-0.000 000 000 742 147 676 636 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 636 2(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
-0.000 000 000 742 147 676 636 2(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:

|-0.000 000 000 742 147 676 636 2| = 0.000 000 000 742 147 676 636 2


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: 0.000 000 000 742 147 676 636 2.

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) 0.000 000 000 742 147 676 636 2 × 2 = 0 + 0.000 000 001 484 295 353 272 4;
  • 2) 0.000 000 001 484 295 353 272 4 × 2 = 0 + 0.000 000 002 968 590 706 544 8;
  • 3) 0.000 000 002 968 590 706 544 8 × 2 = 0 + 0.000 000 005 937 181 413 089 6;
  • 4) 0.000 000 005 937 181 413 089 6 × 2 = 0 + 0.000 000 011 874 362 826 179 2;
  • 5) 0.000 000 011 874 362 826 179 2 × 2 = 0 + 0.000 000 023 748 725 652 358 4;
  • 6) 0.000 000 023 748 725 652 358 4 × 2 = 0 + 0.000 000 047 497 451 304 716 8;
  • 7) 0.000 000 047 497 451 304 716 8 × 2 = 0 + 0.000 000 094 994 902 609 433 6;
  • 8) 0.000 000 094 994 902 609 433 6 × 2 = 0 + 0.000 000 189 989 805 218 867 2;
  • 9) 0.000 000 189 989 805 218 867 2 × 2 = 0 + 0.000 000 379 979 610 437 734 4;
  • 10) 0.000 000 379 979 610 437 734 4 × 2 = 0 + 0.000 000 759 959 220 875 468 8;
  • 11) 0.000 000 759 959 220 875 468 8 × 2 = 0 + 0.000 001 519 918 441 750 937 6;
  • 12) 0.000 001 519 918 441 750 937 6 × 2 = 0 + 0.000 003 039 836 883 501 875 2;
  • 13) 0.000 003 039 836 883 501 875 2 × 2 = 0 + 0.000 006 079 673 767 003 750 4;
  • 14) 0.000 006 079 673 767 003 750 4 × 2 = 0 + 0.000 012 159 347 534 007 500 8;
  • 15) 0.000 012 159 347 534 007 500 8 × 2 = 0 + 0.000 024 318 695 068 015 001 6;
  • 16) 0.000 024 318 695 068 015 001 6 × 2 = 0 + 0.000 048 637 390 136 030 003 2;
  • 17) 0.000 048 637 390 136 030 003 2 × 2 = 0 + 0.000 097 274 780 272 060 006 4;
  • 18) 0.000 097 274 780 272 060 006 4 × 2 = 0 + 0.000 194 549 560 544 120 012 8;
  • 19) 0.000 194 549 560 544 120 012 8 × 2 = 0 + 0.000 389 099 121 088 240 025 6;
  • 20) 0.000 389 099 121 088 240 025 6 × 2 = 0 + 0.000 778 198 242 176 480 051 2;
  • 21) 0.000 778 198 242 176 480 051 2 × 2 = 0 + 0.001 556 396 484 352 960 102 4;
  • 22) 0.001 556 396 484 352 960 102 4 × 2 = 0 + 0.003 112 792 968 705 920 204 8;
  • 23) 0.003 112 792 968 705 920 204 8 × 2 = 0 + 0.006 225 585 937 411 840 409 6;
  • 24) 0.006 225 585 937 411 840 409 6 × 2 = 0 + 0.012 451 171 874 823 680 819 2;
  • 25) 0.012 451 171 874 823 680 819 2 × 2 = 0 + 0.024 902 343 749 647 361 638 4;
  • 26) 0.024 902 343 749 647 361 638 4 × 2 = 0 + 0.049 804 687 499 294 723 276 8;
  • 27) 0.049 804 687 499 294 723 276 8 × 2 = 0 + 0.099 609 374 998 589 446 553 6;
  • 28) 0.099 609 374 998 589 446 553 6 × 2 = 0 + 0.199 218 749 997 178 893 107 2;
  • 29) 0.199 218 749 997 178 893 107 2 × 2 = 0 + 0.398 437 499 994 357 786 214 4;
  • 30) 0.398 437 499 994 357 786 214 4 × 2 = 0 + 0.796 874 999 988 715 572 428 8;
  • 31) 0.796 874 999 988 715 572 428 8 × 2 = 1 + 0.593 749 999 977 431 144 857 6;
  • 32) 0.593 749 999 977 431 144 857 6 × 2 = 1 + 0.187 499 999 954 862 289 715 2;
  • 33) 0.187 499 999 954 862 289 715 2 × 2 = 0 + 0.374 999 999 909 724 579 430 4;
  • 34) 0.374 999 999 909 724 579 430 4 × 2 = 0 + 0.749 999 999 819 449 158 860 8;
  • 35) 0.749 999 999 819 449 158 860 8 × 2 = 1 + 0.499 999 999 638 898 317 721 6;
  • 36) 0.499 999 999 638 898 317 721 6 × 2 = 0 + 0.999 999 999 277 796 635 443 2;
  • 37) 0.999 999 999 277 796 635 443 2 × 2 = 1 + 0.999 999 998 555 593 270 886 4;
  • 38) 0.999 999 998 555 593 270 886 4 × 2 = 1 + 0.999 999 997 111 186 541 772 8;
  • 39) 0.999 999 997 111 186 541 772 8 × 2 = 1 + 0.999 999 994 222 373 083 545 6;
  • 40) 0.999 999 994 222 373 083 545 6 × 2 = 1 + 0.999 999 988 444 746 167 091 2;
  • 41) 0.999 999 988 444 746 167 091 2 × 2 = 1 + 0.999 999 976 889 492 334 182 4;
  • 42) 0.999 999 976 889 492 334 182 4 × 2 = 1 + 0.999 999 953 778 984 668 364 8;
  • 43) 0.999 999 953 778 984 668 364 8 × 2 = 1 + 0.999 999 907 557 969 336 729 6;
  • 44) 0.999 999 907 557 969 336 729 6 × 2 = 1 + 0.999 999 815 115 938 673 459 2;
  • 45) 0.999 999 815 115 938 673 459 2 × 2 = 1 + 0.999 999 630 231 877 346 918 4;
  • 46) 0.999 999 630 231 877 346 918 4 × 2 = 1 + 0.999 999 260 463 754 693 836 8;
  • 47) 0.999 999 260 463 754 693 836 8 × 2 = 1 + 0.999 998 520 927 509 387 673 6;
  • 48) 0.999 998 520 927 509 387 673 6 × 2 = 1 + 0.999 997 041 855 018 775 347 2;
  • 49) 0.999 997 041 855 018 775 347 2 × 2 = 1 + 0.999 994 083 710 037 550 694 4;
  • 50) 0.999 994 083 710 037 550 694 4 × 2 = 1 + 0.999 988 167 420 075 101 388 8;
  • 51) 0.999 988 167 420 075 101 388 8 × 2 = 1 + 0.999 976 334 840 150 202 777 6;
  • 52) 0.999 976 334 840 150 202 777 6 × 2 = 1 + 0.999 952 669 680 300 405 555 2;
  • 53) 0.999 952 669 680 300 405 555 2 × 2 = 1 + 0.999 905 339 360 600 811 110 4;
  • 54) 0.999 905 339 360 600 811 110 4 × 2 = 1 + 0.999 810 678 721 201 622 220 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:


0.000 000 000 742 147 676 636 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 636 2(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:


0.000 000 000 742 147 676 636 2(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 -0.000 000 000 742 147 676 636 2 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