-0.000 000 000 742 147 676 646 678 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 678 8(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
-0.000 000 000 742 147 676 646 678 8(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:

|-0.000 000 000 742 147 676 646 678 8| = 0.000 000 000 742 147 676 646 678 8


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: 0.000 000 000 742 147 676 646 678 8.

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) 0.000 000 000 742 147 676 646 678 8 × 2 = 0 + 0.000 000 001 484 295 353 293 357 6;
  • 2) 0.000 000 001 484 295 353 293 357 6 × 2 = 0 + 0.000 000 002 968 590 706 586 715 2;
  • 3) 0.000 000 002 968 590 706 586 715 2 × 2 = 0 + 0.000 000 005 937 181 413 173 430 4;
  • 4) 0.000 000 005 937 181 413 173 430 4 × 2 = 0 + 0.000 000 011 874 362 826 346 860 8;
  • 5) 0.000 000 011 874 362 826 346 860 8 × 2 = 0 + 0.000 000 023 748 725 652 693 721 6;
  • 6) 0.000 000 023 748 725 652 693 721 6 × 2 = 0 + 0.000 000 047 497 451 305 387 443 2;
  • 7) 0.000 000 047 497 451 305 387 443 2 × 2 = 0 + 0.000 000 094 994 902 610 774 886 4;
  • 8) 0.000 000 094 994 902 610 774 886 4 × 2 = 0 + 0.000 000 189 989 805 221 549 772 8;
  • 9) 0.000 000 189 989 805 221 549 772 8 × 2 = 0 + 0.000 000 379 979 610 443 099 545 6;
  • 10) 0.000 000 379 979 610 443 099 545 6 × 2 = 0 + 0.000 000 759 959 220 886 199 091 2;
  • 11) 0.000 000 759 959 220 886 199 091 2 × 2 = 0 + 0.000 001 519 918 441 772 398 182 4;
  • 12) 0.000 001 519 918 441 772 398 182 4 × 2 = 0 + 0.000 003 039 836 883 544 796 364 8;
  • 13) 0.000 003 039 836 883 544 796 364 8 × 2 = 0 + 0.000 006 079 673 767 089 592 729 6;
  • 14) 0.000 006 079 673 767 089 592 729 6 × 2 = 0 + 0.000 012 159 347 534 179 185 459 2;
  • 15) 0.000 012 159 347 534 179 185 459 2 × 2 = 0 + 0.000 024 318 695 068 358 370 918 4;
  • 16) 0.000 024 318 695 068 358 370 918 4 × 2 = 0 + 0.000 048 637 390 136 716 741 836 8;
  • 17) 0.000 048 637 390 136 716 741 836 8 × 2 = 0 + 0.000 097 274 780 273 433 483 673 6;
  • 18) 0.000 097 274 780 273 433 483 673 6 × 2 = 0 + 0.000 194 549 560 546 866 967 347 2;
  • 19) 0.000 194 549 560 546 866 967 347 2 × 2 = 0 + 0.000 389 099 121 093 733 934 694 4;
  • 20) 0.000 389 099 121 093 733 934 694 4 × 2 = 0 + 0.000 778 198 242 187 467 869 388 8;
  • 21) 0.000 778 198 242 187 467 869 388 8 × 2 = 0 + 0.001 556 396 484 374 935 738 777 6;
  • 22) 0.001 556 396 484 374 935 738 777 6 × 2 = 0 + 0.003 112 792 968 749 871 477 555 2;
  • 23) 0.003 112 792 968 749 871 477 555 2 × 2 = 0 + 0.006 225 585 937 499 742 955 110 4;
  • 24) 0.006 225 585 937 499 742 955 110 4 × 2 = 0 + 0.012 451 171 874 999 485 910 220 8;
  • 25) 0.012 451 171 874 999 485 910 220 8 × 2 = 0 + 0.024 902 343 749 998 971 820 441 6;
  • 26) 0.024 902 343 749 998 971 820 441 6 × 2 = 0 + 0.049 804 687 499 997 943 640 883 2;
  • 27) 0.049 804 687 499 997 943 640 883 2 × 2 = 0 + 0.099 609 374 999 995 887 281 766 4;
  • 28) 0.099 609 374 999 995 887 281 766 4 × 2 = 0 + 0.199 218 749 999 991 774 563 532 8;
  • 29) 0.199 218 749 999 991 774 563 532 8 × 2 = 0 + 0.398 437 499 999 983 549 127 065 6;
  • 30) 0.398 437 499 999 983 549 127 065 6 × 2 = 0 + 0.796 874 999 999 967 098 254 131 2;
  • 31) 0.796 874 999 999 967 098 254 131 2 × 2 = 1 + 0.593 749 999 999 934 196 508 262 4;
  • 32) 0.593 749 999 999 934 196 508 262 4 × 2 = 1 + 0.187 499 999 999 868 393 016 524 8;
  • 33) 0.187 499 999 999 868 393 016 524 8 × 2 = 0 + 0.374 999 999 999 736 786 033 049 6;
  • 34) 0.374 999 999 999 736 786 033 049 6 × 2 = 0 + 0.749 999 999 999 473 572 066 099 2;
  • 35) 0.749 999 999 999 473 572 066 099 2 × 2 = 1 + 0.499 999 999 998 947 144 132 198 4;
  • 36) 0.499 999 999 998 947 144 132 198 4 × 2 = 0 + 0.999 999 999 997 894 288 264 396 8;
  • 37) 0.999 999 999 997 894 288 264 396 8 × 2 = 1 + 0.999 999 999 995 788 576 528 793 6;
  • 38) 0.999 999 999 995 788 576 528 793 6 × 2 = 1 + 0.999 999 999 991 577 153 057 587 2;
  • 39) 0.999 999 999 991 577 153 057 587 2 × 2 = 1 + 0.999 999 999 983 154 306 115 174 4;
  • 40) 0.999 999 999 983 154 306 115 174 4 × 2 = 1 + 0.999 999 999 966 308 612 230 348 8;
  • 41) 0.999 999 999 966 308 612 230 348 8 × 2 = 1 + 0.999 999 999 932 617 224 460 697 6;
  • 42) 0.999 999 999 932 617 224 460 697 6 × 2 = 1 + 0.999 999 999 865 234 448 921 395 2;
  • 43) 0.999 999 999 865 234 448 921 395 2 × 2 = 1 + 0.999 999 999 730 468 897 842 790 4;
  • 44) 0.999 999 999 730 468 897 842 790 4 × 2 = 1 + 0.999 999 999 460 937 795 685 580 8;
  • 45) 0.999 999 999 460 937 795 685 580 8 × 2 = 1 + 0.999 999 998 921 875 591 371 161 6;
  • 46) 0.999 999 998 921 875 591 371 161 6 × 2 = 1 + 0.999 999 997 843 751 182 742 323 2;
  • 47) 0.999 999 997 843 751 182 742 323 2 × 2 = 1 + 0.999 999 995 687 502 365 484 646 4;
  • 48) 0.999 999 995 687 502 365 484 646 4 × 2 = 1 + 0.999 999 991 375 004 730 969 292 8;
  • 49) 0.999 999 991 375 004 730 969 292 8 × 2 = 1 + 0.999 999 982 750 009 461 938 585 6;
  • 50) 0.999 999 982 750 009 461 938 585 6 × 2 = 1 + 0.999 999 965 500 018 923 877 171 2;
  • 51) 0.999 999 965 500 018 923 877 171 2 × 2 = 1 + 0.999 999 931 000 037 847 754 342 4;
  • 52) 0.999 999 931 000 037 847 754 342 4 × 2 = 1 + 0.999 999 862 000 075 695 508 684 8;
  • 53) 0.999 999 862 000 075 695 508 684 8 × 2 = 1 + 0.999 999 724 000 151 391 017 369 6;
  • 54) 0.999 999 724 000 151 391 017 369 6 × 2 = 1 + 0.999 999 448 000 302 782 034 739 2;

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:


0.000 000 000 742 147 676 646 678 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 678 8(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:


0.000 000 000 742 147 676 646 678 8(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 -0.000 000 000 742 147 676 646 678 8 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