-0.000 000 000 742 147 676 644 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 644 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
-0.000 000 000 742 147 676 644 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:

|-0.000 000 000 742 147 676 644 2| = 0.000 000 000 742 147 676 644 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: 0.000 000 000 742 147 676 644 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) 0.000 000 000 742 147 676 644 2 × 2 = 0 + 0.000 000 001 484 295 353 288 4;
  • 2) 0.000 000 001 484 295 353 288 4 × 2 = 0 + 0.000 000 002 968 590 706 576 8;
  • 3) 0.000 000 002 968 590 706 576 8 × 2 = 0 + 0.000 000 005 937 181 413 153 6;
  • 4) 0.000 000 005 937 181 413 153 6 × 2 = 0 + 0.000 000 011 874 362 826 307 2;
  • 5) 0.000 000 011 874 362 826 307 2 × 2 = 0 + 0.000 000 023 748 725 652 614 4;
  • 6) 0.000 000 023 748 725 652 614 4 × 2 = 0 + 0.000 000 047 497 451 305 228 8;
  • 7) 0.000 000 047 497 451 305 228 8 × 2 = 0 + 0.000 000 094 994 902 610 457 6;
  • 8) 0.000 000 094 994 902 610 457 6 × 2 = 0 + 0.000 000 189 989 805 220 915 2;
  • 9) 0.000 000 189 989 805 220 915 2 × 2 = 0 + 0.000 000 379 979 610 441 830 4;
  • 10) 0.000 000 379 979 610 441 830 4 × 2 = 0 + 0.000 000 759 959 220 883 660 8;
  • 11) 0.000 000 759 959 220 883 660 8 × 2 = 0 + 0.000 001 519 918 441 767 321 6;
  • 12) 0.000 001 519 918 441 767 321 6 × 2 = 0 + 0.000 003 039 836 883 534 643 2;
  • 13) 0.000 003 039 836 883 534 643 2 × 2 = 0 + 0.000 006 079 673 767 069 286 4;
  • 14) 0.000 006 079 673 767 069 286 4 × 2 = 0 + 0.000 012 159 347 534 138 572 8;
  • 15) 0.000 012 159 347 534 138 572 8 × 2 = 0 + 0.000 024 318 695 068 277 145 6;
  • 16) 0.000 024 318 695 068 277 145 6 × 2 = 0 + 0.000 048 637 390 136 554 291 2;
  • 17) 0.000 048 637 390 136 554 291 2 × 2 = 0 + 0.000 097 274 780 273 108 582 4;
  • 18) 0.000 097 274 780 273 108 582 4 × 2 = 0 + 0.000 194 549 560 546 217 164 8;
  • 19) 0.000 194 549 560 546 217 164 8 × 2 = 0 + 0.000 389 099 121 092 434 329 6;
  • 20) 0.000 389 099 121 092 434 329 6 × 2 = 0 + 0.000 778 198 242 184 868 659 2;
  • 21) 0.000 778 198 242 184 868 659 2 × 2 = 0 + 0.001 556 396 484 369 737 318 4;
  • 22) 0.001 556 396 484 369 737 318 4 × 2 = 0 + 0.003 112 792 968 739 474 636 8;
  • 23) 0.003 112 792 968 739 474 636 8 × 2 = 0 + 0.006 225 585 937 478 949 273 6;
  • 24) 0.006 225 585 937 478 949 273 6 × 2 = 0 + 0.012 451 171 874 957 898 547 2;
  • 25) 0.012 451 171 874 957 898 547 2 × 2 = 0 + 0.024 902 343 749 915 797 094 4;
  • 26) 0.024 902 343 749 915 797 094 4 × 2 = 0 + 0.049 804 687 499 831 594 188 8;
  • 27) 0.049 804 687 499 831 594 188 8 × 2 = 0 + 0.099 609 374 999 663 188 377 6;
  • 28) 0.099 609 374 999 663 188 377 6 × 2 = 0 + 0.199 218 749 999 326 376 755 2;
  • 29) 0.199 218 749 999 326 376 755 2 × 2 = 0 + 0.398 437 499 998 652 753 510 4;
  • 30) 0.398 437 499 998 652 753 510 4 × 2 = 0 + 0.796 874 999 997 305 507 020 8;
  • 31) 0.796 874 999 997 305 507 020 8 × 2 = 1 + 0.593 749 999 994 611 014 041 6;
  • 32) 0.593 749 999 994 611 014 041 6 × 2 = 1 + 0.187 499 999 989 222 028 083 2;
  • 33) 0.187 499 999 989 222 028 083 2 × 2 = 0 + 0.374 999 999 978 444 056 166 4;
  • 34) 0.374 999 999 978 444 056 166 4 × 2 = 0 + 0.749 999 999 956 888 112 332 8;
  • 35) 0.749 999 999 956 888 112 332 8 × 2 = 1 + 0.499 999 999 913 776 224 665 6;
  • 36) 0.499 999 999 913 776 224 665 6 × 2 = 0 + 0.999 999 999 827 552 449 331 2;
  • 37) 0.999 999 999 827 552 449 331 2 × 2 = 1 + 0.999 999 999 655 104 898 662 4;
  • 38) 0.999 999 999 655 104 898 662 4 × 2 = 1 + 0.999 999 999 310 209 797 324 8;
  • 39) 0.999 999 999 310 209 797 324 8 × 2 = 1 + 0.999 999 998 620 419 594 649 6;
  • 40) 0.999 999 998 620 419 594 649 6 × 2 = 1 + 0.999 999 997 240 839 189 299 2;
  • 41) 0.999 999 997 240 839 189 299 2 × 2 = 1 + 0.999 999 994 481 678 378 598 4;
  • 42) 0.999 999 994 481 678 378 598 4 × 2 = 1 + 0.999 999 988 963 356 757 196 8;
  • 43) 0.999 999 988 963 356 757 196 8 × 2 = 1 + 0.999 999 977 926 713 514 393 6;
  • 44) 0.999 999 977 926 713 514 393 6 × 2 = 1 + 0.999 999 955 853 427 028 787 2;
  • 45) 0.999 999 955 853 427 028 787 2 × 2 = 1 + 0.999 999 911 706 854 057 574 4;
  • 46) 0.999 999 911 706 854 057 574 4 × 2 = 1 + 0.999 999 823 413 708 115 148 8;
  • 47) 0.999 999 823 413 708 115 148 8 × 2 = 1 + 0.999 999 646 827 416 230 297 6;
  • 48) 0.999 999 646 827 416 230 297 6 × 2 = 1 + 0.999 999 293 654 832 460 595 2;
  • 49) 0.999 999 293 654 832 460 595 2 × 2 = 1 + 0.999 998 587 309 664 921 190 4;
  • 50) 0.999 998 587 309 664 921 190 4 × 2 = 1 + 0.999 997 174 619 329 842 380 8;
  • 51) 0.999 997 174 619 329 842 380 8 × 2 = 1 + 0.999 994 349 238 659 684 761 6;
  • 52) 0.999 994 349 238 659 684 761 6 × 2 = 1 + 0.999 988 698 477 319 369 523 2;
  • 53) 0.999 988 698 477 319 369 523 2 × 2 = 1 + 0.999 977 396 954 638 739 046 4;
  • 54) 0.999 977 396 954 638 739 046 4 × 2 = 1 + 0.999 954 793 909 277 478 092 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:


0.000 000 000 742 147 676 644 2(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 644 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:


0.000 000 000 742 147 676 644 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 -0.000 000 000 742 147 676 644 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