-0.000 000 000 742 147 676 653 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 653 4(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
-0.000 000 000 742 147 676 653 4(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:

|-0.000 000 000 742 147 676 653 4| = 0.000 000 000 742 147 676 653 4


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: 0.000 000 000 742 147 676 653 4.

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) 0.000 000 000 742 147 676 653 4 × 2 = 0 + 0.000 000 001 484 295 353 306 8;
  • 2) 0.000 000 001 484 295 353 306 8 × 2 = 0 + 0.000 000 002 968 590 706 613 6;
  • 3) 0.000 000 002 968 590 706 613 6 × 2 = 0 + 0.000 000 005 937 181 413 227 2;
  • 4) 0.000 000 005 937 181 413 227 2 × 2 = 0 + 0.000 000 011 874 362 826 454 4;
  • 5) 0.000 000 011 874 362 826 454 4 × 2 = 0 + 0.000 000 023 748 725 652 908 8;
  • 6) 0.000 000 023 748 725 652 908 8 × 2 = 0 + 0.000 000 047 497 451 305 817 6;
  • 7) 0.000 000 047 497 451 305 817 6 × 2 = 0 + 0.000 000 094 994 902 611 635 2;
  • 8) 0.000 000 094 994 902 611 635 2 × 2 = 0 + 0.000 000 189 989 805 223 270 4;
  • 9) 0.000 000 189 989 805 223 270 4 × 2 = 0 + 0.000 000 379 979 610 446 540 8;
  • 10) 0.000 000 379 979 610 446 540 8 × 2 = 0 + 0.000 000 759 959 220 893 081 6;
  • 11) 0.000 000 759 959 220 893 081 6 × 2 = 0 + 0.000 001 519 918 441 786 163 2;
  • 12) 0.000 001 519 918 441 786 163 2 × 2 = 0 + 0.000 003 039 836 883 572 326 4;
  • 13) 0.000 003 039 836 883 572 326 4 × 2 = 0 + 0.000 006 079 673 767 144 652 8;
  • 14) 0.000 006 079 673 767 144 652 8 × 2 = 0 + 0.000 012 159 347 534 289 305 6;
  • 15) 0.000 012 159 347 534 289 305 6 × 2 = 0 + 0.000 024 318 695 068 578 611 2;
  • 16) 0.000 024 318 695 068 578 611 2 × 2 = 0 + 0.000 048 637 390 137 157 222 4;
  • 17) 0.000 048 637 390 137 157 222 4 × 2 = 0 + 0.000 097 274 780 274 314 444 8;
  • 18) 0.000 097 274 780 274 314 444 8 × 2 = 0 + 0.000 194 549 560 548 628 889 6;
  • 19) 0.000 194 549 560 548 628 889 6 × 2 = 0 + 0.000 389 099 121 097 257 779 2;
  • 20) 0.000 389 099 121 097 257 779 2 × 2 = 0 + 0.000 778 198 242 194 515 558 4;
  • 21) 0.000 778 198 242 194 515 558 4 × 2 = 0 + 0.001 556 396 484 389 031 116 8;
  • 22) 0.001 556 396 484 389 031 116 8 × 2 = 0 + 0.003 112 792 968 778 062 233 6;
  • 23) 0.003 112 792 968 778 062 233 6 × 2 = 0 + 0.006 225 585 937 556 124 467 2;
  • 24) 0.006 225 585 937 556 124 467 2 × 2 = 0 + 0.012 451 171 875 112 248 934 4;
  • 25) 0.012 451 171 875 112 248 934 4 × 2 = 0 + 0.024 902 343 750 224 497 868 8;
  • 26) 0.024 902 343 750 224 497 868 8 × 2 = 0 + 0.049 804 687 500 448 995 737 6;
  • 27) 0.049 804 687 500 448 995 737 6 × 2 = 0 + 0.099 609 375 000 897 991 475 2;
  • 28) 0.099 609 375 000 897 991 475 2 × 2 = 0 + 0.199 218 750 001 795 982 950 4;
  • 29) 0.199 218 750 001 795 982 950 4 × 2 = 0 + 0.398 437 500 003 591 965 900 8;
  • 30) 0.398 437 500 003 591 965 900 8 × 2 = 0 + 0.796 875 000 007 183 931 801 6;
  • 31) 0.796 875 000 007 183 931 801 6 × 2 = 1 + 0.593 750 000 014 367 863 603 2;
  • 32) 0.593 750 000 014 367 863 603 2 × 2 = 1 + 0.187 500 000 028 735 727 206 4;
  • 33) 0.187 500 000 028 735 727 206 4 × 2 = 0 + 0.375 000 000 057 471 454 412 8;
  • 34) 0.375 000 000 057 471 454 412 8 × 2 = 0 + 0.750 000 000 114 942 908 825 6;
  • 35) 0.750 000 000 114 942 908 825 6 × 2 = 1 + 0.500 000 000 229 885 817 651 2;
  • 36) 0.500 000 000 229 885 817 651 2 × 2 = 1 + 0.000 000 000 459 771 635 302 4;
  • 37) 0.000 000 000 459 771 635 302 4 × 2 = 0 + 0.000 000 000 919 543 270 604 8;
  • 38) 0.000 000 000 919 543 270 604 8 × 2 = 0 + 0.000 000 001 839 086 541 209 6;
  • 39) 0.000 000 001 839 086 541 209 6 × 2 = 0 + 0.000 000 003 678 173 082 419 2;
  • 40) 0.000 000 003 678 173 082 419 2 × 2 = 0 + 0.000 000 007 356 346 164 838 4;
  • 41) 0.000 000 007 356 346 164 838 4 × 2 = 0 + 0.000 000 014 712 692 329 676 8;
  • 42) 0.000 000 014 712 692 329 676 8 × 2 = 0 + 0.000 000 029 425 384 659 353 6;
  • 43) 0.000 000 029 425 384 659 353 6 × 2 = 0 + 0.000 000 058 850 769 318 707 2;
  • 44) 0.000 000 058 850 769 318 707 2 × 2 = 0 + 0.000 000 117 701 538 637 414 4;
  • 45) 0.000 000 117 701 538 637 414 4 × 2 = 0 + 0.000 000 235 403 077 274 828 8;
  • 46) 0.000 000 235 403 077 274 828 8 × 2 = 0 + 0.000 000 470 806 154 549 657 6;
  • 47) 0.000 000 470 806 154 549 657 6 × 2 = 0 + 0.000 000 941 612 309 099 315 2;
  • 48) 0.000 000 941 612 309 099 315 2 × 2 = 0 + 0.000 001 883 224 618 198 630 4;
  • 49) 0.000 001 883 224 618 198 630 4 × 2 = 0 + 0.000 003 766 449 236 397 260 8;
  • 50) 0.000 003 766 449 236 397 260 8 × 2 = 0 + 0.000 007 532 898 472 794 521 6;
  • 51) 0.000 007 532 898 472 794 521 6 × 2 = 0 + 0.000 015 065 796 945 589 043 2;
  • 52) 0.000 015 065 796 945 589 043 2 × 2 = 0 + 0.000 030 131 593 891 178 086 4;
  • 53) 0.000 030 131 593 891 178 086 4 × 2 = 0 + 0.000 060 263 187 782 356 172 8;
  • 54) 0.000 060 263 187 782 356 172 8 × 2 = 0 + 0.000 120 526 375 564 712 345 6;

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:


0.000 000 000 742 147 676 653 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 653 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 653 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 653 4 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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