-0.000 000 000 742 147 676 645 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 645 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
-0.000 000 000 742 147 676 645 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:

|-0.000 000 000 742 147 676 645 7| = 0.000 000 000 742 147 676 645 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: 0.000 000 000 742 147 676 645 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) 0.000 000 000 742 147 676 645 7 × 2 = 0 + 0.000 000 001 484 295 353 291 4;
  • 2) 0.000 000 001 484 295 353 291 4 × 2 = 0 + 0.000 000 002 968 590 706 582 8;
  • 3) 0.000 000 002 968 590 706 582 8 × 2 = 0 + 0.000 000 005 937 181 413 165 6;
  • 4) 0.000 000 005 937 181 413 165 6 × 2 = 0 + 0.000 000 011 874 362 826 331 2;
  • 5) 0.000 000 011 874 362 826 331 2 × 2 = 0 + 0.000 000 023 748 725 652 662 4;
  • 6) 0.000 000 023 748 725 652 662 4 × 2 = 0 + 0.000 000 047 497 451 305 324 8;
  • 7) 0.000 000 047 497 451 305 324 8 × 2 = 0 + 0.000 000 094 994 902 610 649 6;
  • 8) 0.000 000 094 994 902 610 649 6 × 2 = 0 + 0.000 000 189 989 805 221 299 2;
  • 9) 0.000 000 189 989 805 221 299 2 × 2 = 0 + 0.000 000 379 979 610 442 598 4;
  • 10) 0.000 000 379 979 610 442 598 4 × 2 = 0 + 0.000 000 759 959 220 885 196 8;
  • 11) 0.000 000 759 959 220 885 196 8 × 2 = 0 + 0.000 001 519 918 441 770 393 6;
  • 12) 0.000 001 519 918 441 770 393 6 × 2 = 0 + 0.000 003 039 836 883 540 787 2;
  • 13) 0.000 003 039 836 883 540 787 2 × 2 = 0 + 0.000 006 079 673 767 081 574 4;
  • 14) 0.000 006 079 673 767 081 574 4 × 2 = 0 + 0.000 012 159 347 534 163 148 8;
  • 15) 0.000 012 159 347 534 163 148 8 × 2 = 0 + 0.000 024 318 695 068 326 297 6;
  • 16) 0.000 024 318 695 068 326 297 6 × 2 = 0 + 0.000 048 637 390 136 652 595 2;
  • 17) 0.000 048 637 390 136 652 595 2 × 2 = 0 + 0.000 097 274 780 273 305 190 4;
  • 18) 0.000 097 274 780 273 305 190 4 × 2 = 0 + 0.000 194 549 560 546 610 380 8;
  • 19) 0.000 194 549 560 546 610 380 8 × 2 = 0 + 0.000 389 099 121 093 220 761 6;
  • 20) 0.000 389 099 121 093 220 761 6 × 2 = 0 + 0.000 778 198 242 186 441 523 2;
  • 21) 0.000 778 198 242 186 441 523 2 × 2 = 0 + 0.001 556 396 484 372 883 046 4;
  • 22) 0.001 556 396 484 372 883 046 4 × 2 = 0 + 0.003 112 792 968 745 766 092 8;
  • 23) 0.003 112 792 968 745 766 092 8 × 2 = 0 + 0.006 225 585 937 491 532 185 6;
  • 24) 0.006 225 585 937 491 532 185 6 × 2 = 0 + 0.012 451 171 874 983 064 371 2;
  • 25) 0.012 451 171 874 983 064 371 2 × 2 = 0 + 0.024 902 343 749 966 128 742 4;
  • 26) 0.024 902 343 749 966 128 742 4 × 2 = 0 + 0.049 804 687 499 932 257 484 8;
  • 27) 0.049 804 687 499 932 257 484 8 × 2 = 0 + 0.099 609 374 999 864 514 969 6;
  • 28) 0.099 609 374 999 864 514 969 6 × 2 = 0 + 0.199 218 749 999 729 029 939 2;
  • 29) 0.199 218 749 999 729 029 939 2 × 2 = 0 + 0.398 437 499 999 458 059 878 4;
  • 30) 0.398 437 499 999 458 059 878 4 × 2 = 0 + 0.796 874 999 998 916 119 756 8;
  • 31) 0.796 874 999 998 916 119 756 8 × 2 = 1 + 0.593 749 999 997 832 239 513 6;
  • 32) 0.593 749 999 997 832 239 513 6 × 2 = 1 + 0.187 499 999 995 664 479 027 2;
  • 33) 0.187 499 999 995 664 479 027 2 × 2 = 0 + 0.374 999 999 991 328 958 054 4;
  • 34) 0.374 999 999 991 328 958 054 4 × 2 = 0 + 0.749 999 999 982 657 916 108 8;
  • 35) 0.749 999 999 982 657 916 108 8 × 2 = 1 + 0.499 999 999 965 315 832 217 6;
  • 36) 0.499 999 999 965 315 832 217 6 × 2 = 0 + 0.999 999 999 930 631 664 435 2;
  • 37) 0.999 999 999 930 631 664 435 2 × 2 = 1 + 0.999 999 999 861 263 328 870 4;
  • 38) 0.999 999 999 861 263 328 870 4 × 2 = 1 + 0.999 999 999 722 526 657 740 8;
  • 39) 0.999 999 999 722 526 657 740 8 × 2 = 1 + 0.999 999 999 445 053 315 481 6;
  • 40) 0.999 999 999 445 053 315 481 6 × 2 = 1 + 0.999 999 998 890 106 630 963 2;
  • 41) 0.999 999 998 890 106 630 963 2 × 2 = 1 + 0.999 999 997 780 213 261 926 4;
  • 42) 0.999 999 997 780 213 261 926 4 × 2 = 1 + 0.999 999 995 560 426 523 852 8;
  • 43) 0.999 999 995 560 426 523 852 8 × 2 = 1 + 0.999 999 991 120 853 047 705 6;
  • 44) 0.999 999 991 120 853 047 705 6 × 2 = 1 + 0.999 999 982 241 706 095 411 2;
  • 45) 0.999 999 982 241 706 095 411 2 × 2 = 1 + 0.999 999 964 483 412 190 822 4;
  • 46) 0.999 999 964 483 412 190 822 4 × 2 = 1 + 0.999 999 928 966 824 381 644 8;
  • 47) 0.999 999 928 966 824 381 644 8 × 2 = 1 + 0.999 999 857 933 648 763 289 6;
  • 48) 0.999 999 857 933 648 763 289 6 × 2 = 1 + 0.999 999 715 867 297 526 579 2;
  • 49) 0.999 999 715 867 297 526 579 2 × 2 = 1 + 0.999 999 431 734 595 053 158 4;
  • 50) 0.999 999 431 734 595 053 158 4 × 2 = 1 + 0.999 998 863 469 190 106 316 8;
  • 51) 0.999 998 863 469 190 106 316 8 × 2 = 1 + 0.999 997 726 938 380 212 633 6;
  • 52) 0.999 997 726 938 380 212 633 6 × 2 = 1 + 0.999 995 453 876 760 425 267 2;
  • 53) 0.999 995 453 876 760 425 267 2 × 2 = 1 + 0.999 990 907 753 520 850 534 4;
  • 54) 0.999 990 907 753 520 850 534 4 × 2 = 1 + 0.999 981 815 507 041 701 068 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:


0.000 000 000 742 147 676 645 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 645 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:


0.000 000 000 742 147 676 645 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 -0.000 000 000 742 147 676 645 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