-0.000 000 000 742 147 676 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 1(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 1(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 1| = 0.000 000 000 742 147 676 1


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 1.

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 1 × 2 = 0 + 0.000 000 001 484 295 352 2;
  • 2) 0.000 000 001 484 295 352 2 × 2 = 0 + 0.000 000 002 968 590 704 4;
  • 3) 0.000 000 002 968 590 704 4 × 2 = 0 + 0.000 000 005 937 181 408 8;
  • 4) 0.000 000 005 937 181 408 8 × 2 = 0 + 0.000 000 011 874 362 817 6;
  • 5) 0.000 000 011 874 362 817 6 × 2 = 0 + 0.000 000 023 748 725 635 2;
  • 6) 0.000 000 023 748 725 635 2 × 2 = 0 + 0.000 000 047 497 451 270 4;
  • 7) 0.000 000 047 497 451 270 4 × 2 = 0 + 0.000 000 094 994 902 540 8;
  • 8) 0.000 000 094 994 902 540 8 × 2 = 0 + 0.000 000 189 989 805 081 6;
  • 9) 0.000 000 189 989 805 081 6 × 2 = 0 + 0.000 000 379 979 610 163 2;
  • 10) 0.000 000 379 979 610 163 2 × 2 = 0 + 0.000 000 759 959 220 326 4;
  • 11) 0.000 000 759 959 220 326 4 × 2 = 0 + 0.000 001 519 918 440 652 8;
  • 12) 0.000 001 519 918 440 652 8 × 2 = 0 + 0.000 003 039 836 881 305 6;
  • 13) 0.000 003 039 836 881 305 6 × 2 = 0 + 0.000 006 079 673 762 611 2;
  • 14) 0.000 006 079 673 762 611 2 × 2 = 0 + 0.000 012 159 347 525 222 4;
  • 15) 0.000 012 159 347 525 222 4 × 2 = 0 + 0.000 024 318 695 050 444 8;
  • 16) 0.000 024 318 695 050 444 8 × 2 = 0 + 0.000 048 637 390 100 889 6;
  • 17) 0.000 048 637 390 100 889 6 × 2 = 0 + 0.000 097 274 780 201 779 2;
  • 18) 0.000 097 274 780 201 779 2 × 2 = 0 + 0.000 194 549 560 403 558 4;
  • 19) 0.000 194 549 560 403 558 4 × 2 = 0 + 0.000 389 099 120 807 116 8;
  • 20) 0.000 389 099 120 807 116 8 × 2 = 0 + 0.000 778 198 241 614 233 6;
  • 21) 0.000 778 198 241 614 233 6 × 2 = 0 + 0.001 556 396 483 228 467 2;
  • 22) 0.001 556 396 483 228 467 2 × 2 = 0 + 0.003 112 792 966 456 934 4;
  • 23) 0.003 112 792 966 456 934 4 × 2 = 0 + 0.006 225 585 932 913 868 8;
  • 24) 0.006 225 585 932 913 868 8 × 2 = 0 + 0.012 451 171 865 827 737 6;
  • 25) 0.012 451 171 865 827 737 6 × 2 = 0 + 0.024 902 343 731 655 475 2;
  • 26) 0.024 902 343 731 655 475 2 × 2 = 0 + 0.049 804 687 463 310 950 4;
  • 27) 0.049 804 687 463 310 950 4 × 2 = 0 + 0.099 609 374 926 621 900 8;
  • 28) 0.099 609 374 926 621 900 8 × 2 = 0 + 0.199 218 749 853 243 801 6;
  • 29) 0.199 218 749 853 243 801 6 × 2 = 0 + 0.398 437 499 706 487 603 2;
  • 30) 0.398 437 499 706 487 603 2 × 2 = 0 + 0.796 874 999 412 975 206 4;
  • 31) 0.796 874 999 412 975 206 4 × 2 = 1 + 0.593 749 998 825 950 412 8;
  • 32) 0.593 749 998 825 950 412 8 × 2 = 1 + 0.187 499 997 651 900 825 6;
  • 33) 0.187 499 997 651 900 825 6 × 2 = 0 + 0.374 999 995 303 801 651 2;
  • 34) 0.374 999 995 303 801 651 2 × 2 = 0 + 0.749 999 990 607 603 302 4;
  • 35) 0.749 999 990 607 603 302 4 × 2 = 1 + 0.499 999 981 215 206 604 8;
  • 36) 0.499 999 981 215 206 604 8 × 2 = 0 + 0.999 999 962 430 413 209 6;
  • 37) 0.999 999 962 430 413 209 6 × 2 = 1 + 0.999 999 924 860 826 419 2;
  • 38) 0.999 999 924 860 826 419 2 × 2 = 1 + 0.999 999 849 721 652 838 4;
  • 39) 0.999 999 849 721 652 838 4 × 2 = 1 + 0.999 999 699 443 305 676 8;
  • 40) 0.999 999 699 443 305 676 8 × 2 = 1 + 0.999 999 398 886 611 353 6;
  • 41) 0.999 999 398 886 611 353 6 × 2 = 1 + 0.999 998 797 773 222 707 2;
  • 42) 0.999 998 797 773 222 707 2 × 2 = 1 + 0.999 997 595 546 445 414 4;
  • 43) 0.999 997 595 546 445 414 4 × 2 = 1 + 0.999 995 191 092 890 828 8;
  • 44) 0.999 995 191 092 890 828 8 × 2 = 1 + 0.999 990 382 185 781 657 6;
  • 45) 0.999 990 382 185 781 657 6 × 2 = 1 + 0.999 980 764 371 563 315 2;
  • 46) 0.999 980 764 371 563 315 2 × 2 = 1 + 0.999 961 528 743 126 630 4;
  • 47) 0.999 961 528 743 126 630 4 × 2 = 1 + 0.999 923 057 486 253 260 8;
  • 48) 0.999 923 057 486 253 260 8 × 2 = 1 + 0.999 846 114 972 506 521 6;
  • 49) 0.999 846 114 972 506 521 6 × 2 = 1 + 0.999 692 229 945 013 043 2;
  • 50) 0.999 692 229 945 013 043 2 × 2 = 1 + 0.999 384 459 890 026 086 4;
  • 51) 0.999 384 459 890 026 086 4 × 2 = 1 + 0.998 768 919 780 052 172 8;
  • 52) 0.998 768 919 780 052 172 8 × 2 = 1 + 0.997 537 839 560 104 345 6;
  • 53) 0.997 537 839 560 104 345 6 × 2 = 1 + 0.995 075 679 120 208 691 2;
  • 54) 0.995 075 679 120 208 691 2 × 2 = 1 + 0.990 151 358 240 417 382 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:


0.000 000 000 742 147 676 1(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 1(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 1(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 1 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