + 0.420 521 617 385 062 4;
  • 50) 0.420 521 617 385 062 4 × 2 = 0 + 0.841 043 234 770 124 8;
  • 51) 0.841 043 234 770 124 8 × 2 = 1 + 0.682 086 469 540 249 6;
  • 52) 0.682 086 469 540 249 6 × 2 = 1 + 0.364 172 939 080 499 2;
  • 53) 0.364 172 939 080 499 2 × 2 = 0 + 0.728 345 878 160 998 4;
  • 54) 0.728 345 878 160 998 4 × 2 = 1 + 0.456 691 756 321 996 8;

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:


0.000 000 000 742 150 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1011 01(2)

6. Positive number before normalization:

0.000 000 000 742 150 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1011 01(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:


0.000 000 000 742 150 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1011 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 1011 01(2) × 20 =

1.1001 1000 0000 0000 0101 101(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 0101 101


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 0010 1101 =

100 1100 0000 0000 0010 1101


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 0010 1101


Decimal number -0.000 000 000 742 150 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0010 1101

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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