-0.000 000 000 742 147 676 646 691 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 691 7(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
-0.000 000 000 742 147 676 646 691 7(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:

|-0.000 000 000 742 147 676 646 691 7| = 0.000 000 000 742 147 676 646 691 7


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: 0.000 000 000 742 147 676 646 691 7.

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) 0.000 000 000 742 147 676 646 691 7 × 2 = 0 + 0.000 000 001 484 295 353 293 383 4;
  • 2) 0.000 000 001 484 295 353 293 383 4 × 2 = 0 + 0.000 000 002 968 590 706 586 766 8;
  • 3) 0.000 000 002 968 590 706 586 766 8 × 2 = 0 + 0.000 000 005 937 181 413 173 533 6;
  • 4) 0.000 000 005 937 181 413 173 533 6 × 2 = 0 + 0.000 000 011 874 362 826 347 067 2;
  • 5) 0.000 000 011 874 362 826 347 067 2 × 2 = 0 + 0.000 000 023 748 725 652 694 134 4;
  • 6) 0.000 000 023 748 725 652 694 134 4 × 2 = 0 + 0.000 000 047 497 451 305 388 268 8;
  • 7) 0.000 000 047 497 451 305 388 268 8 × 2 = 0 + 0.000 000 094 994 902 610 776 537 6;
  • 8) 0.000 000 094 994 902 610 776 537 6 × 2 = 0 + 0.000 000 189 989 805 221 553 075 2;
  • 9) 0.000 000 189 989 805 221 553 075 2 × 2 = 0 + 0.000 000 379 979 610 443 106 150 4;
  • 10) 0.000 000 379 979 610 443 106 150 4 × 2 = 0 + 0.000 000 759 959 220 886 212 300 8;
  • 11) 0.000 000 759 959 220 886 212 300 8 × 2 = 0 + 0.000 001 519 918 441 772 424 601 6;
  • 12) 0.000 001 519 918 441 772 424 601 6 × 2 = 0 + 0.000 003 039 836 883 544 849 203 2;
  • 13) 0.000 003 039 836 883 544 849 203 2 × 2 = 0 + 0.000 006 079 673 767 089 698 406 4;
  • 14) 0.000 006 079 673 767 089 698 406 4 × 2 = 0 + 0.000 012 159 347 534 179 396 812 8;
  • 15) 0.000 012 159 347 534 179 396 812 8 × 2 = 0 + 0.000 024 318 695 068 358 793 625 6;
  • 16) 0.000 024 318 695 068 358 793 625 6 × 2 = 0 + 0.000 048 637 390 136 717 587 251 2;
  • 17) 0.000 048 637 390 136 717 587 251 2 × 2 = 0 + 0.000 097 274 780 273 435 174 502 4;
  • 18) 0.000 097 274 780 273 435 174 502 4 × 2 = 0 + 0.000 194 549 560 546 870 349 004 8;
  • 19) 0.000 194 549 560 546 870 349 004 8 × 2 = 0 + 0.000 389 099 121 093 740 698 009 6;
  • 20) 0.000 389 099 121 093 740 698 009 6 × 2 = 0 + 0.000 778 198 242 187 481 396 019 2;
  • 21) 0.000 778 198 242 187 481 396 019 2 × 2 = 0 + 0.001 556 396 484 374 962 792 038 4;
  • 22) 0.001 556 396 484 374 962 792 038 4 × 2 = 0 + 0.003 112 792 968 749 925 584 076 8;
  • 23) 0.003 112 792 968 749 925 584 076 8 × 2 = 0 + 0.006 225 585 937 499 851 168 153 6;
  • 24) 0.006 225 585 937 499 851 168 153 6 × 2 = 0 + 0.012 451 171 874 999 702 336 307 2;
  • 25) 0.012 451 171 874 999 702 336 307 2 × 2 = 0 + 0.024 902 343 749 999 404 672 614 4;
  • 26) 0.024 902 343 749 999 404 672 614 4 × 2 = 0 + 0.049 804 687 499 998 809 345 228 8;
  • 27) 0.049 804 687 499 998 809 345 228 8 × 2 = 0 + 0.099 609 374 999 997 618 690 457 6;
  • 28) 0.099 609 374 999 997 618 690 457 6 × 2 = 0 + 0.199 218 749 999 995 237 380 915 2;
  • 29) 0.199 218 749 999 995 237 380 915 2 × 2 = 0 + 0.398 437 499 999 990 474 761 830 4;
  • 30) 0.398 437 499 999 990 474 761 830 4 × 2 = 0 + 0.796 874 999 999 980 949 523 660 8;
  • 31) 0.796 874 999 999 980 949 523 660 8 × 2 = 1 + 0.593 749 999 999 961 899 047 321 6;
  • 32) 0.593 749 999 999 961 899 047 321 6 × 2 = 1 + 0.187 499 999 999 923 798 094 643 2;
  • 33) 0.187 499 999 999 923 798 094 643 2 × 2 = 0 + 0.374 999 999 999 847 596 189 286 4;
  • 34) 0.374 999 999 999 847 596 189 286 4 × 2 = 0 + 0.749 999 999 999 695 192 378 572 8;
  • 35) 0.749 999 999 999 695 192 378 572 8 × 2 = 1 + 0.499 999 999 999 390 384 757 145 6;
  • 36) 0.499 999 999 999 390 384 757 145 6 × 2 = 0 + 0.999 999 999 998 780 769 514 291 2;
  • 37) 0.999 999 999 998 780 769 514 291 2 × 2 = 1 + 0.999 999 999 997 561 539 028 582 4;
  • 38) 0.999 999 999 997 561 539 028 582 4 × 2 = 1 + 0.999 999 999 995 123 078 057 164 8;
  • 39) 0.999 999 999 995 123 078 057 164 8 × 2 = 1 + 0.999 999 999 990 246 156 114 329 6;
  • 40) 0.999 999 999 990 246 156 114 329 6 × 2 = 1 + 0.999 999 999 980 492 312 228 659 2;
  • 41) 0.999 999 999 980 492 312 228 659 2 × 2 = 1 + 0.999 999 999 960 984 624 457 318 4;
  • 42) 0.999 999 999 960 984 624 457 318 4 × 2 = 1 + 0.999 999 999 921 969 248 914 636 8;
  • 43) 0.999 999 999 921 969 248 914 636 8 × 2 = 1 + 0.999 999 999 843 938 497 829 273 6;
  • 44) 0.999 999 999 843 938 497 829 273 6 × 2 = 1 + 0.999 999 999 687 876 995 658 547 2;
  • 45) 0.999 999 999 687 876 995 658 547 2 × 2 = 1 + 0.999 999 999 375 753 991 317 094 4;
  • 46) 0.999 999 999 375 753 991 317 094 4 × 2 = 1 + 0.999 999 998 751 507 982 634 188 8;
  • 47) 0.999 999 998 751 507 982 634 188 8 × 2 = 1 + 0.999 999 997 503 015 965 268 377 6;
  • 48) 0.999 999 997 503 015 965 268 377 6 × 2 = 1 + 0.999 999 995 006 031 930 536 755 2;
  • 49) 0.999 999 995 006 031 930 536 755 2 × 2 = 1 + 0.999 999 990 012 063 861 073 510 4;
  • 50) 0.999 999 990 012 063 861 073 510 4 × 2 = 1 + 0.999 999 980 024 127 722 147 020 8;
  • 51) 0.999 999 980 024 127 722 147 020 8 × 2 = 1 + 0.999 999 960 048 255 444 294 041 6;
  • 52) 0.999 999 960 048 255 444 294 041 6 × 2 = 1 + 0.999 999 920 096 510 888 588 083 2;
  • 53) 0.999 999 920 096 510 888 588 083 2 × 2 = 1 + 0.999 999 840 193 021 777 176 166 4;
  • 54) 0.999 999 840 193 021 777 176 166 4 × 2 = 1 + 0.999 999 680 386 043 554 352 332 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:


0.000 000 000 742 147 676 646 691 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 691 7(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:


0.000 000 000 742 147 676 646 691 7(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 -0.000 000 000 742 147 676 646 691 7 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