-0.000 000 000 742 147 676 614 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 614 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
-0.000 000 000 742 147 676 614 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:

|-0.000 000 000 742 147 676 614 8| = 0.000 000 000 742 147 676 614 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: 0.000 000 000 742 147 676 614 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) 0.000 000 000 742 147 676 614 8 × 2 = 0 + 0.000 000 001 484 295 353 229 6;
  • 2) 0.000 000 001 484 295 353 229 6 × 2 = 0 + 0.000 000 002 968 590 706 459 2;
  • 3) 0.000 000 002 968 590 706 459 2 × 2 = 0 + 0.000 000 005 937 181 412 918 4;
  • 4) 0.000 000 005 937 181 412 918 4 × 2 = 0 + 0.000 000 011 874 362 825 836 8;
  • 5) 0.000 000 011 874 362 825 836 8 × 2 = 0 + 0.000 000 023 748 725 651 673 6;
  • 6) 0.000 000 023 748 725 651 673 6 × 2 = 0 + 0.000 000 047 497 451 303 347 2;
  • 7) 0.000 000 047 497 451 303 347 2 × 2 = 0 + 0.000 000 094 994 902 606 694 4;
  • 8) 0.000 000 094 994 902 606 694 4 × 2 = 0 + 0.000 000 189 989 805 213 388 8;
  • 9) 0.000 000 189 989 805 213 388 8 × 2 = 0 + 0.000 000 379 979 610 426 777 6;
  • 10) 0.000 000 379 979 610 426 777 6 × 2 = 0 + 0.000 000 759 959 220 853 555 2;
  • 11) 0.000 000 759 959 220 853 555 2 × 2 = 0 + 0.000 001 519 918 441 707 110 4;
  • 12) 0.000 001 519 918 441 707 110 4 × 2 = 0 + 0.000 003 039 836 883 414 220 8;
  • 13) 0.000 003 039 836 883 414 220 8 × 2 = 0 + 0.000 006 079 673 766 828 441 6;
  • 14) 0.000 006 079 673 766 828 441 6 × 2 = 0 + 0.000 012 159 347 533 656 883 2;
  • 15) 0.000 012 159 347 533 656 883 2 × 2 = 0 + 0.000 024 318 695 067 313 766 4;
  • 16) 0.000 024 318 695 067 313 766 4 × 2 = 0 + 0.000 048 637 390 134 627 532 8;
  • 17) 0.000 048 637 390 134 627 532 8 × 2 = 0 + 0.000 097 274 780 269 255 065 6;
  • 18) 0.000 097 274 780 269 255 065 6 × 2 = 0 + 0.000 194 549 560 538 510 131 2;
  • 19) 0.000 194 549 560 538 510 131 2 × 2 = 0 + 0.000 389 099 121 077 020 262 4;
  • 20) 0.000 389 099 121 077 020 262 4 × 2 = 0 + 0.000 778 198 242 154 040 524 8;
  • 21) 0.000 778 198 242 154 040 524 8 × 2 = 0 + 0.001 556 396 484 308 081 049 6;
  • 22) 0.001 556 396 484 308 081 049 6 × 2 = 0 + 0.003 112 792 968 616 162 099 2;
  • 23) 0.003 112 792 968 616 162 099 2 × 2 = 0 + 0.006 225 585 937 232 324 198 4;
  • 24) 0.006 225 585 937 232 324 198 4 × 2 = 0 + 0.012 451 171 874 464 648 396 8;
  • 25) 0.012 451 171 874 464 648 396 8 × 2 = 0 + 0.024 902 343 748 929 296 793 6;
  • 26) 0.024 902 343 748 929 296 793 6 × 2 = 0 + 0.049 804 687 497 858 593 587 2;
  • 27) 0.049 804 687 497 858 593 587 2 × 2 = 0 + 0.099 609 374 995 717 187 174 4;
  • 28) 0.099 609 374 995 717 187 174 4 × 2 = 0 + 0.199 218 749 991 434 374 348 8;
  • 29) 0.199 218 749 991 434 374 348 8 × 2 = 0 + 0.398 437 499 982 868 748 697 6;
  • 30) 0.398 437 499 982 868 748 697 6 × 2 = 0 + 0.796 874 999 965 737 497 395 2;
  • 31) 0.796 874 999 965 737 497 395 2 × 2 = 1 + 0.593 749 999 931 474 994 790 4;
  • 32) 0.593 749 999 931 474 994 790 4 × 2 = 1 + 0.187 499 999 862 949 989 580 8;
  • 33) 0.187 499 999 862 949 989 580 8 × 2 = 0 + 0.374 999 999 725 899 979 161 6;
  • 34) 0.374 999 999 725 899 979 161 6 × 2 = 0 + 0.749 999 999 451 799 958 323 2;
  • 35) 0.749 999 999 451 799 958 323 2 × 2 = 1 + 0.499 999 998 903 599 916 646 4;
  • 36) 0.499 999 998 903 599 916 646 4 × 2 = 0 + 0.999 999 997 807 199 833 292 8;
  • 37) 0.999 999 997 807 199 833 292 8 × 2 = 1 + 0.999 999 995 614 399 666 585 6;
  • 38) 0.999 999 995 614 399 666 585 6 × 2 = 1 + 0.999 999 991 228 799 333 171 2;
  • 39) 0.999 999 991 228 799 333 171 2 × 2 = 1 + 0.999 999 982 457 598 666 342 4;
  • 40) 0.999 999 982 457 598 666 342 4 × 2 = 1 + 0.999 999 964 915 197 332 684 8;
  • 41) 0.999 999 964 915 197 332 684 8 × 2 = 1 + 0.999 999 929 830 394 665 369 6;
  • 42) 0.999 999 929 830 394 665 369 6 × 2 = 1 + 0.999 999 859 660 789 330 739 2;
  • 43) 0.999 999 859 660 789 330 739 2 × 2 = 1 + 0.999 999 719 321 578 661 478 4;
  • 44) 0.999 999 719 321 578 661 478 4 × 2 = 1 + 0.999 999 438 643 157 322 956 8;
  • 45) 0.999 999 438 643 157 322 956 8 × 2 = 1 + 0.999 998 877 286 314 645 913 6;
  • 46) 0.999 998 877 286 314 645 913 6 × 2 = 1 + 0.999 997 754 572 629 291 827 2;
  • 47) 0.999 997 754 572 629 291 827 2 × 2 = 1 + 0.999 995 509 145 258 583 654 4;
  • 48) 0.999 995 509 145 258 583 654 4 × 2 = 1 + 0.999 991 018 290 517 167 308 8;
  • 49) 0.999 991 018 290 517 167 308 8 × 2 = 1 + 0.999 982 036 581 034 334 617 6;
  • 50) 0.999 982 036 581 034 334 617 6 × 2 = 1 + 0.999 964 073 162 068 669 235 2;
  • 51) 0.999 964 073 162 068 669 235 2 × 2 = 1 + 0.999 928 146 324 137 338 470 4;
  • 52) 0.999 928 146 324 137 338 470 4 × 2 = 1 + 0.999 856 292 648 274 676 940 8;
  • 53) 0.999 856 292 648 274 676 940 8 × 2 = 1 + 0.999 712 585 296 549 353 881 6;
  • 54) 0.999 712 585 296 549 353 881 6 × 2 = 1 + 0.999 425 170 593 098 707 763 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:


0.000 000 000 742 147 676 614 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 614 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:


0.000 000 000 742 147 676 614 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 -0.000 000 000 742 147 676 614 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