-0.000 000 000 742 147 676 625 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 625 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
-0.000 000 000 742 147 676 625 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:

|-0.000 000 000 742 147 676 625 7| = 0.000 000 000 742 147 676 625 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: 0.000 000 000 742 147 676 625 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) 0.000 000 000 742 147 676 625 7 × 2 = 0 + 0.000 000 001 484 295 353 251 4;
  • 2) 0.000 000 001 484 295 353 251 4 × 2 = 0 + 0.000 000 002 968 590 706 502 8;
  • 3) 0.000 000 002 968 590 706 502 8 × 2 = 0 + 0.000 000 005 937 181 413 005 6;
  • 4) 0.000 000 005 937 181 413 005 6 × 2 = 0 + 0.000 000 011 874 362 826 011 2;
  • 5) 0.000 000 011 874 362 826 011 2 × 2 = 0 + 0.000 000 023 748 725 652 022 4;
  • 6) 0.000 000 023 748 725 652 022 4 × 2 = 0 + 0.000 000 047 497 451 304 044 8;
  • 7) 0.000 000 047 497 451 304 044 8 × 2 = 0 + 0.000 000 094 994 902 608 089 6;
  • 8) 0.000 000 094 994 902 608 089 6 × 2 = 0 + 0.000 000 189 989 805 216 179 2;
  • 9) 0.000 000 189 989 805 216 179 2 × 2 = 0 + 0.000 000 379 979 610 432 358 4;
  • 10) 0.000 000 379 979 610 432 358 4 × 2 = 0 + 0.000 000 759 959 220 864 716 8;
  • 11) 0.000 000 759 959 220 864 716 8 × 2 = 0 + 0.000 001 519 918 441 729 433 6;
  • 12) 0.000 001 519 918 441 729 433 6 × 2 = 0 + 0.000 003 039 836 883 458 867 2;
  • 13) 0.000 003 039 836 883 458 867 2 × 2 = 0 + 0.000 006 079 673 766 917 734 4;
  • 14) 0.000 006 079 673 766 917 734 4 × 2 = 0 + 0.000 012 159 347 533 835 468 8;
  • 15) 0.000 012 159 347 533 835 468 8 × 2 = 0 + 0.000 024 318 695 067 670 937 6;
  • 16) 0.000 024 318 695 067 670 937 6 × 2 = 0 + 0.000 048 637 390 135 341 875 2;
  • 17) 0.000 048 637 390 135 341 875 2 × 2 = 0 + 0.000 097 274 780 270 683 750 4;
  • 18) 0.000 097 274 780 270 683 750 4 × 2 = 0 + 0.000 194 549 560 541 367 500 8;
  • 19) 0.000 194 549 560 541 367 500 8 × 2 = 0 + 0.000 389 099 121 082 735 001 6;
  • 20) 0.000 389 099 121 082 735 001 6 × 2 = 0 + 0.000 778 198 242 165 470 003 2;
  • 21) 0.000 778 198 242 165 470 003 2 × 2 = 0 + 0.001 556 396 484 330 940 006 4;
  • 22) 0.001 556 396 484 330 940 006 4 × 2 = 0 + 0.003 112 792 968 661 880 012 8;
  • 23) 0.003 112 792 968 661 880 012 8 × 2 = 0 + 0.006 225 585 937 323 760 025 6;
  • 24) 0.006 225 585 937 323 760 025 6 × 2 = 0 + 0.012 451 171 874 647 520 051 2;
  • 25) 0.012 451 171 874 647 520 051 2 × 2 = 0 + 0.024 902 343 749 295 040 102 4;
  • 26) 0.024 902 343 749 295 040 102 4 × 2 = 0 + 0.049 804 687 498 590 080 204 8;
  • 27) 0.049 804 687 498 590 080 204 8 × 2 = 0 + 0.099 609 374 997 180 160 409 6;
  • 28) 0.099 609 374 997 180 160 409 6 × 2 = 0 + 0.199 218 749 994 360 320 819 2;
  • 29) 0.199 218 749 994 360 320 819 2 × 2 = 0 + 0.398 437 499 988 720 641 638 4;
  • 30) 0.398 437 499 988 720 641 638 4 × 2 = 0 + 0.796 874 999 977 441 283 276 8;
  • 31) 0.796 874 999 977 441 283 276 8 × 2 = 1 + 0.593 749 999 954 882 566 553 6;
  • 32) 0.593 749 999 954 882 566 553 6 × 2 = 1 + 0.187 499 999 909 765 133 107 2;
  • 33) 0.187 499 999 909 765 133 107 2 × 2 = 0 + 0.374 999 999 819 530 266 214 4;
  • 34) 0.374 999 999 819 530 266 214 4 × 2 = 0 + 0.749 999 999 639 060 532 428 8;
  • 35) 0.749 999 999 639 060 532 428 8 × 2 = 1 + 0.499 999 999 278 121 064 857 6;
  • 36) 0.499 999 999 278 121 064 857 6 × 2 = 0 + 0.999 999 998 556 242 129 715 2;
  • 37) 0.999 999 998 556 242 129 715 2 × 2 = 1 + 0.999 999 997 112 484 259 430 4;
  • 38) 0.999 999 997 112 484 259 430 4 × 2 = 1 + 0.999 999 994 224 968 518 860 8;
  • 39) 0.999 999 994 224 968 518 860 8 × 2 = 1 + 0.999 999 988 449 937 037 721 6;
  • 40) 0.999 999 988 449 937 037 721 6 × 2 = 1 + 0.999 999 976 899 874 075 443 2;
  • 41) 0.999 999 976 899 874 075 443 2 × 2 = 1 + 0.999 999 953 799 748 150 886 4;
  • 42) 0.999 999 953 799 748 150 886 4 × 2 = 1 + 0.999 999 907 599 496 301 772 8;
  • 43) 0.999 999 907 599 496 301 772 8 × 2 = 1 + 0.999 999 815 198 992 603 545 6;
  • 44) 0.999 999 815 198 992 603 545 6 × 2 = 1 + 0.999 999 630 397 985 207 091 2;
  • 45) 0.999 999 630 397 985 207 091 2 × 2 = 1 + 0.999 999 260 795 970 414 182 4;
  • 46) 0.999 999 260 795 970 414 182 4 × 2 = 1 + 0.999 998 521 591 940 828 364 8;
  • 47) 0.999 998 521 591 940 828 364 8 × 2 = 1 + 0.999 997 043 183 881 656 729 6;
  • 48) 0.999 997 043 183 881 656 729 6 × 2 = 1 + 0.999 994 086 367 763 313 459 2;
  • 49) 0.999 994 086 367 763 313 459 2 × 2 = 1 + 0.999 988 172 735 526 626 918 4;
  • 50) 0.999 988 172 735 526 626 918 4 × 2 = 1 + 0.999 976 345 471 053 253 836 8;
  • 51) 0.999 976 345 471 053 253 836 8 × 2 = 1 + 0.999 952 690 942 106 507 673 6;
  • 52) 0.999 952 690 942 106 507 673 6 × 2 = 1 + 0.999 905 381 884 213 015 347 2;
  • 53) 0.999 905 381 884 213 015 347 2 × 2 = 1 + 0.999 810 763 768 426 030 694 4;
  • 54) 0.999 810 763 768 426 030 694 4 × 2 = 1 + 0.999 621 527 536 852 061 388 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:


0.000 000 000 742 147 676 625 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 625 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:


0.000 000 000 742 147 676 625 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 -0.000 000 000 742 147 676 625 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