-0.000 000 000 742 147 676 643 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 643 6(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
-0.000 000 000 742 147 676 643 6(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:

|-0.000 000 000 742 147 676 643 6| = 0.000 000 000 742 147 676 643 6


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: 0.000 000 000 742 147 676 643 6.

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) 0.000 000 000 742 147 676 643 6 × 2 = 0 + 0.000 000 001 484 295 353 287 2;
  • 2) 0.000 000 001 484 295 353 287 2 × 2 = 0 + 0.000 000 002 968 590 706 574 4;
  • 3) 0.000 000 002 968 590 706 574 4 × 2 = 0 + 0.000 000 005 937 181 413 148 8;
  • 4) 0.000 000 005 937 181 413 148 8 × 2 = 0 + 0.000 000 011 874 362 826 297 6;
  • 5) 0.000 000 011 874 362 826 297 6 × 2 = 0 + 0.000 000 023 748 725 652 595 2;
  • 6) 0.000 000 023 748 725 652 595 2 × 2 = 0 + 0.000 000 047 497 451 305 190 4;
  • 7) 0.000 000 047 497 451 305 190 4 × 2 = 0 + 0.000 000 094 994 902 610 380 8;
  • 8) 0.000 000 094 994 902 610 380 8 × 2 = 0 + 0.000 000 189 989 805 220 761 6;
  • 9) 0.000 000 189 989 805 220 761 6 × 2 = 0 + 0.000 000 379 979 610 441 523 2;
  • 10) 0.000 000 379 979 610 441 523 2 × 2 = 0 + 0.000 000 759 959 220 883 046 4;
  • 11) 0.000 000 759 959 220 883 046 4 × 2 = 0 + 0.000 001 519 918 441 766 092 8;
  • 12) 0.000 001 519 918 441 766 092 8 × 2 = 0 + 0.000 003 039 836 883 532 185 6;
  • 13) 0.000 003 039 836 883 532 185 6 × 2 = 0 + 0.000 006 079 673 767 064 371 2;
  • 14) 0.000 006 079 673 767 064 371 2 × 2 = 0 + 0.000 012 159 347 534 128 742 4;
  • 15) 0.000 012 159 347 534 128 742 4 × 2 = 0 + 0.000 024 318 695 068 257 484 8;
  • 16) 0.000 024 318 695 068 257 484 8 × 2 = 0 + 0.000 048 637 390 136 514 969 6;
  • 17) 0.000 048 637 390 136 514 969 6 × 2 = 0 + 0.000 097 274 780 273 029 939 2;
  • 18) 0.000 097 274 780 273 029 939 2 × 2 = 0 + 0.000 194 549 560 546 059 878 4;
  • 19) 0.000 194 549 560 546 059 878 4 × 2 = 0 + 0.000 389 099 121 092 119 756 8;
  • 20) 0.000 389 099 121 092 119 756 8 × 2 = 0 + 0.000 778 198 242 184 239 513 6;
  • 21) 0.000 778 198 242 184 239 513 6 × 2 = 0 + 0.001 556 396 484 368 479 027 2;
  • 22) 0.001 556 396 484 368 479 027 2 × 2 = 0 + 0.003 112 792 968 736 958 054 4;
  • 23) 0.003 112 792 968 736 958 054 4 × 2 = 0 + 0.006 225 585 937 473 916 108 8;
  • 24) 0.006 225 585 937 473 916 108 8 × 2 = 0 + 0.012 451 171 874 947 832 217 6;
  • 25) 0.012 451 171 874 947 832 217 6 × 2 = 0 + 0.024 902 343 749 895 664 435 2;
  • 26) 0.024 902 343 749 895 664 435 2 × 2 = 0 + 0.049 804 687 499 791 328 870 4;
  • 27) 0.049 804 687 499 791 328 870 4 × 2 = 0 + 0.099 609 374 999 582 657 740 8;
  • 28) 0.099 609 374 999 582 657 740 8 × 2 = 0 + 0.199 218 749 999 165 315 481 6;
  • 29) 0.199 218 749 999 165 315 481 6 × 2 = 0 + 0.398 437 499 998 330 630 963 2;
  • 30) 0.398 437 499 998 330 630 963 2 × 2 = 0 + 0.796 874 999 996 661 261 926 4;
  • 31) 0.796 874 999 996 661 261 926 4 × 2 = 1 + 0.593 749 999 993 322 523 852 8;
  • 32) 0.593 749 999 993 322 523 852 8 × 2 = 1 + 0.187 499 999 986 645 047 705 6;
  • 33) 0.187 499 999 986 645 047 705 6 × 2 = 0 + 0.374 999 999 973 290 095 411 2;
  • 34) 0.374 999 999 973 290 095 411 2 × 2 = 0 + 0.749 999 999 946 580 190 822 4;
  • 35) 0.749 999 999 946 580 190 822 4 × 2 = 1 + 0.499 999 999 893 160 381 644 8;
  • 36) 0.499 999 999 893 160 381 644 8 × 2 = 0 + 0.999 999 999 786 320 763 289 6;
  • 37) 0.999 999 999 786 320 763 289 6 × 2 = 1 + 0.999 999 999 572 641 526 579 2;
  • 38) 0.999 999 999 572 641 526 579 2 × 2 = 1 + 0.999 999 999 145 283 053 158 4;
  • 39) 0.999 999 999 145 283 053 158 4 × 2 = 1 + 0.999 999 998 290 566 106 316 8;
  • 40) 0.999 999 998 290 566 106 316 8 × 2 = 1 + 0.999 999 996 581 132 212 633 6;
  • 41) 0.999 999 996 581 132 212 633 6 × 2 = 1 + 0.999 999 993 162 264 425 267 2;
  • 42) 0.999 999 993 162 264 425 267 2 × 2 = 1 + 0.999 999 986 324 528 850 534 4;
  • 43) 0.999 999 986 324 528 850 534 4 × 2 = 1 + 0.999 999 972 649 057 701 068 8;
  • 44) 0.999 999 972 649 057 701 068 8 × 2 = 1 + 0.999 999 945 298 115 402 137 6;
  • 45) 0.999 999 945 298 115 402 137 6 × 2 = 1 + 0.999 999 890 596 230 804 275 2;
  • 46) 0.999 999 890 596 230 804 275 2 × 2 = 1 + 0.999 999 781 192 461 608 550 4;
  • 47) 0.999 999 781 192 461 608 550 4 × 2 = 1 + 0.999 999 562 384 923 217 100 8;
  • 48) 0.999 999 562 384 923 217 100 8 × 2 = 1 + 0.999 999 124 769 846 434 201 6;
  • 49) 0.999 999 124 769 846 434 201 6 × 2 = 1 + 0.999 998 249 539 692 868 403 2;
  • 50) 0.999 998 249 539 692 868 403 2 × 2 = 1 + 0.999 996 499 079 385 736 806 4;
  • 51) 0.999 996 499 079 385 736 806 4 × 2 = 1 + 0.999 992 998 158 771 473 612 8;
  • 52) 0.999 992 998 158 771 473 612 8 × 2 = 1 + 0.999 985 996 317 542 947 225 6;
  • 53) 0.999 985 996 317 542 947 225 6 × 2 = 1 + 0.999 971 992 635 085 894 451 2;
  • 54) 0.999 971 992 635 085 894 451 2 × 2 = 1 + 0.999 943 985 270 171 788 902 4;

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:


0.000 000 000 742 147 676 643 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 643 6(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:


0.000 000 000 742 147 676 643 6(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 -0.000 000 000 742 147 676 643 6 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