-0.000 000 000 742 147 676 646 405 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 405(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
-0.000 000 000 742 147 676 646 405(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:

|-0.000 000 000 742 147 676 646 405| = 0.000 000 000 742 147 676 646 405


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: 0.000 000 000 742 147 676 646 405.

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) 0.000 000 000 742 147 676 646 405 × 2 = 0 + 0.000 000 001 484 295 353 292 81;
  • 2) 0.000 000 001 484 295 353 292 81 × 2 = 0 + 0.000 000 002 968 590 706 585 62;
  • 3) 0.000 000 002 968 590 706 585 62 × 2 = 0 + 0.000 000 005 937 181 413 171 24;
  • 4) 0.000 000 005 937 181 413 171 24 × 2 = 0 + 0.000 000 011 874 362 826 342 48;
  • 5) 0.000 000 011 874 362 826 342 48 × 2 = 0 + 0.000 000 023 748 725 652 684 96;
  • 6) 0.000 000 023 748 725 652 684 96 × 2 = 0 + 0.000 000 047 497 451 305 369 92;
  • 7) 0.000 000 047 497 451 305 369 92 × 2 = 0 + 0.000 000 094 994 902 610 739 84;
  • 8) 0.000 000 094 994 902 610 739 84 × 2 = 0 + 0.000 000 189 989 805 221 479 68;
  • 9) 0.000 000 189 989 805 221 479 68 × 2 = 0 + 0.000 000 379 979 610 442 959 36;
  • 10) 0.000 000 379 979 610 442 959 36 × 2 = 0 + 0.000 000 759 959 220 885 918 72;
  • 11) 0.000 000 759 959 220 885 918 72 × 2 = 0 + 0.000 001 519 918 441 771 837 44;
  • 12) 0.000 001 519 918 441 771 837 44 × 2 = 0 + 0.000 003 039 836 883 543 674 88;
  • 13) 0.000 003 039 836 883 543 674 88 × 2 = 0 + 0.000 006 079 673 767 087 349 76;
  • 14) 0.000 006 079 673 767 087 349 76 × 2 = 0 + 0.000 012 159 347 534 174 699 52;
  • 15) 0.000 012 159 347 534 174 699 52 × 2 = 0 + 0.000 024 318 695 068 349 399 04;
  • 16) 0.000 024 318 695 068 349 399 04 × 2 = 0 + 0.000 048 637 390 136 698 798 08;
  • 17) 0.000 048 637 390 136 698 798 08 × 2 = 0 + 0.000 097 274 780 273 397 596 16;
  • 18) 0.000 097 274 780 273 397 596 16 × 2 = 0 + 0.000 194 549 560 546 795 192 32;
  • 19) 0.000 194 549 560 546 795 192 32 × 2 = 0 + 0.000 389 099 121 093 590 384 64;
  • 20) 0.000 389 099 121 093 590 384 64 × 2 = 0 + 0.000 778 198 242 187 180 769 28;
  • 21) 0.000 778 198 242 187 180 769 28 × 2 = 0 + 0.001 556 396 484 374 361 538 56;
  • 22) 0.001 556 396 484 374 361 538 56 × 2 = 0 + 0.003 112 792 968 748 723 077 12;
  • 23) 0.003 112 792 968 748 723 077 12 × 2 = 0 + 0.006 225 585 937 497 446 154 24;
  • 24) 0.006 225 585 937 497 446 154 24 × 2 = 0 + 0.012 451 171 874 994 892 308 48;
  • 25) 0.012 451 171 874 994 892 308 48 × 2 = 0 + 0.024 902 343 749 989 784 616 96;
  • 26) 0.024 902 343 749 989 784 616 96 × 2 = 0 + 0.049 804 687 499 979 569 233 92;
  • 27) 0.049 804 687 499 979 569 233 92 × 2 = 0 + 0.099 609 374 999 959 138 467 84;
  • 28) 0.099 609 374 999 959 138 467 84 × 2 = 0 + 0.199 218 749 999 918 276 935 68;
  • 29) 0.199 218 749 999 918 276 935 68 × 2 = 0 + 0.398 437 499 999 836 553 871 36;
  • 30) 0.398 437 499 999 836 553 871 36 × 2 = 0 + 0.796 874 999 999 673 107 742 72;
  • 31) 0.796 874 999 999 673 107 742 72 × 2 = 1 + 0.593 749 999 999 346 215 485 44;
  • 32) 0.593 749 999 999 346 215 485 44 × 2 = 1 + 0.187 499 999 998 692 430 970 88;
  • 33) 0.187 499 999 998 692 430 970 88 × 2 = 0 + 0.374 999 999 997 384 861 941 76;
  • 34) 0.374 999 999 997 384 861 941 76 × 2 = 0 + 0.749 999 999 994 769 723 883 52;
  • 35) 0.749 999 999 994 769 723 883 52 × 2 = 1 + 0.499 999 999 989 539 447 767 04;
  • 36) 0.499 999 999 989 539 447 767 04 × 2 = 0 + 0.999 999 999 979 078 895 534 08;
  • 37) 0.999 999 999 979 078 895 534 08 × 2 = 1 + 0.999 999 999 958 157 791 068 16;
  • 38) 0.999 999 999 958 157 791 068 16 × 2 = 1 + 0.999 999 999 916 315 582 136 32;
  • 39) 0.999 999 999 916 315 582 136 32 × 2 = 1 + 0.999 999 999 832 631 164 272 64;
  • 40) 0.999 999 999 832 631 164 272 64 × 2 = 1 + 0.999 999 999 665 262 328 545 28;
  • 41) 0.999 999 999 665 262 328 545 28 × 2 = 1 + 0.999 999 999 330 524 657 090 56;
  • 42) 0.999 999 999 330 524 657 090 56 × 2 = 1 + 0.999 999 998 661 049 314 181 12;
  • 43) 0.999 999 998 661 049 314 181 12 × 2 = 1 + 0.999 999 997 322 098 628 362 24;
  • 44) 0.999 999 997 322 098 628 362 24 × 2 = 1 + 0.999 999 994 644 197 256 724 48;
  • 45) 0.999 999 994 644 197 256 724 48 × 2 = 1 + 0.999 999 989 288 394 513 448 96;
  • 46) 0.999 999 989 288 394 513 448 96 × 2 = 1 + 0.999 999 978 576 789 026 897 92;
  • 47) 0.999 999 978 576 789 026 897 92 × 2 = 1 + 0.999 999 957 153 578 053 795 84;
  • 48) 0.999 999 957 153 578 053 795 84 × 2 = 1 + 0.999 999 914 307 156 107 591 68;
  • 49) 0.999 999 914 307 156 107 591 68 × 2 = 1 + 0.999 999 828 614 312 215 183 36;
  • 50) 0.999 999 828 614 312 215 183 36 × 2 = 1 + 0.999 999 657 228 624 430 366 72;
  • 51) 0.999 999 657 228 624 430 366 72 × 2 = 1 + 0.999 999 314 457 248 860 733 44;
  • 52) 0.999 999 314 457 248 860 733 44 × 2 = 1 + 0.999 998 628 914 497 721 466 88;
  • 53) 0.999 998 628 914 497 721 466 88 × 2 = 1 + 0.999 997 257 828 995 442 933 76;
  • 54) 0.999 997 257 828 995 442 933 76 × 2 = 1 + 0.999 994 515 657 990 885 867 52;

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:


0.000 000 000 742 147 676 646 405(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 405(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:


0.000 000 000 742 147 676 646 405(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 -0.000 000 000 742 147 676 646 405 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