-0.000 000 000 742 147 676 668 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 668 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
-0.000 000 000 742 147 676 668 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:

|-0.000 000 000 742 147 676 668 7| = 0.000 000 000 742 147 676 668 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: 0.000 000 000 742 147 676 668 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) 0.000 000 000 742 147 676 668 7 × 2 = 0 + 0.000 000 001 484 295 353 337 4;
  • 2) 0.000 000 001 484 295 353 337 4 × 2 = 0 + 0.000 000 002 968 590 706 674 8;
  • 3) 0.000 000 002 968 590 706 674 8 × 2 = 0 + 0.000 000 005 937 181 413 349 6;
  • 4) 0.000 000 005 937 181 413 349 6 × 2 = 0 + 0.000 000 011 874 362 826 699 2;
  • 5) 0.000 000 011 874 362 826 699 2 × 2 = 0 + 0.000 000 023 748 725 653 398 4;
  • 6) 0.000 000 023 748 725 653 398 4 × 2 = 0 + 0.000 000 047 497 451 306 796 8;
  • 7) 0.000 000 047 497 451 306 796 8 × 2 = 0 + 0.000 000 094 994 902 613 593 6;
  • 8) 0.000 000 094 994 902 613 593 6 × 2 = 0 + 0.000 000 189 989 805 227 187 2;
  • 9) 0.000 000 189 989 805 227 187 2 × 2 = 0 + 0.000 000 379 979 610 454 374 4;
  • 10) 0.000 000 379 979 610 454 374 4 × 2 = 0 + 0.000 000 759 959 220 908 748 8;
  • 11) 0.000 000 759 959 220 908 748 8 × 2 = 0 + 0.000 001 519 918 441 817 497 6;
  • 12) 0.000 001 519 918 441 817 497 6 × 2 = 0 + 0.000 003 039 836 883 634 995 2;
  • 13) 0.000 003 039 836 883 634 995 2 × 2 = 0 + 0.000 006 079 673 767 269 990 4;
  • 14) 0.000 006 079 673 767 269 990 4 × 2 = 0 + 0.000 012 159 347 534 539 980 8;
  • 15) 0.000 012 159 347 534 539 980 8 × 2 = 0 + 0.000 024 318 695 069 079 961 6;
  • 16) 0.000 024 318 695 069 079 961 6 × 2 = 0 + 0.000 048 637 390 138 159 923 2;
  • 17) 0.000 048 637 390 138 159 923 2 × 2 = 0 + 0.000 097 274 780 276 319 846 4;
  • 18) 0.000 097 274 780 276 319 846 4 × 2 = 0 + 0.000 194 549 560 552 639 692 8;
  • 19) 0.000 194 549 560 552 639 692 8 × 2 = 0 + 0.000 389 099 121 105 279 385 6;
  • 20) 0.000 389 099 121 105 279 385 6 × 2 = 0 + 0.000 778 198 242 210 558 771 2;
  • 21) 0.000 778 198 242 210 558 771 2 × 2 = 0 + 0.001 556 396 484 421 117 542 4;
  • 22) 0.001 556 396 484 421 117 542 4 × 2 = 0 + 0.003 112 792 968 842 235 084 8;
  • 23) 0.003 112 792 968 842 235 084 8 × 2 = 0 + 0.006 225 585 937 684 470 169 6;
  • 24) 0.006 225 585 937 684 470 169 6 × 2 = 0 + 0.012 451 171 875 368 940 339 2;
  • 25) 0.012 451 171 875 368 940 339 2 × 2 = 0 + 0.024 902 343 750 737 880 678 4;
  • 26) 0.024 902 343 750 737 880 678 4 × 2 = 0 + 0.049 804 687 501 475 761 356 8;
  • 27) 0.049 804 687 501 475 761 356 8 × 2 = 0 + 0.099 609 375 002 951 522 713 6;
  • 28) 0.099 609 375 002 951 522 713 6 × 2 = 0 + 0.199 218 750 005 903 045 427 2;
  • 29) 0.199 218 750 005 903 045 427 2 × 2 = 0 + 0.398 437 500 011 806 090 854 4;
  • 30) 0.398 437 500 011 806 090 854 4 × 2 = 0 + 0.796 875 000 023 612 181 708 8;
  • 31) 0.796 875 000 023 612 181 708 8 × 2 = 1 + 0.593 750 000 047 224 363 417 6;
  • 32) 0.593 750 000 047 224 363 417 6 × 2 = 1 + 0.187 500 000 094 448 726 835 2;
  • 33) 0.187 500 000 094 448 726 835 2 × 2 = 0 + 0.375 000 000 188 897 453 670 4;
  • 34) 0.375 000 000 188 897 453 670 4 × 2 = 0 + 0.750 000 000 377 794 907 340 8;
  • 35) 0.750 000 000 377 794 907 340 8 × 2 = 1 + 0.500 000 000 755 589 814 681 6;
  • 36) 0.500 000 000 755 589 814 681 6 × 2 = 1 + 0.000 000 001 511 179 629 363 2;
  • 37) 0.000 000 001 511 179 629 363 2 × 2 = 0 + 0.000 000 003 022 359 258 726 4;
  • 38) 0.000 000 003 022 359 258 726 4 × 2 = 0 + 0.000 000 006 044 718 517 452 8;
  • 39) 0.000 000 006 044 718 517 452 8 × 2 = 0 + 0.000 000 012 089 437 034 905 6;
  • 40) 0.000 000 012 089 437 034 905 6 × 2 = 0 + 0.000 000 024 178 874 069 811 2;
  • 41) 0.000 000 024 178 874 069 811 2 × 2 = 0 + 0.000 000 048 357 748 139 622 4;
  • 42) 0.000 000 048 357 748 139 622 4 × 2 = 0 + 0.000 000 096 715 496 279 244 8;
  • 43) 0.000 000 096 715 496 279 244 8 × 2 = 0 + 0.000 000 193 430 992 558 489 6;
  • 44) 0.000 000 193 430 992 558 489 6 × 2 = 0 + 0.000 000 386 861 985 116 979 2;
  • 45) 0.000 000 386 861 985 116 979 2 × 2 = 0 + 0.000 000 773 723 970 233 958 4;
  • 46) 0.000 000 773 723 970 233 958 4 × 2 = 0 + 0.000 001 547 447 940 467 916 8;
  • 47) 0.000 001 547 447 940 467 916 8 × 2 = 0 + 0.000 003 094 895 880 935 833 6;
  • 48) 0.000 003 094 895 880 935 833 6 × 2 = 0 + 0.000 006 189 791 761 871 667 2;
  • 49) 0.000 006 189 791 761 871 667 2 × 2 = 0 + 0.000 012 379 583 523 743 334 4;
  • 50) 0.000 012 379 583 523 743 334 4 × 2 = 0 + 0.000 024 759 167 047 486 668 8;
  • 51) 0.000 024 759 167 047 486 668 8 × 2 = 0 + 0.000 049 518 334 094 973 337 6;
  • 52) 0.000 049 518 334 094 973 337 6 × 2 = 0 + 0.000 099 036 668 189 946 675 2;
  • 53) 0.000 099 036 668 189 946 675 2 × 2 = 0 + 0.000 198 073 336 379 893 350 4;
  • 54) 0.000 198 073 336 379 893 350 4 × 2 = 0 + 0.000 396 146 672 759 786 700 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:


0.000 000 000 742 147 676 668 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 668 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 668 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 668 7 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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