-0.000 000 000 742 147 451 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 451(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
-0.000 000 000 742 147 451(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:

|-0.000 000 000 742 147 451| = 0.000 000 000 742 147 451


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: 0.000 000 000 742 147 451.

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) 0.000 000 000 742 147 451 × 2 = 0 + 0.000 000 001 484 294 902;
  • 2) 0.000 000 001 484 294 902 × 2 = 0 + 0.000 000 002 968 589 804;
  • 3) 0.000 000 002 968 589 804 × 2 = 0 + 0.000 000 005 937 179 608;
  • 4) 0.000 000 005 937 179 608 × 2 = 0 + 0.000 000 011 874 359 216;
  • 5) 0.000 000 011 874 359 216 × 2 = 0 + 0.000 000 023 748 718 432;
  • 6) 0.000 000 023 748 718 432 × 2 = 0 + 0.000 000 047 497 436 864;
  • 7) 0.000 000 047 497 436 864 × 2 = 0 + 0.000 000 094 994 873 728;
  • 8) 0.000 000 094 994 873 728 × 2 = 0 + 0.000 000 189 989 747 456;
  • 9) 0.000 000 189 989 747 456 × 2 = 0 + 0.000 000 379 979 494 912;
  • 10) 0.000 000 379 979 494 912 × 2 = 0 + 0.000 000 759 958 989 824;
  • 11) 0.000 000 759 958 989 824 × 2 = 0 + 0.000 001 519 917 979 648;
  • 12) 0.000 001 519 917 979 648 × 2 = 0 + 0.000 003 039 835 959 296;
  • 13) 0.000 003 039 835 959 296 × 2 = 0 + 0.000 006 079 671 918 592;
  • 14) 0.000 006 079 671 918 592 × 2 = 0 + 0.000 012 159 343 837 184;
  • 15) 0.000 012 159 343 837 184 × 2 = 0 + 0.000 024 318 687 674 368;
  • 16) 0.000 024 318 687 674 368 × 2 = 0 + 0.000 048 637 375 348 736;
  • 17) 0.000 048 637 375 348 736 × 2 = 0 + 0.000 097 274 750 697 472;
  • 18) 0.000 097 274 750 697 472 × 2 = 0 + 0.000 194 549 501 394 944;
  • 19) 0.000 194 549 501 394 944 × 2 = 0 + 0.000 389 099 002 789 888;
  • 20) 0.000 389 099 002 789 888 × 2 = 0 + 0.000 778 198 005 579 776;
  • 21) 0.000 778 198 005 579 776 × 2 = 0 + 0.001 556 396 011 159 552;
  • 22) 0.001 556 396 011 159 552 × 2 = 0 + 0.003 112 792 022 319 104;
  • 23) 0.003 112 792 022 319 104 × 2 = 0 + 0.006 225 584 044 638 208;
  • 24) 0.006 225 584 044 638 208 × 2 = 0 + 0.012 451 168 089 276 416;
  • 25) 0.012 451 168 089 276 416 × 2 = 0 + 0.024 902 336 178 552 832;
  • 26) 0.024 902 336 178 552 832 × 2 = 0 + 0.049 804 672 357 105 664;
  • 27) 0.049 804 672 357 105 664 × 2 = 0 + 0.099 609 344 714 211 328;
  • 28) 0.099 609 344 714 211 328 × 2 = 0 + 0.199 218 689 428 422 656;
  • 29) 0.199 218 689 428 422 656 × 2 = 0 + 0.398 437 378 856 845 312;
  • 30) 0.398 437 378 856 845 312 × 2 = 0 + 0.796 874 757 713 690 624;
  • 31) 0.796 874 757 713 690 624 × 2 = 1 + 0.593 749 515 427 381 248;
  • 32) 0.593 749 515 427 381 248 × 2 = 1 + 0.187 499 030 854 762 496;
  • 33) 0.187 499 030 854 762 496 × 2 = 0 + 0.374 998 061 709 524 992;
  • 34) 0.374 998 061 709 524 992 × 2 = 0 + 0.749 996 123 419 049 984;
  • 35) 0.749 996 123 419 049 984 × 2 = 1 + 0.499 992 246 838 099 968;
  • 36) 0.499 992 246 838 099 968 × 2 = 0 + 0.999 984 493 676 199 936;
  • 37) 0.999 984 493 676 199 936 × 2 = 1 + 0.999 968 987 352 399 872;
  • 38) 0.999 968 987 352 399 872 × 2 = 1 + 0.999 937 974 704 799 744;
  • 39) 0.999 937 974 704 799 744 × 2 = 1 + 0.999 875 949 409 599 488;
  • 40) 0.999 875 949 409 599 488 × 2 = 1 + 0.999 751 898 819 198 976;
  • 41) 0.999 751 898 819 198 976 × 2 = 1 + 0.999 503 797 638 397 952;
  • 42) 0.999 503 797 638 397 952 × 2 = 1 + 0.999 007 595 276 795 904;
  • 43) 0.999 007 595 276 795 904 × 2 = 1 + 0.998 015 190 553 591 808;
  • 44) 0.998 015 190 553 591 808 × 2 = 1 + 0.996 030 381 107 183 616;
  • 45) 0.996 030 381 107 183 616 × 2 = 1 + 0.992 060 762 214 367 232;
  • 46) 0.992 060 762 214 367 232 × 2 = 1 + 0.984 121 524 428 734 464;
  • 47) 0.984 121 524 428 734 464 × 2 = 1 + 0.968 243 048 857 468 928;
  • 48) 0.968 243 048 857 468 928 × 2 = 1 + 0.936 486 097 714 937 856;
  • 49) 0.936 486 097 714 937 856 × 2 = 1 + 0.872 972 195 429 875 712;
  • 50) 0.872 972 195 429 875 712 × 2 = 1 + 0.745 944 390 859 751 424;
  • 51) 0.745 944 390 859 751 424 × 2 = 1 + 0.491 888 781 719 502 848;
  • 52) 0.491 888 781 719 502 848 × 2 = 0 + 0.983 777 563 439 005 696;
  • 53) 0.983 777 563 439 005 696 × 2 = 1 + 0.967 555 126 878 011 392;
  • 54) 0.967 555 126 878 011 392 × 2 = 1 + 0.935 110 253 756 022 784;

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:


0.000 000 000 742 147 451(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 451(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 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:


0.000 000 000 742 147 451(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1110 11(2) × 20 =

1.1001 0111 1111 1111 1111 011(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 011


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 1011 =

100 1011 1111 1111 1111 1011


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 1011


Decimal number -0.000 000 000 742 147 451 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1011

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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