-0.000 000 000 742 147 676 629 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 629 3(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
-0.000 000 000 742 147 676 629 3(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:

|-0.000 000 000 742 147 676 629 3| = 0.000 000 000 742 147 676 629 3


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: 0.000 000 000 742 147 676 629 3.

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) 0.000 000 000 742 147 676 629 3 × 2 = 0 + 0.000 000 001 484 295 353 258 6;
  • 2) 0.000 000 001 484 295 353 258 6 × 2 = 0 + 0.000 000 002 968 590 706 517 2;
  • 3) 0.000 000 002 968 590 706 517 2 × 2 = 0 + 0.000 000 005 937 181 413 034 4;
  • 4) 0.000 000 005 937 181 413 034 4 × 2 = 0 + 0.000 000 011 874 362 826 068 8;
  • 5) 0.000 000 011 874 362 826 068 8 × 2 = 0 + 0.000 000 023 748 725 652 137 6;
  • 6) 0.000 000 023 748 725 652 137 6 × 2 = 0 + 0.000 000 047 497 451 304 275 2;
  • 7) 0.000 000 047 497 451 304 275 2 × 2 = 0 + 0.000 000 094 994 902 608 550 4;
  • 8) 0.000 000 094 994 902 608 550 4 × 2 = 0 + 0.000 000 189 989 805 217 100 8;
  • 9) 0.000 000 189 989 805 217 100 8 × 2 = 0 + 0.000 000 379 979 610 434 201 6;
  • 10) 0.000 000 379 979 610 434 201 6 × 2 = 0 + 0.000 000 759 959 220 868 403 2;
  • 11) 0.000 000 759 959 220 868 403 2 × 2 = 0 + 0.000 001 519 918 441 736 806 4;
  • 12) 0.000 001 519 918 441 736 806 4 × 2 = 0 + 0.000 003 039 836 883 473 612 8;
  • 13) 0.000 003 039 836 883 473 612 8 × 2 = 0 + 0.000 006 079 673 766 947 225 6;
  • 14) 0.000 006 079 673 766 947 225 6 × 2 = 0 + 0.000 012 159 347 533 894 451 2;
  • 15) 0.000 012 159 347 533 894 451 2 × 2 = 0 + 0.000 024 318 695 067 788 902 4;
  • 16) 0.000 024 318 695 067 788 902 4 × 2 = 0 + 0.000 048 637 390 135 577 804 8;
  • 17) 0.000 048 637 390 135 577 804 8 × 2 = 0 + 0.000 097 274 780 271 155 609 6;
  • 18) 0.000 097 274 780 271 155 609 6 × 2 = 0 + 0.000 194 549 560 542 311 219 2;
  • 19) 0.000 194 549 560 542 311 219 2 × 2 = 0 + 0.000 389 099 121 084 622 438 4;
  • 20) 0.000 389 099 121 084 622 438 4 × 2 = 0 + 0.000 778 198 242 169 244 876 8;
  • 21) 0.000 778 198 242 169 244 876 8 × 2 = 0 + 0.001 556 396 484 338 489 753 6;
  • 22) 0.001 556 396 484 338 489 753 6 × 2 = 0 + 0.003 112 792 968 676 979 507 2;
  • 23) 0.003 112 792 968 676 979 507 2 × 2 = 0 + 0.006 225 585 937 353 959 014 4;
  • 24) 0.006 225 585 937 353 959 014 4 × 2 = 0 + 0.012 451 171 874 707 918 028 8;
  • 25) 0.012 451 171 874 707 918 028 8 × 2 = 0 + 0.024 902 343 749 415 836 057 6;
  • 26) 0.024 902 343 749 415 836 057 6 × 2 = 0 + 0.049 804 687 498 831 672 115 2;
  • 27) 0.049 804 687 498 831 672 115 2 × 2 = 0 + 0.099 609 374 997 663 344 230 4;
  • 28) 0.099 609 374 997 663 344 230 4 × 2 = 0 + 0.199 218 749 995 326 688 460 8;
  • 29) 0.199 218 749 995 326 688 460 8 × 2 = 0 + 0.398 437 499 990 653 376 921 6;
  • 30) 0.398 437 499 990 653 376 921 6 × 2 = 0 + 0.796 874 999 981 306 753 843 2;
  • 31) 0.796 874 999 981 306 753 843 2 × 2 = 1 + 0.593 749 999 962 613 507 686 4;
  • 32) 0.593 749 999 962 613 507 686 4 × 2 = 1 + 0.187 499 999 925 227 015 372 8;
  • 33) 0.187 499 999 925 227 015 372 8 × 2 = 0 + 0.374 999 999 850 454 030 745 6;
  • 34) 0.374 999 999 850 454 030 745 6 × 2 = 0 + 0.749 999 999 700 908 061 491 2;
  • 35) 0.749 999 999 700 908 061 491 2 × 2 = 1 + 0.499 999 999 401 816 122 982 4;
  • 36) 0.499 999 999 401 816 122 982 4 × 2 = 0 + 0.999 999 998 803 632 245 964 8;
  • 37) 0.999 999 998 803 632 245 964 8 × 2 = 1 + 0.999 999 997 607 264 491 929 6;
  • 38) 0.999 999 997 607 264 491 929 6 × 2 = 1 + 0.999 999 995 214 528 983 859 2;
  • 39) 0.999 999 995 214 528 983 859 2 × 2 = 1 + 0.999 999 990 429 057 967 718 4;
  • 40) 0.999 999 990 429 057 967 718 4 × 2 = 1 + 0.999 999 980 858 115 935 436 8;
  • 41) 0.999 999 980 858 115 935 436 8 × 2 = 1 + 0.999 999 961 716 231 870 873 6;
  • 42) 0.999 999 961 716 231 870 873 6 × 2 = 1 + 0.999 999 923 432 463 741 747 2;
  • 43) 0.999 999 923 432 463 741 747 2 × 2 = 1 + 0.999 999 846 864 927 483 494 4;
  • 44) 0.999 999 846 864 927 483 494 4 × 2 = 1 + 0.999 999 693 729 854 966 988 8;
  • 45) 0.999 999 693 729 854 966 988 8 × 2 = 1 + 0.999 999 387 459 709 933 977 6;
  • 46) 0.999 999 387 459 709 933 977 6 × 2 = 1 + 0.999 998 774 919 419 867 955 2;
  • 47) 0.999 998 774 919 419 867 955 2 × 2 = 1 + 0.999 997 549 838 839 735 910 4;
  • 48) 0.999 997 549 838 839 735 910 4 × 2 = 1 + 0.999 995 099 677 679 471 820 8;
  • 49) 0.999 995 099 677 679 471 820 8 × 2 = 1 + 0.999 990 199 355 358 943 641 6;
  • 50) 0.999 990 199 355 358 943 641 6 × 2 = 1 + 0.999 980 398 710 717 887 283 2;
  • 51) 0.999 980 398 710 717 887 283 2 × 2 = 1 + 0.999 960 797 421 435 774 566 4;
  • 52) 0.999 960 797 421 435 774 566 4 × 2 = 1 + 0.999 921 594 842 871 549 132 8;
  • 53) 0.999 921 594 842 871 549 132 8 × 2 = 1 + 0.999 843 189 685 743 098 265 6;
  • 54) 0.999 843 189 685 743 098 265 6 × 2 = 1 + 0.999 686 379 371 486 196 531 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:


0.000 000 000 742 147 676 629 3(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 629 3(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:


0.000 000 000 742 147 676 629 3(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 -0.000 000 000 742 147 676 629 3 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