-0.000 000 000 742 147 676 646 677 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 677 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
-0.000 000 000 742 147 676 646 677 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:

|-0.000 000 000 742 147 676 646 677 1| = 0.000 000 000 742 147 676 646 677 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: 0.000 000 000 742 147 676 646 677 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) 0.000 000 000 742 147 676 646 677 1 × 2 = 0 + 0.000 000 001 484 295 353 293 354 2;
  • 2) 0.000 000 001 484 295 353 293 354 2 × 2 = 0 + 0.000 000 002 968 590 706 586 708 4;
  • 3) 0.000 000 002 968 590 706 586 708 4 × 2 = 0 + 0.000 000 005 937 181 413 173 416 8;
  • 4) 0.000 000 005 937 181 413 173 416 8 × 2 = 0 + 0.000 000 011 874 362 826 346 833 6;
  • 5) 0.000 000 011 874 362 826 346 833 6 × 2 = 0 + 0.000 000 023 748 725 652 693 667 2;
  • 6) 0.000 000 023 748 725 652 693 667 2 × 2 = 0 + 0.000 000 047 497 451 305 387 334 4;
  • 7) 0.000 000 047 497 451 305 387 334 4 × 2 = 0 + 0.000 000 094 994 902 610 774 668 8;
  • 8) 0.000 000 094 994 902 610 774 668 8 × 2 = 0 + 0.000 000 189 989 805 221 549 337 6;
  • 9) 0.000 000 189 989 805 221 549 337 6 × 2 = 0 + 0.000 000 379 979 610 443 098 675 2;
  • 10) 0.000 000 379 979 610 443 098 675 2 × 2 = 0 + 0.000 000 759 959 220 886 197 350 4;
  • 11) 0.000 000 759 959 220 886 197 350 4 × 2 = 0 + 0.000 001 519 918 441 772 394 700 8;
  • 12) 0.000 001 519 918 441 772 394 700 8 × 2 = 0 + 0.000 003 039 836 883 544 789 401 6;
  • 13) 0.000 003 039 836 883 544 789 401 6 × 2 = 0 + 0.000 006 079 673 767 089 578 803 2;
  • 14) 0.000 006 079 673 767 089 578 803 2 × 2 = 0 + 0.000 012 159 347 534 179 157 606 4;
  • 15) 0.000 012 159 347 534 179 157 606 4 × 2 = 0 + 0.000 024 318 695 068 358 315 212 8;
  • 16) 0.000 024 318 695 068 358 315 212 8 × 2 = 0 + 0.000 048 637 390 136 716 630 425 6;
  • 17) 0.000 048 637 390 136 716 630 425 6 × 2 = 0 + 0.000 097 274 780 273 433 260 851 2;
  • 18) 0.000 097 274 780 273 433 260 851 2 × 2 = 0 + 0.000 194 549 560 546 866 521 702 4;
  • 19) 0.000 194 549 560 546 866 521 702 4 × 2 = 0 + 0.000 389 099 121 093 733 043 404 8;
  • 20) 0.000 389 099 121 093 733 043 404 8 × 2 = 0 + 0.000 778 198 242 187 466 086 809 6;
  • 21) 0.000 778 198 242 187 466 086 809 6 × 2 = 0 + 0.001 556 396 484 374 932 173 619 2;
  • 22) 0.001 556 396 484 374 932 173 619 2 × 2 = 0 + 0.003 112 792 968 749 864 347 238 4;
  • 23) 0.003 112 792 968 749 864 347 238 4 × 2 = 0 + 0.006 225 585 937 499 728 694 476 8;
  • 24) 0.006 225 585 937 499 728 694 476 8 × 2 = 0 + 0.012 451 171 874 999 457 388 953 6;
  • 25) 0.012 451 171 874 999 457 388 953 6 × 2 = 0 + 0.024 902 343 749 998 914 777 907 2;
  • 26) 0.024 902 343 749 998 914 777 907 2 × 2 = 0 + 0.049 804 687 499 997 829 555 814 4;
  • 27) 0.049 804 687 499 997 829 555 814 4 × 2 = 0 + 0.099 609 374 999 995 659 111 628 8;
  • 28) 0.099 609 374 999 995 659 111 628 8 × 2 = 0 + 0.199 218 749 999 991 318 223 257 6;
  • 29) 0.199 218 749 999 991 318 223 257 6 × 2 = 0 + 0.398 437 499 999 982 636 446 515 2;
  • 30) 0.398 437 499 999 982 636 446 515 2 × 2 = 0 + 0.796 874 999 999 965 272 893 030 4;
  • 31) 0.796 874 999 999 965 272 893 030 4 × 2 = 1 + 0.593 749 999 999 930 545 786 060 8;
  • 32) 0.593 749 999 999 930 545 786 060 8 × 2 = 1 + 0.187 499 999 999 861 091 572 121 6;
  • 33) 0.187 499 999 999 861 091 572 121 6 × 2 = 0 + 0.374 999 999 999 722 183 144 243 2;
  • 34) 0.374 999 999 999 722 183 144 243 2 × 2 = 0 + 0.749 999 999 999 444 366 288 486 4;
  • 35) 0.749 999 999 999 444 366 288 486 4 × 2 = 1 + 0.499 999 999 998 888 732 576 972 8;
  • 36) 0.499 999 999 998 888 732 576 972 8 × 2 = 0 + 0.999 999 999 997 777 465 153 945 6;
  • 37) 0.999 999 999 997 777 465 153 945 6 × 2 = 1 + 0.999 999 999 995 554 930 307 891 2;
  • 38) 0.999 999 999 995 554 930 307 891 2 × 2 = 1 + 0.999 999 999 991 109 860 615 782 4;
  • 39) 0.999 999 999 991 109 860 615 782 4 × 2 = 1 + 0.999 999 999 982 219 721 231 564 8;
  • 40) 0.999 999 999 982 219 721 231 564 8 × 2 = 1 + 0.999 999 999 964 439 442 463 129 6;
  • 41) 0.999 999 999 964 439 442 463 129 6 × 2 = 1 + 0.999 999 999 928 878 884 926 259 2;
  • 42) 0.999 999 999 928 878 884 926 259 2 × 2 = 1 + 0.999 999 999 857 757 769 852 518 4;
  • 43) 0.999 999 999 857 757 769 852 518 4 × 2 = 1 + 0.999 999 999 715 515 539 705 036 8;
  • 44) 0.999 999 999 715 515 539 705 036 8 × 2 = 1 + 0.999 999 999 431 031 079 410 073 6;
  • 45) 0.999 999 999 431 031 079 410 073 6 × 2 = 1 + 0.999 999 998 862 062 158 820 147 2;
  • 46) 0.999 999 998 862 062 158 820 147 2 × 2 = 1 + 0.999 999 997 724 124 317 640 294 4;
  • 47) 0.999 999 997 724 124 317 640 294 4 × 2 = 1 + 0.999 999 995 448 248 635 280 588 8;
  • 48) 0.999 999 995 448 248 635 280 588 8 × 2 = 1 + 0.999 999 990 896 497 270 561 177 6;
  • 49) 0.999 999 990 896 497 270 561 177 6 × 2 = 1 + 0.999 999 981 792 994 541 122 355 2;
  • 50) 0.999 999 981 792 994 541 122 355 2 × 2 = 1 + 0.999 999 963 585 989 082 244 710 4;
  • 51) 0.999 999 963 585 989 082 244 710 4 × 2 = 1 + 0.999 999 927 171 978 164 489 420 8;
  • 52) 0.999 999 927 171 978 164 489 420 8 × 2 = 1 + 0.999 999 854 343 956 328 978 841 6;
  • 53) 0.999 999 854 343 956 328 978 841 6 × 2 = 1 + 0.999 999 708 687 912 657 957 683 2;
  • 54) 0.999 999 708 687 912 657 957 683 2 × 2 = 1 + 0.999 999 417 375 825 315 915 366 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:


0.000 000 000 742 147 676 646 677 1(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 677 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:


0.000 000 000 742 147 676 646 677 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 -0.000 000 000 742 147 676 646 677 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